hypothesis testing 0.05

Statistics Made Easy

How to Interpret a P-Value Less Than 0.05 (With Examples)

A hypothesis test is used to test whether or not some hypothesis about a population parameter is true.

Whenever we perform a hypothesis test, we always define a null and alternative hypothesis:

Null Hypothesis (H 0 ): The sample data occurs purely from chance.
Alternative Hypothesis (H A ): The sample data is influenced by some non-random cause.

If the p-value of the hypothesis test is less than some significance level (e.g. α = .05), then we can reject the null hypothesis and conclude that we have sufficient evidence to say that the alternative hypothesis is true.

If the p-value is not less than .05, then we fail to reject the null hypothesis and conclude that we do not have sufficient evidence to say that the alternative hypothesis is true.

The following examples explain how to interpret a p-value less than .05 and how to interpret a p-value greater than .05 in practice.

Example: Interpret a P-Value Less Than 0.05

Suppose a factory claims that they produce tires that each weigh 200 pounds.

An auditor comes in and tests the null hypothesis that the mean weight of a tire is 200 pounds against the alternative hypothesis that the mean weight of a tire is not 200 pounds, using a 0.05 level of significance.

The null hypothesis (H 0 ): μ = 200

The alternative hypothesis: (H A ): μ ≠ 200

Upon conducting a hypothesis test for a mean, the auditor gets a p-value of 0.0154 .

Since the p-value of 0.0154 is less than the significance level of 0.05 , the auditor rejects the null hypothesis and concludes that there is sufficient evidence to say that the true average weight of a tire is not 200 pounds.

Example: Interpret a P-Value Greater Than 0.05

Suppose a biologist believes that a certain fertilizer will cause plants to grow more during a three-month period than they normally do, which is currently 20 inches. To test this, she applies the fertilizer to each of the plants in her laboratory for three months.

She then performs a hypothesis test using the following hypotheses:

The null hypothesis (H 0 ): μ = 20 inches (the fertilizer will have no effect on the mean plant growth)

The alternative hypothesis: (H A ): μ > 20 inches (the fertilizer will cause mean plant growth to increase)

Upon conducting a hypothesis test for a mean, the biologist gets a p-value of 0.2338 .

Since the p-value of 0.2338 is greater than the significance level of 0.05 , the biologist fails to reject the null hypothesis and concludes that there is not sufficient evidence to say that the fertilizer leads to increased plant growth.

Additional Resources

An Explanation of P-Values and Statistical Significance Statistical vs. Practical Significance P-Value vs. Alpha: What’s the Difference?

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Hey there. My name is Zach Bobbitt. I have a Master of Science degree in Applied Statistics and I’ve worked on machine learning algorithms for professional businesses in both healthcare and retail. I’m passionate about statistics, machine learning, and data visualization and I created Statology to be a resource for both students and teachers alike. My goal with this site is to help you learn statistics through using simple terms, plenty of real-world examples, and helpful illustrations.

One Reply to “How to Interpret a P-Value Less Than 0.05 (With Examples)”

in case 2 where p-value is greater than 0.5, it means we should reject the null hypothesis, which means we reject this statement : “the fertilizer will have no effect on the mean plant growth”

which means : the fertilizer will have effect on the mean plant growth

why do you conclude that ” that there is not sufficient evidence to say that the fertilizer leads to increased plant growth”?

P-Value And Statistical Significance: What It Is & Why It Matters

Saul Mcleod, PhD

Editor-in-Chief for Simply Psychology

BSc (Hons) Psychology, MRes, PhD, University of Manchester

Saul Mcleod, PhD., is a qualified psychology teacher with over 18 years of experience in further and higher education. He has been published in peer-reviewed journals, including the Journal of Clinical Psychology.

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The p-value in statistics quantifies the evidence against a null hypothesis. A low p-value suggests data is inconsistent with the null, potentially favoring an alternative hypothesis. Common significance thresholds are 0.05 or 0.01.

P-Value Explained in Normal Distribution

Hypothesis testing

When you perform a statistical test, a p-value helps you determine the significance of your results in relation to the null hypothesis.

The null hypothesis (H0) states no relationship exists between the two variables being studied (one variable does not affect the other). It states the results are due to chance and are not significant in supporting the idea being investigated. Thus, the null hypothesis assumes that whatever you try to prove did not happen.

The alternative hypothesis (Ha or H1) is the one you would believe if the null hypothesis is concluded to be untrue.

The alternative hypothesis states that the independent variable affected the dependent variable, and the results are significant in supporting the theory being investigated (i.e., the results are not due to random chance).

What a p-value tells you

A p-value, or probability value, is a number describing how likely it is that your data would have occurred by random chance (i.e., that the null hypothesis is true).

The level of statistical significance is often expressed as a p-value between 0 and 1.

The smaller the p -value, the less likely the results occurred by random chance, and the stronger the evidence that you should reject the null hypothesis.

Remember, a p-value doesn’t tell you if the null hypothesis is true or false. It just tells you how likely you’d see the data you observed (or more extreme data) if the null hypothesis was true. It’s a piece of evidence, not a definitive proof.

Example: Test Statistic and p-Value

Suppose you’re conducting a study to determine whether a new drug has an effect on pain relief compared to a placebo. If the new drug has no impact, your test statistic will be close to the one predicted by the null hypothesis (no difference between the drug and placebo groups), and the resulting p-value will be close to 1. It may not be precisely 1 because real-world variations may exist. Conversely, if the new drug indeed reduces pain significantly, your test statistic will diverge further from what’s expected under the null hypothesis, and the p-value will decrease. The p-value will never reach zero because there’s always a slim possibility, though highly improbable, that the observed results occurred by random chance.

P-value interpretation

The significance level (alpha) is a set probability threshold (often 0.05), while the p-value is the probability you calculate based on your study or analysis.

A p-value less than or equal to your significance level (typically ≤ 0.05) is statistically significant.

A p-value less than or equal to a predetermined significance level (often 0.05 or 0.01) indicates a statistically significant result, meaning the observed data provide strong evidence against the null hypothesis.

This suggests the effect under study likely represents a real relationship rather than just random chance.

For instance, if you set α = 0.05, you would reject the null hypothesis if your p -value ≤ 0.05.

It indicates strong evidence against the null hypothesis, as there is less than a 5% probability the null is correct (and the results are random).

Therefore, we reject the null hypothesis and accept the alternative hypothesis.

Example: Statistical Significance

Upon analyzing the pain relief effects of the new drug compared to the placebo, the computed p-value is less than 0.01, which falls well below the predetermined alpha value of 0.05. Consequently, you conclude that there is a statistically significant difference in pain relief between the new drug and the placebo.

What does a p-value of 0.001 mean?

A p-value of 0.001 is highly statistically significant beyond the commonly used 0.05 threshold. It indicates strong evidence of a real effect or difference, rather than just random variation.

Specifically, a p-value of 0.001 means there is only a 0.1% chance of obtaining a result at least as extreme as the one observed, assuming the null hypothesis is correct.

Such a small p-value provides strong evidence against the null hypothesis, leading to rejecting the null in favor of the alternative hypothesis.

A p-value more than the significance level (typically p > 0.05) is not statistically significant and indicates strong evidence for the null hypothesis.

This means we retain the null hypothesis and reject the alternative hypothesis. You should note that you cannot accept the null hypothesis; we can only reject it or fail to reject it.

Note : when the p-value is above your threshold of significance, it does not mean that there is a 95% probability that the alternative hypothesis is true.

One-Tailed Test

Probability and statistical significance in ab testing. Statistical significance in a b experiments

Two-Tailed Test

How do you calculate the p-value ?

Most statistical software packages like R, SPSS, and others automatically calculate your p-value. This is the easiest and most common way.

Online resources and tables are available to estimate the p-value based on your test statistic and degrees of freedom.

These tables help you understand how often you would expect to see your test statistic under the null hypothesis.

Understanding the Statistical Test:

Different statistical tests are designed to answer specific research questions or hypotheses. Each test has its own underlying assumptions and characteristics.

For example, you might use a t-test to compare means, a chi-squared test for categorical data, or a correlation test to measure the strength of a relationship between variables.

Be aware that the number of independent variables you include in your analysis can influence the magnitude of the test statistic needed to produce the same p-value.

This factor is particularly important to consider when comparing results across different analyses.

Example: Choosing a Statistical Test

If you’re comparing the effectiveness of just two different drugs in pain relief, a two-sample t-test is a suitable choice for comparing these two groups. However, when you’re examining the impact of three or more drugs, it’s more appropriate to employ an Analysis of Variance ( ANOVA) . Utilizing multiple pairwise comparisons in such cases can lead to artificially low p-values and an overestimation of the significance of differences between the drug groups.

How to report

A statistically significant result cannot prove that a research hypothesis is correct (which implies 100% certainty).

Instead, we may state our results “provide support for” or “give evidence for” our research hypothesis (as there is still a slight probability that the results occurred by chance and the null hypothesis was correct – e.g., less than 5%).

Example: Reporting the results

In our comparison of the pain relief effects of the new drug and the placebo, we observed that participants in the drug group experienced a significant reduction in pain ( M = 3.5; SD = 0.8) compared to those in the placebo group ( M = 5.2; SD = 0.7), resulting in an average difference of 1.7 points on the pain scale (t(98) = -9.36; p < 0.001).

The 6th edition of the APA style manual (American Psychological Association, 2010) states the following on the topic of reporting p-values:

“When reporting p values, report exact p values (e.g., p = .031) to two or three decimal places. However, report p values less than .001 as p < .001.

The tradition of reporting p values in the form p < .10, p < .05, p < .01, and so forth, was appropriate in a time when only limited tables of critical values were available.” (p. 114)

Do not use 0 before the decimal point for the statistical value p as it cannot equal 1. In other words, write p = .001 instead of p = 0.001.
Please pay attention to issues of italics ( p is always italicized) and spacing (either side of the = sign).
p = .000 (as outputted by some statistical packages such as SPSS) is impossible and should be written as p < .001.
The opposite of significant is “nonsignificant,” not “insignificant.”

Why is the p -value not enough?

A lower p-value is sometimes interpreted as meaning there is a stronger relationship between two variables.

However, statistical significance means that it is unlikely that the null hypothesis is true (less than 5%).

To understand the strength of the difference between the two groups (control vs. experimental) a researcher needs to calculate the effect size .

When do you reject the null hypothesis?

In statistical hypothesis testing, you reject the null hypothesis when the p-value is less than or equal to the significance level (α) you set before conducting your test. The significance level is the probability of rejecting the null hypothesis when it is true. Commonly used significance levels are 0.01, 0.05, and 0.10.

Remember, rejecting the null hypothesis doesn’t prove the alternative hypothesis; it just suggests that the alternative hypothesis may be plausible given the observed data.

The p -value is conditional upon the null hypothesis being true but is unrelated to the truth or falsity of the alternative hypothesis.

What does p-value of 0.05 mean?

If your p-value is less than or equal to 0.05 (the significance level), you would conclude that your result is statistically significant. This means the evidence is strong enough to reject the null hypothesis in favor of the alternative hypothesis.

Are all p-values below 0.05 considered statistically significant?

No, not all p-values below 0.05 are considered statistically significant. The threshold of 0.05 is commonly used, but it’s just a convention. Statistical significance depends on factors like the study design, sample size, and the magnitude of the observed effect.

A p-value below 0.05 means there is evidence against the null hypothesis, suggesting a real effect. However, it’s essential to consider the context and other factors when interpreting results.

Researchers also look at effect size and confidence intervals to determine the practical significance and reliability of findings.

How does sample size affect the interpretation of p-values?

Sample size can impact the interpretation of p-values. A larger sample size provides more reliable and precise estimates of the population, leading to narrower confidence intervals.

With a larger sample, even small differences between groups or effects can become statistically significant, yielding lower p-values. In contrast, smaller sample sizes may not have enough statistical power to detect smaller effects, resulting in higher p-values.

Therefore, a larger sample size increases the chances of finding statistically significant results when there is a genuine effect, making the findings more trustworthy and robust.

Can a non-significant p-value indicate that there is no effect or difference in the data?

No, a non-significant p-value does not necessarily indicate that there is no effect or difference in the data. It means that the observed data do not provide strong enough evidence to reject the null hypothesis.

There could still be a real effect or difference, but it might be smaller or more variable than the study was able to detect.

Other factors like sample size, study design, and measurement precision can influence the p-value. It’s important to consider the entire body of evidence and not rely solely on p-values when interpreting research findings.

Can P values be exactly zero?

While a p-value can be extremely small, it cannot technically be absolute zero. When a p-value is reported as p = 0.000, the actual p-value is too small for the software to display. This is often interpreted as strong evidence against the null hypothesis. For p values less than 0.001, report as p < .001

Further Information

P-values and significance tests (Kahn Academy)
Hypothesis testing and p-values (Kahn Academy)
Wasserstein, R. L., Schirm, A. L., & Lazar, N. A. (2019). Moving to a world beyond “ p “< 0.05”.
Criticism of using the “ p “< 0.05”.
Publication manual of the American Psychological Association
Statistics for Psychology Book Download

Bland, J. M., & Altman, D. G. (1994). One and two sided tests of significance: Authors’ reply. BMJ: British Medical Journal , 309 (6958), 874.

Goodman, S. N., & Royall, R. (1988). Evidence and scientific research. American Journal of Public Health , 78 (12), 1568-1574.

Goodman, S. (2008, July). A dirty dozen: twelve p-value misconceptions . In Seminars in hematology (Vol. 45, No. 3, pp. 135-140). WB Saunders.

Lang, J. M., Rothman, K. J., & Cann, C. I. (1998). That confounded P-value. Epidemiology (Cambridge, Mass.) , 9 (1), 7-8.

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Understanding Hypothesis Tests: Significance Levels (Alpha) and P values in Statistics

Topics: Hypothesis Testing , Statistics

What do significance levels and P values mean in hypothesis tests? What is statistical significance anyway? In this post, I’ll continue to focus on concepts and graphs to help you gain a more intuitive understanding of how hypothesis tests work in statistics.

To bring it to life, I’ll add the significance level and P value to the graph in my previous post in order to perform a graphical version of the 1 sample t-test. It’s easier to understand when you can see what statistical significance truly means!

Here’s where we left off in my last post . We want to determine whether our sample mean (330.6) indicates that this year's average energy cost is significantly different from last year’s average energy cost of $260.

The probability distribution plot above shows the distribution of sample means we’d obtain under the assumption that the null hypothesis is true (population mean = 260) and we repeatedly drew a large number of random samples.

I left you with a question: where do we draw the line for statistical significance on the graph? Now we'll add in the significance level and the P value, which are the decision-making tools we'll need.

We'll use these tools to test the following hypotheses:

Null hypothesis: The population mean equals the hypothesized mean (260).
Alternative hypothesis: The population mean differs from the hypothesized mean (260).

What Is the Significance Level (Alpha)?

The significance level, also denoted as alpha or α, is the probability of rejecting the null hypothesis when it is true. For example, a significance level of 0.05 indicates a 5% risk of concluding that a difference exists when there is no actual difference.

These types of definitions can be hard to understand because of their technical nature. A picture makes the concepts much easier to comprehend!

The significance level determines how far out from the null hypothesis value we'll draw that line on the graph. To graph a significance level of 0.05, we need to shade the 5% of the distribution that is furthest away from the null hypothesis.

Probability plot that shows the critical regions for a significance level of 0.05

In the graph above, the two shaded areas are equidistant from the null hypothesis value and each area has a probability of 0.025, for a total of 0.05. In statistics, we call these shaded areas the critical region for a two-tailed test. If the population mean is 260, we’d expect to obtain a sample mean that falls in the critical region 5% of the time. The critical region defines how far away our sample statistic must be from the null hypothesis value before we can say it is unusual enough to reject the null hypothesis.

Our sample mean (330.6) falls within the critical region, which indicates it is statistically significant at the 0.05 level.

We can also see if it is statistically significant using the other common significance level of 0.01.

Probability plot that shows the critical regions for a significance level of 0.01

The two shaded areas each have a probability of 0.005, which adds up to a total probability of 0.01. This time our sample mean does not fall within the critical region and we fail to reject the null hypothesis. This comparison shows why you need to choose your significance level before you begin your study. It protects you from choosing a significance level because it conveniently gives you significant results!

Thanks to the graph, we were able to determine that our results are statistically significant at the 0.05 level without using a P value. However, when you use the numeric output produced by statistical software , you’ll need to compare the P value to your significance level to make this determination.

Ready for a demo of Minitab Statistical Software? Just ask!

What Are P values?

P-values are the probability of obtaining an effect at least as extreme as the one in your sample data, assuming the truth of the null hypothesis.

This definition of P values, while technically correct, is a bit convoluted. It’s easier to understand with a graph!

To graph the P value for our example data set, we need to determine the distance between the sample mean and the null hypothesis value (330.6 - 260 = 70.6). Next, we can graph the probability of obtaining a sample mean that is at least as extreme in both tails of the distribution (260 +/- 70.6).

Probability plot that shows the p-value for our sample mean

In the graph above, the two shaded areas each have a probability of 0.01556, for a total probability 0.03112. This probability represents the likelihood of obtaining a sample mean that is at least as extreme as our sample mean in both tails of the distribution if the population mean is 260. That’s our P value!

When a P value is less than or equal to the significance level, you reject the null hypothesis. If we take the P value for our example and compare it to the common significance levels, it matches the previous graphical results. The P value of 0.03112 is statistically significant at an alpha level of 0.05, but not at the 0.01 level.

If we stick to a significance level of 0.05, we can conclude that the average energy cost for the population is greater than 260.

A common mistake is to interpret the P-value as the probability that the null hypothesis is true. To understand why this interpretation is incorrect, please read my blog post How to Correctly Interpret P Values .

Discussion about Statistically Significant Results

A hypothesis test evaluates two mutually exclusive statements about a population to determine which statement is best supported by the sample data. A test result is statistically significant when the sample statistic is unusual enough relative to the null hypothesis that we can reject the null hypothesis for the entire population. “Unusual enough” in a hypothesis test is defined by:

The assumption that the null hypothesis is true—the graphs are centered on the null hypothesis value.
The significance level—how far out do we draw the line for the critical region?
Our sample statistic—does it fall in the critical region?

Keep in mind that there is no magic significance level that distinguishes between the studies that have a true effect and those that don’t with 100% accuracy. The common alpha values of 0.05 and 0.01 are simply based on tradition. For a significance level of 0.05, expect to obtain sample means in the critical region 5% of the time when the null hypothesis is true . In these cases, you won’t know that the null hypothesis is true but you’ll reject it because the sample mean falls in the critical region. That’s why the significance level is also referred to as an error rate!

This type of error doesn’t imply that the experimenter did anything wrong or require any other unusual explanation. The graphs show that when the null hypothesis is true, it is possible to obtain these unusual sample means for no reason other than random sampling error. It’s just luck of the draw.

Significance levels and P values are important tools that help you quantify and control this type of error in a hypothesis test. Using these tools to decide when to reject the null hypothesis increases your chance of making the correct decision.

If you like this post, you might want to read the other posts in this series that use the same graphical framework:

Previous: Why We Need to Use Hypothesis Tests
Next: Confidence Intervals and Confidence Levels

If you'd like to see how I made these graphs, please read: How to Create a Graphical Version of the 1-sample t-Test .

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Hypothesis Testing – A Deep Dive into Hypothesis Testing, The Backbone of Statistical Inference

September 21, 2023

Explore the intricacies of hypothesis testing, a cornerstone of statistical analysis. Dive into methods, interpretations, and applications for making data-driven decisions.

In this Blog post we will learn:

What is Hypothesis Testing?
Steps in Hypothesis Testing 2.1. Set up Hypotheses: Null and Alternative 2.2. Choose a Significance Level (α) 2.3. Calculate a test statistic and P-Value 2.4. Make a Decision
Example : Testing a new drug.
Example in python

1. What is Hypothesis Testing?

In simple terms, hypothesis testing is a method used to make decisions or inferences about population parameters based on sample data. Imagine being handed a dice and asked if it’s biased. By rolling it a few times and analyzing the outcomes, you’d be engaging in the essence of hypothesis testing.

Think of hypothesis testing as the scientific method of the statistics world. Suppose you hear claims like “This new drug works wonders!” or “Our new website design boosts sales.” How do you know if these statements hold water? Enter hypothesis testing.

2. Steps in Hypothesis Testing

Set up Hypotheses : Begin with a null hypothesis (H0) and an alternative hypothesis (Ha).
Choose a Significance Level (α) : Typically 0.05, this is the probability of rejecting the null hypothesis when it’s actually true. Think of it as the chance of accusing an innocent person.
Calculate Test statistic and P-Value : Gather evidence (data) and calculate a test statistic.
p-value : This is the probability of observing the data, given that the null hypothesis is true. A small p-value (typically ≤ 0.05) suggests the data is inconsistent with the null hypothesis.
Decision Rule : If the p-value is less than or equal to α, you reject the null hypothesis in favor of the alternative.

2.1. Set up Hypotheses: Null and Alternative

Before diving into testing, we must formulate hypotheses. The null hypothesis (H0) represents the default assumption, while the alternative hypothesis (H1) challenges it.

For instance, in drug testing, H0 : “The new drug is no better than the existing one,” H1 : “The new drug is superior .”

2.2. Choose a Significance Level (α)

When You collect and analyze data to test H0 and H1 hypotheses. Based on your analysis, you decide whether to reject the null hypothesis in favor of the alternative, or fail to reject / Accept the null hypothesis.

The significance level, often denoted by $α$, represents the probability of rejecting the null hypothesis when it is actually true.

In other words, it’s the risk you’re willing to take of making a Type I error (false positive).

Type I Error (False Positive) :

Symbolized by the Greek letter alpha (α).
Occurs when you incorrectly reject a true null hypothesis . In other words, you conclude that there is an effect or difference when, in reality, there isn’t.
The probability of making a Type I error is denoted by the significance level of a test. Commonly, tests are conducted at the 0.05 significance level , which means there’s a 5% chance of making a Type I error .
Commonly used significance levels are 0.01, 0.05, and 0.10, but the choice depends on the context of the study and the level of risk one is willing to accept.

Example : If a drug is not effective (truth), but a clinical trial incorrectly concludes that it is effective (based on the sample data), then a Type I error has occurred.

Type II Error (False Negative) :

Symbolized by the Greek letter beta (β).
Occurs when you accept a false null hypothesis . This means you conclude there is no effect or difference when, in reality, there is.
The probability of making a Type II error is denoted by β. The power of a test (1 – β) represents the probability of correctly rejecting a false null hypothesis.

Example : If a drug is effective (truth), but a clinical trial incorrectly concludes that it is not effective (based on the sample data), then a Type II error has occurred.

Balancing the Errors :

In practice, there’s a trade-off between Type I and Type II errors. Reducing the risk of one typically increases the risk of the other. For example, if you want to decrease the probability of a Type I error (by setting a lower significance level), you might increase the probability of a Type II error unless you compensate by collecting more data or making other adjustments.

It’s essential to understand the consequences of both types of errors in any given context. In some situations, a Type I error might be more severe, while in others, a Type II error might be of greater concern. This understanding guides researchers in designing their experiments and choosing appropriate significance levels.

2.3. Calculate a test statistic and P-Value

Test statistic : A test statistic is a single number that helps us understand how far our sample data is from what we’d expect under a null hypothesis (a basic assumption we’re trying to test against). Generally, the larger the test statistic, the more evidence we have against our null hypothesis. It helps us decide whether the differences we observe in our data are due to random chance or if there’s an actual effect.

P-value : The P-value tells us how likely we would get our observed results (or something more extreme) if the null hypothesis were true. It’s a value between 0 and 1. – A smaller P-value (typically below 0.05) means that the observation is rare under the null hypothesis, so we might reject the null hypothesis. – A larger P-value suggests that what we observed could easily happen by random chance, so we might not reject the null hypothesis.

2.4. Make a Decision

Relationship between $α$ and P-Value

When conducting a hypothesis test:

We then calculate the p-value from our sample data and the test statistic.

Finally, we compare the p-value to our chosen $α$:

If $p−value≤α$: We reject the null hypothesis in favor of the alternative hypothesis. The result is said to be statistically significant.
If $p−value>α$: We fail to reject the null hypothesis. There isn’t enough statistical evidence to support the alternative hypothesis.

3. Example : Testing a new drug.

Imagine we are investigating whether a new drug is effective at treating headaches faster than drug B.

Setting Up the Experiment : You gather 100 people who suffer from headaches. Half of them (50 people) are given the new drug (let’s call this the ‘Drug Group’), and the other half are given a sugar pill, which doesn’t contain any medication.

Set up Hypotheses : Before starting, you make a prediction:
Null Hypothesis (H0): The new drug has no effect. Any difference in healing time between the two groups is just due to random chance.
Alternative Hypothesis (H1): The new drug does have an effect. The difference in healing time between the two groups is significant and not just by chance.

Calculate Test statistic and P-Value : After the experiment, you analyze the data. The “test statistic” is a number that helps you understand the difference between the two groups in terms of standard units.

For instance, let’s say:

The average healing time in the Drug Group is 2 hours.
The average healing time in the Placebo Group is 3 hours.

The test statistic helps you understand how significant this 1-hour difference is. If the groups are large and the spread of healing times in each group is small, then this difference might be significant. But if there’s a huge variation in healing times, the 1-hour difference might not be so special.

Imagine the P-value as answering this question: “If the new drug had NO real effect, what’s the probability that I’d see a difference as extreme (or more extreme) as the one I found, just by random chance?”

For instance:

P-value of 0.01 means there’s a 1% chance that the observed difference (or a more extreme difference) would occur if the drug had no effect. That’s pretty rare, so we might consider the drug effective.
P-value of 0.5 means there’s a 50% chance you’d see this difference just by chance. That’s pretty high, so we might not be convinced the drug is doing much.
If the P-value is less than ($α$) 0.05: the results are “statistically significant,” and they might reject the null hypothesis , believing the new drug has an effect.
If the P-value is greater than ($α$) 0.05: the results are not statistically significant, and they don’t reject the null hypothesis , remaining unsure if the drug has a genuine effect.

4. Example in python

For simplicity, let’s say we’re using a t-test (common for comparing means). Let’s dive into Python:

Making a Decision : “The results are statistically significant! p-value < 0.05 , The drug seems to have an effect!” If not, we’d say, “Looks like the drug isn’t as miraculous as we thought.”

5. Conclusion

Hypothesis testing is an indispensable tool in data science, allowing us to make data-driven decisions with confidence. By understanding its principles, conducting tests properly, and considering real-world applications, you can harness the power of hypothesis testing to unlock valuable insights from your data.

Correlation – connecting the dots, the role of correlation in data analysis, sampling and sampling distributions – a comprehensive guide on sampling and sampling distributions, law of large numbers – a deep dive into the world of statistics, central limit theorem – a deep dive into central limit theorem and its significance in statistics, skewness and kurtosis – peaks and tails, understanding data through skewness and kurtosis”, similar articles, complete introduction to linear regression in r, how to implement common statistical significance tests and find the p value, logistic regression – a complete tutorial with examples in r.

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Hypothesis Testing

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A hypothesis test is a statistical inference method used to test the significance of a proposed (hypothesized) relation between population statistics (parameters) and their corresponding sample estimators . In other words, hypothesis tests are used to determine if there is enough evidence in a sample to prove a hypothesis true for the entire population.

The test considers two hypotheses: the null hypothesis , which is a statement meant to be tested, usually something like "there is no effect" with the intention of proving this false, and the alternate hypothesis , which is the statement meant to stand after the test is performed. The two hypotheses must be mutually exclusive ; moreover, in most applications, the two are complementary (one being the negation of the other). The test works by comparing the $p$-value to the level of significance (a chosen target). If the $p$-value is less than or equal to the level of significance, then the null hypothesis is rejected.

When analyzing data, only samples of a certain size might be manageable as efficient computations. In some situations the error terms follow a continuous or infinite distribution, hence the use of samples to suggest accuracy of the chosen test statistics. The method of hypothesis testing gives an advantage over guessing what distribution or which parameters the data follows.

Definitions and Methodology

Hypothesis test and confidence intervals.

In statistical inference, properties (parameters) of a population are analyzed by sampling data sets. Given assumptions on the distribution, i.e. a statistical model of the data, certain hypotheses can be deduced from the known behavior of the model. These hypotheses must be tested against sampled data from the population.

The null hypothesis $($denoted $H_0)$ is a statement that is assumed to be true. If the null hypothesis is rejected, then there is enough evidence (statistical significance) to accept the alternate hypothesis $($denoted $H_1).$ Before doing any test for significance, both hypotheses must be clearly stated and non-conflictive, i.e. mutually exclusive, statements. Rejecting the null hypothesis, given that it is true, is called a type I error and it is denoted $\alpha$, which is also its probability of occurrence. Failing to reject the null hypothesis, given that it is false, is called a type II error and it is denoted $\beta$, which is also its probability of occurrence. Also, $\alpha$ is known as the significance level , and $1-\beta$ is known as the power of the test. $H_0$ $\textbf{is true}$$\hspace{15mm}$ $H_0$ $\textbf{is false}$ $\textbf{Reject}$ $H_0$$\hspace{10mm}$ Type I error Correct Decision $\textbf{Reject}$ $H_1$ Correct Decision Type II error The test statistic is the standardized value following the sampled data under the assumption that the null hypothesis is true, and a chosen particular test. These tests depend on the statistic to be studied and the assumed distribution it follows, e.g. the population mean following a normal distribution. The $p$-value is the probability of observing an extreme test statistic in the direction of the alternate hypothesis, given that the null hypothesis is true. The critical value is the value of the assumed distribution of the test statistic such that the probability of making a type I error is small.

Methodologies: Given an estimator $\hat \theta$ of a population statistic $\theta$, following a probability distribution $P(T)$, computed from a sample $\mathcal{S},$ and given a significance level $\alpha$ and test statistic $t^*,$ define $H_0$ and $H_1;$ compute the test statistic $t^*.$ $p$-value Approach (most prevalent): Find the $p$-value using $t^*$ (right-tailed). If the $p$-value is at most $\alpha,$ reject $H_0$. Otherwise, reject $H_1$. Critical Value Approach: Find the critical value solving the equation $P(T\geq t_\alpha)=\alpha$ (right-tailed). If $t^*>t_\alpha$, reject $H_0$. Otherwise, reject $H_1$. Note: Failing to reject $H_0$ only means inability to accept $H_1$, and it does not mean to accept $H_0$.

Assume a normally distributed population has recorded cholesterol levels with various statistics computed. From a sample of 100 subjects in the population, the sample mean was 214.12 mg/dL (milligrams per deciliter), with a sample standard deviation of 45.71 mg/dL. Perform a hypothesis test, with significance level 0.05, to test if there is enough evidence to conclude that the population mean is larger than 200 mg/dL. Hypothesis Test We will perform a hypothesis test using the $p$-value approach with significance level $\alpha=0.05:$ Define $H_0$: $\mu=200$. Define $H_1$: $\mu>200$. Since our values are normally distributed, the test statistic is $z^*=\frac{\bar X - \mu_0}{\frac{s}{\sqrt{n}}}=\frac{214.12 - 200}{\frac{45.71}{\sqrt{100}}}\approx 3.09$. Using a standard normal distribution, we find that our $p$-value is approximately $0.001$. Since the $p$-value is at most $\alpha=0.05,$ we reject $H_0$. Therefore, we can conclude that the test shows sufficient evidence to support the claim that $\mu$ is larger than $200$ mg/dL.

If the sample size was smaller, the normal and $t$-distributions behave differently. Also, the question itself must be managed by a double-tail test instead.

Assume a population's cholesterol levels are recorded and various statistics are computed. From a sample of 25 subjects, the sample mean was 214.12 mg/dL (milligrams per deciliter), with a sample standard deviation of 45.71 mg/dL. Perform a hypothesis test, with significance level 0.05, to test if there is enough evidence to conclude that the population mean is not equal to 200 mg/dL. Hypothesis Test We will perform a hypothesis test using the $p$-value approach with significance level $\alpha=0.05$ and the $t$-distribution with 24 degrees of freedom: Define $H_0$: $\mu=200$. Define $H_1$: $\mu\neq 200$. Using the $t$-distribution, the test statistic is $t^*=\frac{\bar X - \mu_0}{\frac{s}{\sqrt{n}}}=\frac{214.12 - 200}{\frac{45.71}{\sqrt{25}}}\approx 1.54$. Using a $t$-distribution with 24 degrees of freedom, we find that our $p$-value is approximately $2(0.068)=0.136$. We have multiplied by two since this is a two-tailed argument, i.e. the mean can be smaller than or larger than. Since the $p$-value is larger than $\alpha=0.05,$ we fail to reject $H_0$. Therefore, the test does not show sufficient evidence to support the claim that $\mu$ is not equal to $200$ mg/dL.

The complement of the rejection on a two-tailed hypothesis test (with significance level $\alpha$) for a population parameter $\theta$ is equivalent to finding a confidence interval $($with confidence level $1-\alpha)$ for the population parameter $\theta$. If the assumption on the parameter $\theta$ falls inside the confidence interval, then the test has failed to reject the null hypothesis $($with $p$-value greater than $\alpha).$ Otherwise, if $\theta$ does not fall in the confidence interval, then the null hypothesis is rejected in favor of the alternate $($with $p$-value at most $\alpha).$

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Why 0.05? Two Examples That Put Students in the Role of Decision Maker

Leah Dorazio

Any teacher of Introductory Statistics has heard this question more times than they can remember: “Why 0.05?” Here, the value 0.05 refers to the significance level in a hypothesis test. A nice overview of hypothesis tests is described in the Fall/Winter 2015 issue of STN .

In this article, I provide a brief review of the concepts of hypothesis test and significance level. Then, I describe two activities for teachers to proactively address the question of why 0.05 with their students, one tactile and numerical and the other online and visual. These activities challenge students to use their intuition and probability sense to make their own decisions.

In a hypothesis test, there are competing claims, or competing hypotheses. The alternate hypothesis HA asserts that a real change or effect has taken place, while the null hypothesis H0 asserts that no change or effect has taken place. The significance level defines how much evidence we require to reject H0 in favor of HA. It serves as the cutoff. The default cutoff commonly used is 0.05. If the p -value is less than 0.05, we reject H0. If the p -value is greater than 0.05, we do not reject H0.

But why use a cutoff of 0.05? An interesting look back at the thinking of the founders of hypothesis testing is described in a 2013 article in the magazine Significance , but that complicates the issue rather than explains it. A simple response might be to say that it is merely a rule of thumb or a mutually agreed upon value within the statistical community. Students are not satisfied with this answer. One might go on to assert that a value two standard deviations away from the expected is considered unusual. To connect this to the logic of hypothesis testing, as the probability of an event falls below 5%, we become more and more surprised if we witness it occurring. Students get this logic, but they still want to know why 0.05 and not some other number.

The first example involves my favorite way to introduce hypothesis tests and their corresponding logic. The basic idea is described by Stephen Eckert in the Journal of Statistics Education . I have heard of slightly different versions from various teachers over the years. It goes like this. Buy two decks of cards. Create a modified deck of cards that contains all red cards. In class, present the students with a game based on the color of the cards drawn. As an example, say that the first student who draws a black card will get $10. Randomly pick students one by one and have them select a card, showing the card to the class. Put the card back into the deck and reshuffle. (This is important—once, I did the drawing without replacement and two students both got a red Jack of Hearts!). After three red cards are drawn, students will get increasingly excited, thinking that they might be the winner; they are not yet suspicious. One by one, they continue to draw only red cards. Four red cards in a row, five …

At this point, you will begin to hear a lot of grumbling and questioning if the deck is rigged. With an exaggerated show of being offended, ask, “Why would you say that?”After a few responses, someone will eventually say, “Because if the deck were fair, it’s really unlikely that we would get five red cards in a row.” Exactly. Now they have experienced and verbalized the logic of hypothesis testing on their own.

At this point, we can investigate the probabilities. If the deck were fair (p=0.5), the probability of getting four red cards in a row would be: (1/2)(1/2)(1/2)(1/2)=(1/16)=0.0625

This is where students start to get suspicious, but will not yet denounce you. If the deck were fair (p=0.5), the probability of getting five red cards in a row would be: (1/2)(1/2)(1/2)(1/2)(1/2)=(1/32)=0.0315

And yes, at this point, students will be arguing that the deck is unfair (that is, rejecting the null claim of a fair deck). These calculations serve as a first pass in explaining why a cutoff or significance level of 0.05 is reasonable. It is around where our natural suspicion starts to kick in.

There are many variations on how to carry out this demonstration and how to wrap it up. I stop after five draws, saying that I want to stop while I’m ahead. If you continue, students will become too certain that the deck is unfair. With five draws, there is still doubt. At this point, students can choose to believe the deck is unfair or they can choose to believe they were just unlucky. Students will, of course, want to see the deck, which you should by no means show them. As I tell my students—the whole practice of statistics is dealing with uncertainty; there are no correct answers against which you can check the decisions you make. I like to put one black card in the deck. At the end, I will say, “I cannot tell you whether the deck is fair or unfair, but …” and I show them the black card. This gives them a shock and makes them reconsider their conclusion based on this new evidence. If you want to really confound the students, steam open one of the new decks, and, after creating the modified deck, carefully reseal it. This way, you can make a big to-do of showing the class it is a brand new deck (thus increasing the threshold of suspicion).

Figure 1: Three plots from openintro.org/stat/why05

The second example comes from OpenIntro.org , a group (for which I am a volunteer) that produces free and open-source statistics resources, including introductory statistics textbooks and TI and Casio graphing calculator tutorials. Instead of approaching the why 0.05 question from a quantitative angle using a one-sample test for proportions, it approaches it from a visual angle using scatterplots and the hypothesis test for a significant linear relationship. After an introductory video, the visitor is presented with a series of 15 scatterplots (three of which are shown in Figure 1 ). For each one, they must decide whether the graph provides enough evidence of a real, upward trend or not enough evidence for a real, upward trend; that is, whether they reject H0 or do not reject H0. After completing this task, a follow-up video explains the results and then the visitor is presented with their individual results based on the choices they made for the set of scatterplots. Each graph has an associated p -value for the test on the slope of the regression line.

Figure 2: The author’s results from openintro.org/stat/why05

Now the “Aha!” moment. The next screen reveals all 15 graphs with their associated p-values, and for each one whether the visitor rejected H0 or did not reject H0 based on their intuition and their visual inspection. I did this activity, and, as seen in Figure 2 the cutoff for me was 0.06! This means I found evidence for a significant linear relationship for the graphs that had an associated p -value less than 0.06 and did not find evidence for a significant linear relationship for the graphs that had an associated p -value greater than 0.06. Of course, each individual’s cutoff may vary. This activity is great to do in class (the videos can be skipped, with the teacher filling in the necessary explanation). When all the students finish, make a dotplot of the cutoffs each student (subconsciously) used and find the class average. You will find that this value is consistently close to 0.05!

There are limitations to focusing too narrowly on the p -value. The ASA recently published a statement on p -values, stating: “Well-reasoned statistical arguments contain much more than the value of a single number and whether that number exceeds an arbitrary threshold.” As teachers, we must also address issues of experimental design, the question of causality, and the true interpretation of p -value. Nevertheless, these two introductory examples serve to provide students with a strong understanding of the logic of hypothesis testing and a sense of why 0.05 is a reasonable default significance level to use. If a problem does not indicate what significance level to use, students can be allowed to use either their individual cutoff value from the activity or the class average (assuming these are reasonable values). By engaging students in deciding for themselves how much evidence is enough evidence, they come to better understand that evidence presents itself along a continuum, even though, at the end of the day, a decision must be made.

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Why do we so often use 0.05 for hypothesis testing?

In this online exercise, we hope to help you gain an improved understanding of what a significance level is, and why a value in the neighborhood of 0.05 is reasonable as a default.

Watch this video before starting

We consider two competing hypotheses

H 0 : There is no real trend. Any apparent trend is due to chance.
H A : There is a real, upward trend.

Your Natural Threshold

Based on your selections, your "natural" threshold for this exercise is around .

Watch the video below to understand what this number means.

How was it calculated?

We can compute p-values to quantify the probability, when there is no real trend, of making a Type 1 Error and (incorrectly) concluding there is a trend.

H 0 : There is no real trend. Any apparent trend is due to chance. (Response of "No")
H A : There is a real, upward trend. (Response of "Trend")

Below are your decisions alongside the p-values for the corresponding plots:

- 0.014 - 0.031 - 0.041 - 0.056 - 0.064 - 0.096 - 0.105 - 0.153 - 0.167 - 0.253 - 0.327 - 0.436 - 0.514 - 0.598 - 0.823

Please take our 2-minute survey on this exercise.

Full set of plots

If you're curious to see the plots in addition to your decisions and the corresponding p-values, browse them below.

An interesting article on 0.05 as a significance level: Kelly M. Emily Dickinson and monkeys on the stair Or: What is the significance of the 5% significance level? Significance 10:5. 2013.

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S.3.1 hypothesis testing (critical value approach).

The critical value approach involves determining "likely" or "unlikely" by determining whether or not the observed test statistic is more extreme than would be expected if the null hypothesis were true. That is, it entails comparing the observed test statistic to some cutoff value, called the " critical value ." If the test statistic is more extreme than the critical value, then the null hypothesis is rejected in favor of the alternative hypothesis. If the test statistic is not as extreme as the critical value, then the null hypothesis is not rejected.

Specifically, the four steps involved in using the critical value approach to conducting any hypothesis test are:

Specify the null and alternative hypotheses.
Using the sample data and assuming the null hypothesis is true, calculate the value of the test statistic. To conduct the hypothesis test for the population mean μ , we use the t -statistic $t^*=\frac{\bar{x}-\mu}{s/\sqrt{n}}$ which follows a t -distribution with n - 1 degrees of freedom.
Determine the critical value by finding the value of the known distribution of the test statistic such that the probability of making a Type I error — which is denoted $\alpha$ (greek letter "alpha") and is called the " significance level of the test " — is small (typically 0.01, 0.05, or 0.10).
Compare the test statistic to the critical value. If the test statistic is more extreme in the direction of the alternative than the critical value, reject the null hypothesis in favor of the alternative hypothesis. If the test statistic is less extreme than the critical value, do not reject the null hypothesis.

Example S.3.1.1

Mean gpa section .

In our example concerning the mean grade point average, suppose we take a random sample of n = 15 students majoring in mathematics. Since n = 15, our test statistic t * has n - 1 = 14 degrees of freedom. Also, suppose we set our significance level α at 0.05 so that we have only a 5% chance of making a Type I error.

Right-Tailed

The critical value for conducting the right-tailed test H 0 : μ = 3 versus H A : μ > 3 is the t -value, denoted t $\alpha$ , n - 1 , such that the probability to the right of it is $\alpha$. It can be shown using either statistical software or a t -table that the critical value t 0.05,14 is 1.7613. That is, we would reject the null hypothesis H 0 : μ = 3 in favor of the alternative hypothesis H A : μ > 3 if the test statistic t * is greater than 1.7613. Visually, the rejection region is shaded red in the graph.

t distribution graph for a t value of 1.76131

Left-Tailed

The critical value for conducting the left-tailed test H 0 : μ = 3 versus H A : μ < 3 is the t -value, denoted -t ( $\alpha$ , n - 1) , such that the probability to the left of it is $\alpha$. It can be shown using either statistical software or a t -table that the critical value -t 0.05,14 is -1.7613. That is, we would reject the null hypothesis H 0 : μ = 3 in favor of the alternative hypothesis H A : μ < 3 if the test statistic t * is less than -1.7613. Visually, the rejection region is shaded red in the graph.

There are two critical values for the two-tailed test H 0 : μ = 3 versus H A : μ ≠ 3 — one for the left-tail denoted -t ( $\alpha$ / 2, n - 1) and one for the right-tail denoted t ( $\alpha$ / 2, n - 1) . The value - t ( $\alpha$ /2, n - 1) is the t -value such that the probability to the left of it is $\alpha$/2, and the value t ( $\alpha$ /2, n - 1) is the t -value such that the probability to the right of it is $\alpha$/2. It can be shown using either statistical software or a t -table that the critical value -t 0.025,14 is -2.1448 and the critical value t 0.025,14 is 2.1448. That is, we would reject the null hypothesis H 0 : μ = 3 in favor of the alternative hypothesis H A : μ ≠ 3 if the test statistic t * is less than -2.1448 or greater than 2.1448. Visually, the rejection region is shaded red in the graph.

t distribution graph for a two tailed test of 0.05 level of significance

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9.E: Hypothesis Testing with One Sample (Exercises)

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These are homework exercises to accompany the Textmap created for "Introductory Statistics" by OpenStax.

9.1: Introduction

9.2: null and alternative hypotheses.

Some of the following statements refer to the null hypothesis, some to the alternate hypothesis.

State the null hypothesis, $H_{0}$, and the alternative hypothesis. $H_{a}$, in terms of the appropriate parameter $(\mu \text{or} p)$.

The mean number of years Americans work before retiring is 34.
At most 60% of Americans vote in presidential elections.
The mean starting salary for San Jose State University graduates is at least $100,000 per year.
Twenty-nine percent of high school seniors get drunk each month.
Fewer than 5% of adults ride the bus to work in Los Angeles.
The mean number of cars a person owns in her lifetime is not more than ten.
About half of Americans prefer to live away from cities, given the choice.
Europeans have a mean paid vacation each year of six weeks.
The chance of developing breast cancer is under 11% for women.
Private universities' mean tuition cost is more than $20,000 per year.
$H_{0}: \mu = 34; H_{a}: \mu \neq 34$
$H_{0}: p \leq 0.60; H_{a}: p > 0.60$
$H_{0}: \mu \geq 100,000; H_{a}: \mu < 100,000$
$H_{0}: p = 0.29; H_{a}: p \neq 0.29$
$H_{0}: p = 0.05; H_{a}: p < 0.05$
$H_{0}: \mu \leq 10; H_{a}: \mu > 10$
$H_{0}: p = 0.50; H_{a}: p \neq 0.50$
$H_{0}: \mu = 6; H_{a}: \mu \neq 6$
$H_{0}: p ≥ 0.11; H_{a}: p < 0.11$
$H_{0}: \mu \leq 20,000; H_{a}: \mu > 20,000$

Over the past few decades, public health officials have examined the link between weight concerns and teen girls' smoking. Researchers surveyed a group of 273 randomly selected teen girls living in Massachusetts (between 12 and 15 years old). After four years the girls were surveyed again. Sixty-three said they smoked to stay thin. Is there good evidence that more than thirty percent of the teen girls smoke to stay thin? The alternative hypothesis is:

$p < 0.30$
$p \leq 0.30$
$p \geq 0.30$
$p > 0.30$

A statistics instructor believes that fewer than 20% of Evergreen Valley College (EVC) students attended the opening night midnight showing of the latest Harry Potter movie. She surveys 84 of her students and finds that 11 attended the midnight showing. An appropriate alternative hypothesis is:

$p = 0.20$
$p > 0.20$
$p < 0.20$
$p \leq 0.20$

Previously, an organization reported that teenagers spent 4.5 hours per week, on average, on the phone. The organization thinks that, currently, the mean is higher. Fifteen randomly chosen teenagers were asked how many hours per week they spend on the phone. The sample mean was 4.75 hours with a sample standard deviation of 2.0. Conduct a hypothesis test. The null and alternative hypotheses are:

$H_{0}: \bar{x} = 4.5, H_{a}: \bar{x} > 4.5$
$H_{0}: \mu \geq 4.5, H_{a}: \mu < 4.5$
$H_{0}: \mu = 4.75, H_{a}: \mu > 4.75$
$H_{0}: \mu = 4.5, H_{a}: \mu > 4.5$

9.3: Outcomes and the Type I and Type II Errors

State the Type I and Type II errors in complete sentences given the following statements.

The mean number of cars a person owns in his or her lifetime is not more than ten.
Private universities mean tuition cost is more than $20,000 per year.
Type I error: We conclude that the mean is not 34 years, when it really is 34 years. Type II error: We conclude that the mean is 34 years, when in fact it really is not 34 years.
Type I error: We conclude that more than 60% of Americans vote in presidential elections, when the actual percentage is at most 60%.Type II error: We conclude that at most 60% of Americans vote in presidential elections when, in fact, more than 60% do.
Type I error: We conclude that the mean starting salary is less than $100,000, when it really is at least $100,000. Type II error: We conclude that the mean starting salary is at least $100,000 when, in fact, it is less than $100,000.
Type I error: We conclude that the proportion of high school seniors who get drunk each month is not 29%, when it really is 29%. Type II error: We conclude that the proportion of high school seniors who get drunk each month is 29% when, in fact, it is not 29%.
Type I error: We conclude that fewer than 5% of adults ride the bus to work in Los Angeles, when the percentage that do is really 5% or more. Type II error: We conclude that 5% or more adults ride the bus to work in Los Angeles when, in fact, fewer that 5% do.
Type I error: We conclude that the mean number of cars a person owns in his or her lifetime is more than 10, when in reality it is not more than 10. Type II error: We conclude that the mean number of cars a person owns in his or her lifetime is not more than 10 when, in fact, it is more than 10.
Type I error: We conclude that the proportion of Americans who prefer to live away from cities is not about half, though the actual proportion is about half. Type II error: We conclude that the proportion of Americans who prefer to live away from cities is half when, in fact, it is not half.
Type I error: We conclude that the duration of paid vacations each year for Europeans is not six weeks, when in fact it is six weeks. Type II error: We conclude that the duration of paid vacations each year for Europeans is six weeks when, in fact, it is not.
Type I error: We conclude that the proportion is less than 11%, when it is really at least 11%. Type II error: We conclude that the proportion of women who develop breast cancer is at least 11%, when in fact it is less than 11%.
Type I error: We conclude that the average tuition cost at private universities is more than $20,000, though in reality it is at most $20,000. Type II error: We conclude that the average tuition cost at private universities is at most $20,000 when, in fact, it is more than $20,000.

For statements a-j in Exercise 9.109 , answer the following in complete sentences.

State a consequence of committing a Type I error.
State a consequence of committing a Type II error.

When a new drug is created, the pharmaceutical company must subject it to testing before receiving the necessary permission from the Food and Drug Administration (FDA) to market the drug. Suppose the null hypothesis is “the drug is unsafe.” What is the Type II Error?

To conclude the drug is safe when in, fact, it is unsafe.
Not to conclude the drug is safe when, in fact, it is safe.
To conclude the drug is safe when, in fact, it is safe.
Not to conclude the drug is unsafe when, in fact, it is unsafe.

A statistics instructor believes that fewer than 20% of Evergreen Valley College (EVC) students attended the opening midnight showing of the latest Harry Potter movie. She surveys 84 of her students and finds that 11 of them attended the midnight showing. The Type I error is to conclude that the percent of EVC students who attended is ________.

at least 20%, when in fact, it is less than 20%.
20%, when in fact, it is 20%.
less than 20%, when in fact, it is at least 20%.
less than 20%, when in fact, it is less than 20%.

It is believed that Lake Tahoe Community College (LTCC) Intermediate Algebra students get less than seven hours of sleep per night, on average. A survey of 22 LTCC Intermediate Algebra students generated a mean of 7.24 hours with a standard deviation of 1.93 hours. At a level of significance of 5%, do LTCC Intermediate Algebra students get less than seven hours of sleep per night, on average?

The Type II error is not to reject that the mean number of hours of sleep LTCC students get per night is at least seven when, in fact, the mean number of hours

is more than seven hours.
is at most seven hours.
is at least seven hours.
is less than seven hours.

to conclude that the current mean hours per week is higher than 4.5, when in fact, it is higher
to conclude that the current mean hours per week is higher than 4.5, when in fact, it is the same
to conclude that the mean hours per week currently is 4.5, when in fact, it is higher
to conclude that the mean hours per week currently is no higher than 4.5, when in fact, it is not higher

9.4: Distribution Needed for Hypothesis Testing

$N\left(7.24, \frac{1.93}{\sqrt{22}}\right)$
$N\left(7.24, 1.93\right)$

9.5: Rare Events, the Sample, Decision and Conclusion

The National Institute of Mental Health published an article stating that in any one-year period, approximately 9.5 percent of American adults suffer from depression or a depressive illness. Suppose that in a survey of 100 people in a certain town, seven of them suffered from depression or a depressive illness. Conduct a hypothesis test to determine if the true proportion of people in that town suffering from depression or a depressive illness is lower than the percent in the general adult American population.

Is this a test of one mean or proportion?
State the null and alternative hypotheses. $H_{0}$ : ____________________ $H_{a}$ : ____________________
Is this a right-tailed, left-tailed, or two-tailed test?
What symbol represents the random variable for this test?
In words, define the random variable for this test.
$x =$ ________________
$n =$ ________________
$p′ =$ _____________
Calculate $\sigma_{x} =$ __________. Show the formula set-up.
State the distribution to use for the hypothesis test.
Find the $p\text{-value}$.
Reason for the decision:
Conclusion (write out in a complete sentence):

9.6: Additional Information and Full Hypothesis Test Examples

For each of the word problems, use a solution sheet to do the hypothesis test. The solution sheet is found in [link] . Please feel free to make copies of the solution sheets. For the online version of the book, it is suggested that you copy the .doc or the .pdf files.

If you are using a Student's $t$ - distribution for one of the following homework problems, you may assume that the underlying population is normally distributed. (In general, you must first prove that assumption, however.)

A particular brand of tires claims that its deluxe tire averages at least 50,000 miles before it needs to be replaced. From past studies of this tire, the standard deviation is known to be 8,000. A survey of owners of that tire design is conducted. From the 28 tires surveyed, the mean lifespan was 46,500 miles with a standard deviation of 9,800 miles. Using $\alpha = 0.05$, is the data highly inconsistent with the claim?

$H_{0}: \mu \geq 50,000$
$H_{a}: \mu < 50,000$
Let $\bar{X} =$ the average lifespan of a brand of tires.
normal distribution
$z = -2.315$
$p\text{-value} = 0.0103$
Check student’s solution.
alpha: 0.05
Decision: Reject the null hypothesis.
Reason for decision: The $p\text{-value}$ is less than 0.05.
Conclusion: There is sufficient evidence to conclude that the mean lifespan of the tires is less than 50,000 miles.
$(43,537, 49,463)$

From generation to generation, the mean age when smokers first start to smoke varies. However, the standard deviation of that age remains constant of around 2.1 years. A survey of 40 smokers of this generation was done to see if the mean starting age is at least 19. The sample mean was 18.1 with a sample standard deviation of 1.3. Do the data support the claim at the 5% level?

The cost of a daily newspaper varies from city to city. However, the variation among prices remains steady with a standard deviation of 20¢. A study was done to test the claim that the mean cost of a daily newspaper is $1.00. Twelve costs yield a mean cost of 95¢ with a standard deviation of 18¢. Do the data support the claim at the 1% level?

$H_{0}: \mu = $1.00$
$H_{a}: \mu \neq $1.00$
Let $\bar{X} =$ the average cost of a daily newspaper.
$z = –0.866$
$p\text{-value} = 0.3865$
$\alpha: 0.01$
Decision: Do not reject the null hypothesis.
Reason for decision: The $p\text{-value}$ is greater than 0.01.
Conclusion: There is sufficient evidence to support the claim that the mean cost of daily papers is $1. The mean cost could be $1.
$($0.84, $1.06)$

An article in the San Jose Mercury News stated that students in the California state university system take 4.5 years, on average, to finish their undergraduate degrees. Suppose you believe that the mean time is longer. You conduct a survey of 49 students and obtain a sample mean of 5.1 with a sample standard deviation of 1.2. Do the data support your claim at the 1% level?

The mean number of sick days an employee takes per year is believed to be about ten. Members of a personnel department do not believe this figure. They randomly survey eight employees. The number of sick days they took for the past year are as follows: 12; 4; 15; 3; 11; 8; 6; 8. Let $x =$ the number of sick days they took for the past year. Should the personnel team believe that the mean number is ten?

$H_{0}: \mu = 10$
$H_{a}: \mu \neq 10$
Let $\bar{X}$ the mean number of sick days an employee takes per year.
Student’s t -distribution
$t = –1.12$
$p\text{-value} = 0.300$
$\alpha: 0.05$
Reason for decision: The $p\text{-value}$ is greater than 0.05.
Conclusion: At the 5% significance level, there is insufficient evidence to conclude that the mean number of sick days is not ten.
$(4.9443, 11.806)$

In 1955, Life Magazine reported that the 25 year-old mother of three worked, on average, an 80 hour week. Recently, many groups have been studying whether or not the women's movement has, in fact, resulted in an increase in the average work week for women (combining employment and at-home work). Suppose a study was done to determine if the mean work week has increased. 81 women were surveyed with the following results. The sample mean was 83; the sample standard deviation was ten. Does it appear that the mean work week has increased for women at the 5% level?

Your statistics instructor claims that 60 percent of the students who take her Elementary Statistics class go through life feeling more enriched. For some reason that she can't quite figure out, most people don't believe her. You decide to check this out on your own. You randomly survey 64 of her past Elementary Statistics students and find that 34 feel more enriched as a result of her class. Now, what do you think?

$H_{0}: p \geq 0.6$
$H_{a}: p < 0.6$
Let $P′ =$ the proportion of students who feel more enriched as a result of taking Elementary Statistics.
normal for a single proportion
$p\text{-value} = 0.1308$
Conclusion: There is insufficient evidence to conclude that less than 60 percent of her students feel more enriched.

The “plus-4s” confidence interval is $(0.411, 0.648)$

A Nissan Motor Corporation advertisement read, “The average man’s I.Q. is 107. The average brown trout’s I.Q. is 4. So why can’t man catch brown trout?” Suppose you believe that the brown trout’s mean I.Q. is greater than four. You catch 12 brown trout. A fish psychologist determines the I.Q.s as follows: 5; 4; 7; 3; 6; 4; 5; 3; 6; 3; 8; 5. Conduct a hypothesis test of your belief.

Refer to Exercise 9.119 . Conduct a hypothesis test to see if your decision and conclusion would change if your belief were that the brown trout’s mean I.Q. is not four.

$H_{0}: \mu = 4$
$H_{a}: \mu \neq 4$
Let $\bar{X}$ the average I.Q. of a set of brown trout.
two-tailed Student's t-test
$t = 1.95$
$p\text{-value} = 0.076$
Reason for decision: The $p\text{-value}$ is greater than 0.05
Conclusion: There is insufficient evidence to conclude that the average IQ of brown trout is not four.
$(3.8865,5.9468)$

According to an article in Newsweek , the natural ratio of girls to boys is 100:105. In China, the birth ratio is 100: 114 (46.7% girls). Suppose you don’t believe the reported figures of the percent of girls born in China. You conduct a study. In this study, you count the number of girls and boys born in 150 randomly chosen recent births. There are 60 girls and 90 boys born of the 150. Based on your study, do you believe that the percent of girls born in China is 46.7?

A poll done for Newsweek found that 13% of Americans have seen or sensed the presence of an angel. A contingent doubts that the percent is really that high. It conducts its own survey. Out of 76 Americans surveyed, only two had seen or sensed the presence of an angel. As a result of the contingent’s survey, would you agree with the Newsweek poll? In complete sentences, also give three reasons why the two polls might give different results.

$H_{a}: p < 0.13$
Let $P′ =$ the proportion of Americans who have seen or sensed angels
–2.688
$p\text{-value} = 0.0036$
Reason for decision: The $p\text{-value}$e is less than 0.05.
Conclusion: There is sufficient evidence to conclude that the percentage of Americans who have seen or sensed an angel is less than 13%.

The“plus-4s” confidence interval is (0.0022, 0.0978)

The mean work week for engineers in a start-up company is believed to be about 60 hours. A newly hired engineer hopes that it’s shorter. She asks ten engineering friends in start-ups for the lengths of their mean work weeks. Based on the results that follow, should she count on the mean work week to be shorter than 60 hours?

Data (length of mean work week): 70; 45; 55; 60; 65; 55; 55; 60; 50; 55.

Use the “Lap time” data for Lap 4 (see [link] ) to test the claim that Terri finishes Lap 4, on average, in less than 129 seconds. Use all twenty races given.

$H_{0}: \mu \geq 129$
$H_{a}: \mu < 129$
Let $\bar{X} =$ the average time in seconds that Terri finishes Lap 4.
Student's t -distribution
$t = 1.209$
Conclusion: There is insufficient evidence to conclude that Terri’s mean lap time is less than 129 seconds.
$(128.63, 130.37)$

Use the “Initial Public Offering” data (see [link] ) to test the claim that the mean offer price was $18 per share. Do not use all the data. Use your random number generator to randomly survey 15 prices.

The following questions were written by past students. They are excellent problems!

"Asian Family Reunion," by Chau Nguyen

Every two years it comes around.

We all get together from different towns.

In my honest opinion,

It's not a typical family reunion.

Not forty, or fifty, or sixty,

But how about seventy companions!

The kids would play, scream, and shout

One minute they're happy, another they'll pout.

The teenagers would look, stare, and compare

From how they look to what they wear.

The men would chat about their business

That they make more, but never less.

Money is always their subject

And there's always talk of more new projects.

The women get tired from all of the chats

They head to the kitchen to set out the mats.

Some would sit and some would stand

Eating and talking with plates in their hands.

Then come the games and the songs

And suddenly, everyone gets along!

With all that laughter, it's sad to say

That it always ends in the same old way.

They hug and kiss and say "good-bye"

And then they all begin to cry!

I say that 60 percent shed their tears

But my mom counted 35 people this year.

She said that boys and men will always have their pride,

So we won't ever see them cry.

I myself don't think she's correct,

So could you please try this problem to see if you object?

$H_{0}: p = 0.60$
$H_{a}: p < 0.60$
Let $P′ =$ the proportion of family members who shed tears at a reunion.
–1.71
Reason for decision: $p\text{-value} < \alpha$
Conclusion: At the 5% significance level, there is sufficient evidence to conclude that the proportion of family members who shed tears at a reunion is less than 0.60. However, the test is weak because the $p\text{-value}$ and alpha are quite close, so other tests should be done.
We are 95% confident that between 38.29% and 61.71% of family members will shed tears at a family reunion. $(0.3829, 0.6171)$. The“plus-4s” confidence interval (see chapter 8) is $(0.3861, 0.6139)$

Note that here the “large-sample” $1 - \text{PropZTest}$ provides the approximate $p\text{-value}$ of 0.0438. Whenever a $p\text{-value}$ based on a normal approximation is close to the level of significance, the exact $p\text{-value}$ based on binomial probabilities should be calculated whenever possible. This is beyond the scope of this course.

"The Problem with Angels," by Cyndy Dowling

Although this problem is wholly mine,

The catalyst came from the magazine, Time.

On the magazine cover I did find

The realm of angels tickling my mind.

Inside, 69% I found to be

In angels, Americans do believe.

Then, it was time to rise to the task,

Ninety-five high school and college students I did ask.

Viewing all as one group,

Random sampling to get the scoop.

So, I asked each to be true,

"Do you believe in angels?" Tell me, do!

Hypothesizing at the start,

Totally believing in my heart

That the proportion who said yes

Would be equal on this test.

Lo and behold, seventy-three did arrive,

Out of the sample of ninety-five.

Now your job has just begun,

Solve this problem and have some fun.

"Blowing Bubbles," by Sondra Prull

Studying stats just made me tense,

I had to find some sane defense.

Some light and lifting simple play

To float my math anxiety away.

Blowing bubbles lifts me high

Takes my troubles to the sky.

POIK! They're gone, with all my stress

Bubble therapy is the best.

The label said each time I blew

The average number of bubbles would be at least 22.

I blew and blew and this I found

From 64 blows, they all are round!

But the number of bubbles in 64 blows

Varied widely, this I know.

20 per blow became the mean

They deviated by 6, and not 16.

From counting bubbles, I sure did relax

But now I give to you your task.

Was 22 a reasonable guess?

Find the answer and pass this test!

$H_{0}: \mu \geq 22$
$H_{a}: \mu < 22$
Let $\bar{X} =$ the mean number of bubbles per blow.
–2.667
$p\text{-value} = 0.00486$
Conclusion: There is sufficient evidence to conclude that the mean number of bubbles per blow is less than 22.
$(18.501, 21.499)$

"Dalmatian Darnation," by Kathy Sparling

A greedy dog breeder named Spreckles

Bred puppies with numerous freckles

The Dalmatians he sought

Possessed spot upon spot

The more spots, he thought, the more shekels.

His competitors did not agree

That freckles would increase the fee.

They said, “Spots are quite nice

But they don't affect price;

One should breed for improved pedigree.”

The breeders decided to prove

This strategy was a wrong move.

Breeding only for spots

Would wreak havoc, they thought.

His theory they want to disprove.

They proposed a contest to Spreckles

Comparing dog prices to freckles.

In records they looked up

One hundred one pups:

Dalmatians that fetched the most shekels.

They asked Mr. Spreckles to name

An average spot count he'd claim

To bring in big bucks.

Said Spreckles, “Well, shucks,

It's for one hundred one that I aim.”

Said an amateur statistician

Who wanted to help with this mission.

“Twenty-one for the sample

Standard deviation's ample:

They examined one hundred and one

Dalmatians that fetched a good sum.

They counted each spot,

Mark, freckle and dot

And tallied up every one.

Instead of one hundred one spots

They averaged ninety six dots

Can they muzzle Spreckles’

Obsession with freckles

Based on all the dog data they've got?

"Macaroni and Cheese, please!!" by Nedda Misherghi and Rachelle Hall

As a poor starving student I don't have much money to spend for even the bare necessities. So my favorite and main staple food is macaroni and cheese. It's high in taste and low in cost and nutritional value.

One day, as I sat down to determine the meaning of life, I got a serious craving for this, oh, so important, food of my life. So I went down the street to Greatway to get a box of macaroni and cheese, but it was SO expensive! $2.02 !!! Can you believe it? It made me stop and think. The world is changing fast. I had thought that the mean cost of a box (the normal size, not some super-gigantic-family-value-pack) was at most $1, but now I wasn't so sure. However, I was determined to find out. I went to 53 of the closest grocery stores and surveyed the prices of macaroni and cheese. Here are the data I wrote in my notebook:

Price per box of Mac and Cheese:

5 stores @ $2.02
15 stores @ $0.25
3 stores @ $1.29
6 stores @ $0.35
4 stores @ $2.27
7 stores @ $1.50
5 stores @ $1.89
8 stores @ 0.75.

I could see that the cost varied but I had to sit down to figure out whether or not I was right. If it does turn out that this mouth-watering dish is at most $1, then I'll throw a big cheesy party in our next statistics lab, with enough macaroni and cheese for just me. (After all, as a poor starving student I can't be expected to feed our class of animals!)

$H_{0}: \mu \leq 1$
$H_{a}: \mu > 1$
Let $\bar{X} =$ the mean cost in dollars of macaroni and cheese in a certain town.
Student's $t$-distribution
$t = 0.340$
$p\text{-value} = 0.36756$
Conclusion: The mean cost could be $1, or less. At the 5% significance level, there is insufficient evidence to conclude that the mean price of a box of macaroni and cheese is more than $1.
$(0.8291, 1.241)$

"William Shakespeare: The Tragedy of Hamlet, Prince of Denmark," by Jacqueline Ghodsi

THE CHARACTERS (in order of appearance):

HAMLET, Prince of Denmark and student of Statistics
POLONIUS, Hamlet’s tutor
HOROTIO, friend to Hamlet and fellow student

Scene: The great library of the castle, in which Hamlet does his lessons

(The day is fair, but the face of Hamlet is clouded. He paces the large room. His tutor, Polonius, is reprimanding Hamlet regarding the latter’s recent experience. Horatio is seated at the large table at right stage.)

POLONIUS: My Lord, how cans’t thou admit that thou hast seen a ghost! It is but a figment of your imagination!

HAMLET: I beg to differ; I know of a certainty that five-and-seventy in one hundred of us, condemned to the whips and scorns of time as we are, have gazed upon a spirit of health, or goblin damn’d, be their intents wicked or charitable.

POLONIUS If thou doest insist upon thy wretched vision then let me invest your time; be true to thy work and speak to me through the reason of the null and alternate hypotheses. (He turns to Horatio.) Did not Hamlet himself say, “What piece of work is man, how noble in reason, how infinite in faculties? Then let not this foolishness persist. Go, Horatio, make a survey of three-and-sixty and discover what the true proportion be. For my part, I will never succumb to this fantasy, but deem man to be devoid of all reason should thy proposal of at least five-and-seventy in one hundred hold true.

HORATIO (to Hamlet): What should we do, my Lord?

HAMLET: Go to thy purpose, Horatio.

HORATIO: To what end, my Lord?

HAMLET: That you must teach me. But let me conjure you by the rights of our fellowship, by the consonance of our youth, but the obligation of our ever-preserved love, be even and direct with me, whether I am right or no.

(Horatio exits, followed by Polonius, leaving Hamlet to ponder alone.)

(The next day, Hamlet awaits anxiously the presence of his friend, Horatio. Polonius enters and places some books upon the table just a moment before Horatio enters.)

POLONIUS: So, Horatio, what is it thou didst reveal through thy deliberations?

HORATIO: In a random survey, for which purpose thou thyself sent me forth, I did discover that one-and-forty believe fervently that the spirits of the dead walk with us. Before my God, I might not this believe, without the sensible and true avouch of mine own eyes.

POLONIUS: Give thine own thoughts no tongue, Horatio. (Polonius turns to Hamlet.) But look to’t I charge you, my Lord. Come Horatio, let us go together, for this is not our test. (Horatio and Polonius leave together.)

HAMLET: To reject, or not reject, that is the question: whether ‘tis nobler in the mind to suffer the slings and arrows of outrageous statistics, or to take arms against a sea of data, and, by opposing, end them. (Hamlet resignedly attends to his task.)

(Curtain falls)

"Untitled," by Stephen Chen

I've often wondered how software is released and sold to the public. Ironically, I work for a company that sells products with known problems. Unfortunately, most of the problems are difficult to create, which makes them difficult to fix. I usually use the test program X, which tests the product, to try to create a specific problem. When the test program is run to make an error occur, the likelihood of generating an error is 1%.

So, armed with this knowledge, I wrote a new test program Y that will generate the same error that test program X creates, but more often. To find out if my test program is better than the original, so that I can convince the management that I'm right, I ran my test program to find out how often I can generate the same error. When I ran my test program 50 times, I generated the error twice. While this may not seem much better, I think that I can convince the management to use my test program instead of the original test program. Am I right?

$H_{0}: p = 0.01$
$H_{a}: p > 0.01$
Let $P′ =$ the proportion of errors generated
Normal for a single proportion
Decision: Reject the null hypothesis
Conclusion: At the 5% significance level, there is sufficient evidence to conclude that the proportion of errors generated is more than 0.01.

The“plus-4s” confidence interval is $(0.004, 0.144)$.

"Japanese Girls’ Names"

by Kumi Furuichi

It used to be very typical for Japanese girls’ names to end with “ko.” (The trend might have started around my grandmothers’ generation and its peak might have been around my mother’s generation.) “Ko” means “child” in Chinese characters. Parents would name their daughters with “ko” attaching to other Chinese characters which have meanings that they want their daughters to become, such as Sachiko—happy child, Yoshiko—a good child, Yasuko—a healthy child, and so on.

However, I noticed recently that only two out of nine of my Japanese girlfriends at this school have names which end with “ko.” More and more, parents seem to have become creative, modernized, and, sometimes, westernized in naming their children.

I have a feeling that, while 70 percent or more of my mother’s generation would have names with “ko” at the end, the proportion has dropped among my peers. I wrote down all my Japanese friends’, ex-classmates’, co-workers, and acquaintances’ names that I could remember. Following are the names. (Some are repeats.) Test to see if the proportion has dropped for this generation.

Ai, Akemi, Akiko, Ayumi, Chiaki, Chie, Eiko, Eri, Eriko, Fumiko, Harumi, Hitomi, Hiroko, Hiroko, Hidemi, Hisako, Hinako, Izumi, Izumi, Junko, Junko, Kana, Kanako, Kanayo, Kayo, Kayoko, Kazumi, Keiko, Keiko, Kei, Kumi, Kumiko, Kyoko, Kyoko, Madoka, Maho, Mai, Maiko, Maki, Miki, Miki, Mikiko, Mina, Minako, Miyako, Momoko, Nana, Naoko, Naoko, Naoko, Noriko, Rieko, Rika, Rika, Rumiko, Rei, Reiko, Reiko, Sachiko, Sachiko, Sachiyo, Saki, Sayaka, Sayoko, Sayuri, Seiko, Shiho, Shizuka, Sumiko, Takako, Takako, Tomoe, Tomoe, Tomoko, Touko, Yasuko, Yasuko, Yasuyo, Yoko, Yoko, Yoko, Yoshiko, Yoshiko, Yoshiko, Yuka, Yuki, Yuki, Yukiko, Yuko, Yuko.

"Phillip’s Wish," by Suzanne Osorio

My nephew likes to play

Chasing the girls makes his day.

He asked his mother

If it is okay

To get his ear pierced.

She said, “No way!”

To poke a hole through your ear,

Is not what I want for you, dear.

He argued his point quite well,

Says even my macho pal, Mel,

Has gotten this done.

It’s all just for fun.

C’mon please, mom, please, what the hell.

Again Phillip complained to his mother,

Saying half his friends (including their brothers)

Are piercing their ears

And they have no fears

He wants to be like the others.

She said, “I think it’s much less.

We must do a hypothesis test.

And if you are right,

I won’t put up a fight.

But, if not, then my case will rest.”

We proceeded to call fifty guys

To see whose prediction would fly.

Nineteen of the fifty

Said piercing was nifty

And earrings they’d occasionally buy.

Then there’s the other thirty-one,

Who said they’d never have this done.

So now this poem’s finished.

Will his hopes be diminished,

Or will my nephew have his fun?

$H_{0}: p = 0.50$
$H_{a}: p < 0.50$
Let $P′ =$ the proportion of friends that has a pierced ear.
–1.70
$p\text{-value} = 0.0448$
Reason for decision: The $p\text{-value}$ is less than 0.05. (However, they are very close.)
Conclusion: There is sufficient evidence to support the claim that less than 50% of his friends have pierced ears.
Confidence Interval: $(0.245, 0.515)$: The “plus-4s” confidence interval is $(0.259, 0.519)$.

"The Craven," by Mark Salangsang

Once upon a morning dreary

In stats class I was weak and weary.

Pondering over last night’s homework

Whose answers were now on the board

This I did and nothing more.

While I nodded nearly napping

Suddenly, there came a tapping.

As someone gently rapping,

Rapping my head as I snore.

Quoth the teacher, “Sleep no more.”

“In every class you fall asleep,”

The teacher said, his voice was deep.

“So a tally I’ve begun to keep

Of every class you nap and snore.

The percentage being forty-four.”

“My dear teacher I must confess,

While sleeping is what I do best.

The percentage, I think, must be less,

A percentage less than forty-four.”

This I said and nothing more.

“We’ll see,” he said and walked away,

And fifty classes from that day

He counted till the month of May

The classes in which I napped and snored.

The number he found was twenty-four.

At a significance level of 0.05,

Please tell me am I still alive?

Or did my grade just take a dive

Plunging down beneath the floor?

Upon thee I hereby implore.

Toastmasters International cites a report by Gallop Poll that 40% of Americans fear public speaking. A student believes that less than 40% of students at her school fear public speaking. She randomly surveys 361 schoolmates and finds that 135 report they fear public speaking. Conduct a hypothesis test to determine if the percent at her school is less than 40%.

$H_{0}: p = 0.40$
$H_{a}: p < 0.40$
Let $P′ =$ the proportion of schoolmates who fear public speaking.
–1.01
$p\text{-value} = 0.1563$
Conclusion: There is insufficient evidence to support the claim that less than 40% of students at the school fear public speaking.
Confidence Interval: $(0.3241, 0.4240)$: The “plus-4s” confidence interval is $(0.3257, 0.4250)$.

Sixty-eight percent of online courses taught at community colleges nationwide were taught by full-time faculty. To test if 68% also represents California’s percent for full-time faculty teaching the online classes, Long Beach City College (LBCC) in California, was randomly selected for comparison. In the same year, 34 of the 44 online courses LBCC offered were taught by full-time faculty. Conduct a hypothesis test to determine if 68% represents California. NOTE: For more accurate results, use more California community colleges and this past year's data.

According to an article in Bloomberg Businessweek , New York City's most recent adult smoking rate is 14%. Suppose that a survey is conducted to determine this year’s rate. Nine out of 70 randomly chosen N.Y. City residents reply that they smoke. Conduct a hypothesis test to determine if the rate is still 14% or if it has decreased.

$H_{0}: p = 0.14$
$H_{a}: p < 0.14$
Let $P′ =$ the proportion of NYC residents that smoke.
–0.2756
$p\text{-value} = 0.3914$
At the 5% significance level, there is insufficient evidence to conclude that the proportion of NYC residents who smoke is less than 0.14.
Confidence Interval: $(0.0502, 0.2070)$: The “plus-4s” confidence interval (see chapter 8) is $(0.0676, 0.2297)$.

The mean age of De Anza College students in a previous term was 26.6 years old. An instructor thinks the mean age for online students is older than 26.6. She randomly surveys 56 online students and finds that the sample mean is 29.4 with a standard deviation of 2.1. Conduct a hypothesis test.

Registered nurses earned an average annual salary of $69,110. For that same year, a survey was conducted of 41 California registered nurses to determine if the annual salary is higher than $69,110 for California nurses. The sample average was $71,121 with a sample standard deviation of $7,489. Conduct a hypothesis test.

$H_{0}: \mu = 69,110$
$H_{0}: \mu > 69,110$
Let $\bar{X} =$ the mean salary in dollars for California registered nurses.
$t = 1.719$
$p\text{-value}: 0.0466$
Conclusion: At the 5% significance level, there is sufficient evidence to conclude that the mean salary of California registered nurses exceeds $69,110.
$($68,757, $73,485)$

La Leche League International reports that the mean age of weaning a child from breastfeeding is age four to five worldwide. In America, most nursing mothers wean their children much earlier. Suppose a random survey is conducted of 21 U.S. mothers who recently weaned their children. The mean weaning age was nine months (3/4 year) with a standard deviation of 4 months. Conduct a hypothesis test to determine if the mean weaning age in the U.S. is less than four years old.

After conducting the test, your decision and conclusion are

Reject $H_{0}$: There is sufficient evidence to conclude that more than 30% of teen girls smoke to stay thin.
Do not reject $H_{0}$: There is not sufficient evidence to conclude that less than 30% of teen girls smoke to stay thin.
Do not reject $H_{0}$: There is not sufficient evidence to conclude that more than 30% of teen girls smoke to stay thin.
Reject $H_{0}$: There is sufficient evidence to conclude that less than 30% of teen girls smoke to stay thin.

At a 1% level of significance, an appropriate conclusion is:

There is insufficient evidence to conclude that the percent of EVC students who attended the midnight showing of Harry Potter is less than 20%.
There is sufficient evidence to conclude that the percent of EVC students who attended the midnight showing of Harry Potter is more than 20%.
There is sufficient evidence to conclude that the percent of EVC students who attended the midnight showing of Harry Potter is less than 20%.
There is insufficient evidence to conclude that the percent of EVC students who attended the midnight showing of Harry Potter is at least 20%.

At a significance level of $a = 0.05$, what is the correct conclusion?

There is enough evidence to conclude that the mean number of hours is more than 4.75
There is enough evidence to conclude that the mean number of hours is more than 4.5
There is not enough evidence to conclude that the mean number of hours is more than 4.5
There is not enough evidence to conclude that the mean number of hours is more than 4.75

Instructions: For the following ten exercises,

Hypothesis testing: For the following ten exercises, answer each question.

State the null and alternate hypothesis.

State the $p\text{-value}$.

State $\alpha$.

What is your decision?

Write a conclusion.

Answer any other questions asked in the problem.

According to the Center for Disease Control website, in 2011 at least 18% of high school students have smoked a cigarette. An Introduction to Statistics class in Davies County, KY conducted a hypothesis test at the local high school (a medium sized–approximately 1,200 students–small city demographic) to determine if the local high school’s percentage was lower. One hundred fifty students were chosen at random and surveyed. Of the 150 students surveyed, 82 have smoked. Use a significance level of 0.05 and using appropriate statistical evidence, conduct a hypothesis test and state the conclusions.

A recent survey in the N.Y. Times Almanac indicated that 48.8% of families own stock. A broker wanted to determine if this survey could be valid. He surveyed a random sample of 250 families and found that 142 owned some type of stock. At the 0.05 significance level, can the survey be considered to be accurate?

$H_{0}: p = 0.488$ $H_{a}: p \neq 0.488$
$p\text{-value} = 0.0114$
$\alpha = 0.05$
Reject the null hypothesis.
At the 5% level of significance, there is enough evidence to conclude that 48.8% of families own stocks.
The survey does not appear to be accurate.

Driver error can be listed as the cause of approximately 54% of all fatal auto accidents, according to the American Automobile Association. Thirty randomly selected fatal accidents are examined, and it is determined that 14 were caused by driver error. Using $\alpha = 0.05$, is the AAA proportion accurate?

The US Department of Energy reported that 51.7% of homes were heated by natural gas. A random sample of 221 homes in Kentucky found that 115 were heated by natural gas. Does the evidence support the claim for Kentucky at the $\alpha = 0.05$ level in Kentucky? Are the results applicable across the country? Why?

$H_{0}: p = 0.517$ $H_{0}: p \neq 0.517$
$p\text{-value} = 0.9203$.
$\alpha = 0.05$.
Do not reject the null hypothesis.
At the 5% significance level, there is not enough evidence to conclude that the proportion of homes in Kentucky that are heated by natural gas is 0.517.
However, we cannot generalize this result to the entire nation. First, the sample’s population is only the state of Kentucky. Second, it is reasonable to assume that homes in the extreme north and south will have extreme high usage and low usage, respectively. We would need to expand our sample base to include these possibilities if we wanted to generalize this claim to the entire nation.

For Americans using library services, the American Library Association claims that at most 67% of patrons borrow books. The library director in Owensboro, Kentucky feels this is not true, so she asked a local college statistic class to conduct a survey. The class randomly selected 100 patrons and found that 82 borrowed books. Did the class demonstrate that the percentage was higher in Owensboro, KY? Use $\alpha = 0.01$ level of significance. What is the possible proportion of patrons that do borrow books from the Owensboro Library?

The Weather Underground reported that the mean amount of summer rainfall for the northeastern US is at least 11.52 inches. Ten cities in the northeast are randomly selected and the mean rainfall amount is calculated to be 7.42 inches with a standard deviation of 1.3 inches. At the $\alpha = 0.05 level$, can it be concluded that the mean rainfall was below the reported average? What if $\alpha = 0.01$? Assume the amount of summer rainfall follows a normal distribution.

$H_{0}: \mu \geq 11.52$ $H_{a}: \mu < 11.52$
$p\text{-value} = 0.000002$ which is almost 0.
At the 5% significance level, there is enough evidence to conclude that the mean amount of summer rain in the northeaster US is less than 11.52 inches, on average.
We would make the same conclusion if alpha was 1% because the $p\text{-value}$ is almost 0.

A survey in the N.Y. Times Almanac finds the mean commute time (one way) is 25.4 minutes for the 15 largest US cities. The Austin, TX chamber of commerce feels that Austin’s commute time is less and wants to publicize this fact. The mean for 25 randomly selected commuters is 22.1 minutes with a standard deviation of 5.3 minutes. At the $\alpha = 0.10$ level, is the Austin, TX commute significantly less than the mean commute time for the 15 largest US cities?

A report by the Gallup Poll found that a woman visits her doctor, on average, at most 5.8 times each year. A random sample of 20 women results in these yearly visit totals

3; 2; 1; 3; 7; 2; 9; 4; 6; 6; 8; 0; 5; 6; 4; 2; 1; 3; 4; 1

At the $\alpha = 0.05$ level can it be concluded that the sample mean is higher than 5.8 visits per year?

$H_{0}: \mu \leq 5.8$ $H_{a}: \mu > 5.8$
$p\text{-value} = 0.9987$
At the 5% level of significance, there is not enough evidence to conclude that a woman visits her doctor, on average, more than 5.8 times a year.

According to the N.Y. Times Almanac the mean family size in the U.S. is 3.18. A sample of a college math class resulted in the following family sizes:

5; 4; 5; 4; 4; 3; 6; 4; 3; 3; 5; 5; 6; 3; 3; 2; 7; 4; 5; 2; 2; 2; 3; 2

At $\alpha = 0.05$ level, is the class’ mean family size greater than the national average? Does the Almanac result remain valid? Why?

The student academic group on a college campus claims that freshman students study at least 2.5 hours per day, on average. One Introduction to Statistics class was skeptical. The class took a random sample of 30 freshman students and found a mean study time of 137 minutes with a standard deviation of 45 minutes. At α = 0.01 level, is the student academic group’s claim correct?

$H_{0}: \mu \geq 150$ $H_{0}: \mu < 150$
$p\text{-value} = 0.0622$
$\alpha = 0.01$
At the 1% significance level, there is not enough evidence to conclude that freshmen students study less than 2.5 hours per day, on average.
The student academic group’s claim appears to be correct.

9.7: Hypothesis Testing of a Single Mean and Single Proportion

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COMMENTS

How to Interpret a P-Value Less Than 0.05 (With Examples)
The null hypothesis (H0): μ = 200. The alternative hypothesis: (HA): μ ≠ 200. Upon conducting a hypothesis test for a mean, the auditor gets a p-value of 0.0154. Since the p-value of 0.0154 is less than the significance level of 0.05, the auditor rejects the null hypothesis and concludes that there is sufficient evidence to say that the ...
Understanding P-Values and Statistical Significance
In statistical hypothesis testing, you reject the null hypothesis when the p-value is less than or equal to the significance level (α) you set before conducting your test. The significance level is the probability of rejecting the null hypothesis when it is true. Commonly used significance levels are 0.01, 0.05, and 0.10.
Hypothesis Testing
Table of contents. Step 1: State your null and alternate hypothesis. Step 2: Collect data. Step 3: Perform a statistical test. Step 4: Decide whether to reject or fail to reject your null hypothesis. Step 5: Present your findings. Other interesting articles. Frequently asked questions about hypothesis testing.
Understanding Hypothesis Tests: Significance Levels (Alpha) and P
The P value of 0.03112 is statistically significant at an alpha level of 0.05, but not at the 0.01 level. If we stick to a significance level of 0.05, we can conclude that the average energy cost for the population is greater than 260. A common mistake is to interpret the P-value as the probability that the null hypothesis is true.
How Hypothesis Tests Work: Significance Levels (Alpha) and P values
Using P values and Significance Levels Together. If your P value is less than or equal to your alpha level, reject the null hypothesis. The P value results are consistent with our graphical representation. The P value of 0.03112 is significant at the alpha level of 0.05 but not 0.01.
Understanding P-values
The p value is a number, calculated from a statistical test, that describes how likely you are to have found a particular set of observations if the null hypothesis were true. P values are used in hypothesis testing to help decide whether to reject the null hypothesis. The smaller the p value, the more likely you are to reject the null hypothesis.
Interpreting P values
Here is the technical definition of P values: P values are the probability of observing a sample statistic that is at least as extreme as your sample statistic when you assume that the null hypothesis is true. Let's go back to our hypothetical medication study. Suppose the hypothesis test generates a P value of 0.03.
4.4: Hypothesis Testing
Testing Hypotheses using Confidence Intervals. We can start the evaluation of the hypothesis setup by comparing 2006 and 2012 run times using a point estimate from the 2012 sample: x¯12 = 95.61 x ¯ 12 = 95.61 minutes. This estimate suggests the average time is actually longer than the 2006 time, 93.29 minutes.
Hypothesis Testing
The Four Steps in Hypothesis Testing. STEP 1: State the appropriate null and alternative hypotheses, Ho and Ha. STEP 2: Obtain a random sample, collect relevant data, and check whether the data meet the conditions under which the test can be used. If the conditions are met, summarize the data using a test statistic.
Hypothesis Testing
Hypothesis testing is an indispensable tool in data science, allowing us to make data-driven decisions with confidence. By understanding its principles, conducting tests properly, and considering real-world applications, you can harness the power of hypothesis testing to unlock valuable insights from your data.
Hypothesis Testing
A hypothesis test is a statistical inference method used to test the significance of a proposed (hypothesized) relation between population statistics (parameters) and their corresponding sample estimators. In other words, hypothesis tests are used to determine if there is enough evidence in a sample to prove a hypothesis true for the entire population. The test considers two hypotheses: the ...
Hypothesis Testing and Confidence Intervals
The relationship between the confidence level and the significance level for a hypothesis test is as follows: Confidence level = 1 - Significance level (alpha) For example, if your significance level is 0.05, the equivalent confidence level is 95%. Both of the following conditions represent statistically significant results: The P-value in a ...
Why 0.05? Two Examples That Put Students in the Role of Decision Maker
Here, the value 0.05 refers to the significance level in a hypothesis test. A nice overview of hypothesis tests is described in the Fall/Winter 2015 issue of STN. In this article, I provide a brief review of the concepts of hypothesis test and significance level. Then, I describe two activities for teachers to proactively address the question ...
Why do we so often use 0.05 in hypothesis testing?
Why do we so often use 0.05 for hypothesis testing? In this online exercise, we hope to help you gain an improved understanding of what a significance level is, and why a value in the neighborhood of 0.05 is reasonable as a default. Watch this video before starting.
S.3.2 Hypothesis Testing (P-Value Approach)
The P -value is, therefore, the area under a tn - 1 = t14 curve to the left of -2.5 and to the right of 2.5. It can be shown using statistical software that the P -value is 0.0127 + 0.0127, or 0.0254. The graph depicts this visually. Note that the P -value for a two-tailed test is always two times the P -value for either of the one-tailed tests.
S.3.3 Hypothesis Testing Examples
If the biologist set her significance level $\alpha$ at 0.05 and used the critical value approach to conduct her hypothesis test, she would reject the null hypothesis if her test statistic t* were less than -1.6939 (determined using statistical software or a t-table):s-3-3. Since the biologist's test statistic, t* = -4.60, is less than -1.6939, the biologist rejects the null hypothesis.
Hypothesis Testing
Example: Criminal Trial Analogy. First, state 2 hypotheses, the null hypothesis ("H 0 ") and the alternative hypothesis ("H A "). H 0: Defendant is not guilty.; H A: Defendant is guilty.; Usually the H 0 is a statement of "no effect", or "no change", or "chance only" about a population parameter.. While the H A, depending on the situation, is that there is a difference ...
An Easy Introduction to Statistical Significance (With Examples)
The p value determines statistical significance. An extremely low p value indicates high statistical significance, while a high p value means low or no statistical significance. Example: Hypothesis testing. To test your hypothesis, you first collect data from two groups. The experimental group actively smiles, while the control group does not.
Hypothesis Testing
Step 2: State the Alternate Hypothesis. The claim is that the students have above average IQ scores, so: H 1: μ > 100. The fact that we are looking for scores "greater than" a certain point means that this is a one-tailed test. Step 3: Draw a picture to help you visualize the problem. Step 4: State the alpha level.
Hypothesis Testing
Hypothesis testing is a technique that is used to verify whether the results of an experiment are statistically significant. It involves the setting up of a null hypothesis and an alternate hypothesis. There are three types of tests that can be conducted under hypothesis testing - z test, t test, and chi square test.
S.3.1 Hypothesis Testing (Critical Value Approach)
The critical value for conducting the left-tailed test H0 : μ = 3 versus HA : μ < 3 is the t -value, denoted -t( α, n - 1), such that the probability to the left of it is α. It can be shown using either statistical software or a t -table that the critical value -t0.05,14 is -1.7613. That is, we would reject the null hypothesis H0 : μ = 3 ...
9.E: Hypothesis Testing with One Sample (Exercises)
An Introduction to Statistics class in Davies County, KY conducted a hypothesis test at the local high school (a medium sized-approximately 1,200 students-small city demographic) to determine if the local high school's percentage was lower. One hundred fifty students were chosen at random and surveyed.