hypothesis testing population means

Hypothesis Testing for Means & Proportions

Lisa Sullivan, PhD

Professor of Biostatistics

Boston University School of Public Health

Introduction

This is the first of three modules that will addresses the second area of statistical inference, which is hypothesis testing, in which a specific statement or hypothesis is generated about a population parameter, and sample statistics are used to assess the likelihood that the hypothesis is true. The hypothesis is based on available information and the investigator's belief about the population parameters. The process of hypothesis testing involves setting up two competing hypotheses, the null hypothesis and the alternate hypothesis. One selects a random sample (or multiple samples when there are more comparison groups), computes summary statistics and then assesses the likelihood that the sample data support the research or alternative hypothesis. Similar to estimation, the process of hypothesis testing is based on probability theory and the Central Limit Theorem.  

This module will focus on hypothesis testing for means and proportions. The next two modules in this series will address analysis of variance and chi-squared tests. 

Learning Objectives

After completing this module, the student will be able to:

• Define null and research hypothesis, test statistic, level of significance and decision rule
• Distinguish between Type I and Type II errors and discuss the implications of each
• Explain the difference between one and two sided tests of hypothesis
• Estimate and interpret p-values
• Explain the relationship between confidence interval estimates and p-values in drawing inferences
• Differentiate hypothesis testing procedures based on type of outcome variable and number of sample

Introduction to Hypothesis Testing

Techniques for hypothesis testing  .

The techniques for hypothesis testing depend on

• the type of outcome variable being analyzed (continuous, dichotomous, discrete)
• the number of comparison groups in the investigation
• whether the comparison groups are independent (i.e., physically separate such as men versus women) or dependent (i.e., matched or paired such as pre- and post-assessments on the same participants).

In estimation we focused explicitly on techniques for one and two samples and discussed estimation for a specific parameter (e.g., the mean or proportion of a population), for differences (e.g., difference in means, the risk difference) and ratios (e.g., the relative risk and odds ratio). Here we will focus on procedures for one and two samples when the outcome is either continuous (and we focus on means) or dichotomous (and we focus on proportions).

General Approach: A Simple Example

The Centers for Disease Control (CDC) reported on trends in weight, height and body mass index from the 1960's through 2002. 1 The general trend was that Americans were much heavier and slightly taller in 2002 as compared to 1960; both men and women gained approximately 24 pounds, on average, between 1960 and 2002.   In 2002, the mean weight for men was reported at 191 pounds. Suppose that an investigator hypothesizes that weights are even higher in 2006 (i.e., that the trend continued over the subsequent 4 years). The research hypothesis is that the mean weight in men in 2006 is more than 191 pounds. The null hypothesis is that there is no change in weight, and therefore the mean weight is still 191 pounds in 2006.  

In order to test the hypotheses, we select a random sample of American males in 2006 and measure their weights. Suppose we have resources available to recruit n=100 men into our sample. We weigh each participant and compute summary statistics on the sample data. Suppose in the sample we determine the following:

Do the sample data support the null or research hypothesis? The sample mean of 197.1 is numerically higher than 191. However, is this difference more than would be expected by chance? In hypothesis testing, we assume that the null hypothesis holds until proven otherwise. We therefore need to determine the likelihood of observing a sample mean of 197.1 or higher when the true population mean is 191 (i.e., if the null hypothesis is true or under the null hypothesis). We can compute this probability using the Central Limit Theorem. Specifically,

(Notice that we use the sample standard deviation in computing the Z score. This is generally an appropriate substitution as long as the sample size is large, n > 30. Thus, there is less than a 1% probability of observing a sample mean as large as 197.1 when the true population mean is 191. Do you think that the null hypothesis is likely true? Based on how unlikely it is to observe a sample mean of 197.1 under the null hypothesis (i.e., <1% probability), we might infer, from our data, that the null hypothesis is probably not true.

Suppose that the sample data had turned out differently. Suppose that we instead observed the following in 2006:

How likely it is to observe a sample mean of 192.1 or higher when the true population mean is 191 (i.e., if the null hypothesis is true)? We can again compute this probability using the Central Limit Theorem. Specifically,

There is a 33.4% probability of observing a sample mean as large as 192.1 when the true population mean is 191. Do you think that the null hypothesis is likely true?  

Neither of the sample means that we obtained allows us to know with certainty whether the null hypothesis is true or not. However, our computations suggest that, if the null hypothesis were true, the probability of observing a sample mean >197.1 is less than 1%. In contrast, if the null hypothesis were true, the probability of observing a sample mean >192.1 is about 33%. We can't know whether the null hypothesis is true, but the sample that provided a mean value of 197.1 provides much stronger evidence in favor of rejecting the null hypothesis, than the sample that provided a mean value of 192.1. Note that this does not mean that a sample mean of 192.1 indicates that the null hypothesis is true; it just doesn't provide compelling evidence to reject it.

In essence, hypothesis testing is a procedure to compute a probability that reflects the strength of the evidence (based on a given sample) for rejecting the null hypothesis. In hypothesis testing, we determine a threshold or cut-off point (called the critical value) to decide when to believe the null hypothesis and when to believe the research hypothesis. It is important to note that it is possible to observe any sample mean when the true population mean is true (in this example equal to 191), but some sample means are very unlikely. Based on the two samples above it would seem reasonable to believe the research hypothesis when x̄ = 197.1, but to believe the null hypothesis when x̄ =192.1. What we need is a threshold value such that if x̄ is above that threshold then we believe that H 1 is true and if x̄ is below that threshold then we believe that H 0 is true. The difficulty in determining a threshold for x̄ is that it depends on the scale of measurement. In this example, the threshold, sometimes called the critical value, might be 195 (i.e., if the sample mean is 195 or more then we believe that H 1 is true and if the sample mean is less than 195 then we believe that H 0 is true). Suppose we are interested in assessing an increase in blood pressure over time, the critical value will be different because blood pressures are measured in millimeters of mercury (mmHg) as opposed to in pounds. In the following we will explain how the critical value is determined and how we handle the issue of scale.

First, to address the issue of scale in determining the critical value, we convert our sample data (in particular the sample mean) into a Z score. We know from the module on probability that the center of the Z distribution is zero and extreme values are those that exceed 2 or fall below -2. Z scores above 2 and below -2 represent approximately 5% of all Z values. If the observed sample mean is close to the mean specified in H 0 (here m =191), then Z will be close to zero. If the observed sample mean is much larger than the mean specified in H 0 , then Z will be large.  

In hypothesis testing, we select a critical value from the Z distribution. This is done by first determining what is called the level of significance, denoted α ("alpha"). What we are doing here is drawing a line at extreme values. The level of significance is the probability that we reject the null hypothesis (in favor of the alternative) when it is actually true and is also called the Type I error rate.

α = Level of significance = P(Type I error) = P(Reject H 0 | H 0 is true).

Because α is a probability, it ranges between 0 and 1. The most commonly used value in the medical literature for α is 0.05, or 5%. Thus, if an investigator selects α=0.05, then they are allowing a 5% probability of incorrectly rejecting the null hypothesis in favor of the alternative when the null is in fact true. Depending on the circumstances, one might choose to use a level of significance of 1% or 10%. For example, if an investigator wanted to reject the null only if there were even stronger evidence than that ensured with α=0.05, they could choose a =0.01as their level of significance. The typical values for α are 0.01, 0.05 and 0.10, with α=0.05 the most commonly used value.  

Suppose in our weight study we select α=0.05. We need to determine the value of Z that holds 5% of the values above it (see below).

The critical value of Z for α =0.05 is Z = 1.645 (i.e., 5% of the distribution is above Z=1.645). With this value we can set up what is called our decision rule for the test. The rule is to reject H 0 if the Z score is 1.645 or more.  

With the first sample we have

Because 2.38 > 1.645, we reject the null hypothesis. (The same conclusion can be drawn by comparing the 0.0087 probability of observing a sample mean as extreme as 197.1 to the level of significance of 0.05. If the observed probability is smaller than the level of significance we reject H 0 ). Because the Z score exceeds the critical value, we conclude that the mean weight for men in 2006 is more than 191 pounds, the value reported in 2002. If we observed the second sample (i.e., sample mean =192.1), we would not be able to reject the null hypothesis because the Z score is 0.43 which is not in the rejection region (i.e., the region in the tail end of the curve above 1.645). With the second sample we do not have sufficient evidence (because we set our level of significance at 5%) to conclude that weights have increased. Again, the same conclusion can be reached by comparing probabilities. The probability of observing a sample mean as extreme as 192.1 is 33.4% which is not below our 5% level of significance.

Hypothesis Testing: Upper-, Lower, and Two Tailed Tests

The procedure for hypothesis testing is based on the ideas described above. Specifically, we set up competing hypotheses, select a random sample from the population of interest and compute summary statistics. We then determine whether the sample data supports the null or alternative hypotheses. The procedure can be broken down into the following five steps.  

• Step 1. Set up hypotheses and select the level of significance α.

H 0 : Null hypothesis (no change, no difference);  

H 1 : Research hypothesis (investigator's belief); α =0.05

• Step 2. Select the appropriate test statistic.  

The test statistic is a single number that summarizes the sample information.   An example of a test statistic is the Z statistic computed as follows:

When the sample size is small, we will use t statistics (just as we did when constructing confidence intervals for small samples). As we present each scenario, alternative test statistics are provided along with conditions for their appropriate use.

• Step 3.  Set up decision rule.  

The decision rule is a statement that tells under what circumstances to reject the null hypothesis. The decision rule is based on specific values of the test statistic (e.g., reject H 0 if Z > 1.645). The decision rule for a specific test depends on 3 factors: the research or alternative hypothesis, the test statistic and the level of significance. Each is discussed below.

• The decision rule depends on whether an upper-tailed, lower-tailed, or two-tailed test is proposed. In an upper-tailed test the decision rule has investigators reject H 0 if the test statistic is larger than the critical value. In a lower-tailed test the decision rule has investigators reject H 0 if the test statistic is smaller than the critical value.  In a two-tailed test the decision rule has investigators reject H 0 if the test statistic is extreme, either larger than an upper critical value or smaller than a lower critical value.
• The exact form of the test statistic is also important in determining the decision rule. If the test statistic follows the standard normal distribution (Z), then the decision rule will be based on the standard normal distribution. If the test statistic follows the t distribution, then the decision rule will be based on the t distribution. The appropriate critical value will be selected from the t distribution again depending on the specific alternative hypothesis and the level of significance.  
• The third factor is the level of significance. The level of significance which is selected in Step 1 (e.g., α =0.05) dictates the critical value.   For example, in an upper tailed Z test, if α =0.05 then the critical value is Z=1.645.  

The following figures illustrate the rejection regions defined by the decision rule for upper-, lower- and two-tailed Z tests with α=0.05. Notice that the rejection regions are in the upper, lower and both tails of the curves, respectively. The decision rules are written below each figure.

Rejection Region for Lower-Tailed Z Test (H 1 : μ < μ 0 ) with α =0.05

The decision rule is: Reject H 0 if Z < 1.645.

Rejection Region for Two-Tailed Z Test (H 1 : μ ≠ μ 0 ) with α =0.05

The decision rule is: Reject H 0 if Z < -1.960 or if Z > 1.960.

The complete table of critical values of Z for upper, lower and two-tailed tests can be found in the table of Z values to the right in "Other Resources."

Critical values of t for upper, lower and two-tailed tests can be found in the table of t values in "Other Resources."

• Step 4. Compute the test statistic.  

Here we compute the test statistic by substituting the observed sample data into the test statistic identified in Step 2.

• Step 5. Conclusion.  

The final conclusion is made by comparing the test statistic (which is a summary of the information observed in the sample) to the decision rule. The final conclusion will be either to reject the null hypothesis (because the sample data are very unlikely if the null hypothesis is true) or not to reject the null hypothesis (because the sample data are not very unlikely).  

If the null hypothesis is rejected, then an exact significance level is computed to describe the likelihood of observing the sample data assuming that the null hypothesis is true. The exact level of significance is called the p-value and it will be less than the chosen level of significance if we reject H 0 .

Statistical computing packages provide exact p-values as part of their standard output for hypothesis tests. In fact, when using a statistical computing package, the steps outlined about can be abbreviated. The hypotheses (step 1) should always be set up in advance of any analysis and the significance criterion should also be determined (e.g., α =0.05). Statistical computing packages will produce the test statistic (usually reporting the test statistic as t) and a p-value. The investigator can then determine statistical significance using the following: If p < α then reject H 0 .  

• Step 1. Set up hypotheses and determine level of significance

H 0 : μ = 191 H 1 : μ > 191                 α =0.05

The research hypothesis is that weights have increased, and therefore an upper tailed test is used.

• Step 2. Select the appropriate test statistic.

Because the sample size is large (n > 30) the appropriate test statistic is

• Step 3. Set up decision rule.  

In this example, we are performing an upper tailed test (H 1 : μ> 191), with a Z test statistic and selected α =0.05.   Reject H 0 if Z > 1.645.

We now substitute the sample data into the formula for the test statistic identified in Step 2.  

We reject H 0 because 2.38 > 1.645. We have statistically significant evidence at a =0.05, to show that the mean weight in men in 2006 is more than 191 pounds. Because we rejected the null hypothesis, we now approximate the p-value which is the likelihood of observing the sample data if the null hypothesis is true. An alternative definition of the p-value is the smallest level of significance where we can still reject H 0 . In this example, we observed Z=2.38 and for α=0.05, the critical value was 1.645. Because 2.38 exceeded 1.645 we rejected H 0 . In our conclusion we reported a statistically significant increase in mean weight at a 5% level of significance. Using the table of critical values for upper tailed tests, we can approximate the p-value. If we select α=0.025, the critical value is 1.96, and we still reject H 0 because 2.38 > 1.960. If we select α=0.010 the critical value is 2.326, and we still reject H 0 because 2.38 > 2.326. However, if we select α=0.005, the critical value is 2.576, and we cannot reject H 0 because 2.38 < 2.576. Therefore, the smallest α where we still reject H 0 is 0.010. This is the p-value. A statistical computing package would produce a more precise p-value which would be in between 0.005 and 0.010. Here we are approximating the p-value and would report p < 0.010.                  

Type I and Type II Errors

In all tests of hypothesis, there are two types of errors that can be committed. The first is called a Type I error and refers to the situation where we incorrectly reject H 0 when in fact it is true. This is also called a false positive result (as we incorrectly conclude that the research hypothesis is true when in fact it is not). When we run a test of hypothesis and decide to reject H 0 (e.g., because the test statistic exceeds the critical value in an upper tailed test) then either we make a correct decision because the research hypothesis is true or we commit a Type I error. The different conclusions are summarized in the table below. Note that we will never know whether the null hypothesis is really true or false (i.e., we will never know which row of the following table reflects reality).

Table - Conclusions in Test of Hypothesis

In the first step of the hypothesis test, we select a level of significance, α, and α= P(Type I error). Because we purposely select a small value for α, we control the probability of committing a Type I error. For example, if we select α=0.05, and our test tells us to reject H 0 , then there is a 5% probability that we commit a Type I error. Most investigators are very comfortable with this and are confident when rejecting H 0 that the research hypothesis is true (as it is the more likely scenario when we reject H 0 ).

When we run a test of hypothesis and decide not to reject H 0 (e.g., because the test statistic is below the critical value in an upper tailed test) then either we make a correct decision because the null hypothesis is true or we commit a Type II error. Beta (β) represents the probability of a Type II error and is defined as follows: β=P(Type II error) = P(Do not Reject H 0 | H 0 is false). Unfortunately, we cannot choose β to be small (e.g., 0.05) to control the probability of committing a Type II error because β depends on several factors including the sample size, α, and the research hypothesis. When we do not reject H 0 , it may be very likely that we are committing a Type II error (i.e., failing to reject H 0 when in fact it is false). Therefore, when tests are run and the null hypothesis is not rejected we often make a weak concluding statement allowing for the possibility that we might be committing a Type II error. If we do not reject H 0 , we conclude that we do not have significant evidence to show that H 1 is true. We do not conclude that H 0 is true.

 The most common reason for a Type II error is a small sample size.

Tests with One Sample, Continuous Outcome

Hypothesis testing applications with a continuous outcome variable in a single population are performed according to the five-step procedure outlined above. A key component is setting up the null and research hypotheses. The objective is to compare the mean in a single population to known mean (μ 0 ). The known value is generally derived from another study or report, for example a study in a similar, but not identical, population or a study performed some years ago. The latter is called a historical control. It is important in setting up the hypotheses in a one sample test that the mean specified in the null hypothesis is a fair and reasonable comparator. This will be discussed in the examples that follow.

Test Statistics for Testing H 0 : μ= μ 0

• if n > 30
• if n < 30

Note that statistical computing packages will use the t statistic exclusively and make the necessary adjustments for comparing the test statistic to appropriate values from probability tables to produce a p-value. 

The National Center for Health Statistics (NCHS) published a report in 2005 entitled Health, United States, containing extensive information on major trends in the health of Americans. Data are provided for the US population as a whole and for specific ages, sexes and races.  The NCHS report indicated that in 2002 Americans paid an average of $3,302 per year on health care and prescription drugs. An investigator hypothesizes that in 2005 expenditures have decreased primarily due to the availability of generic drugs. To test the hypothesis, a sample of 100 Americans are selected and their expenditures on health care and prescription drugs in 2005 are measured.   The sample data are summarized as follows: n=100, x̄

=$3,190 and s=$890. Is there statistical evidence of a reduction in expenditures on health care and prescription drugs in 2005? Is the sample mean of $3,190 evidence of a true reduction in the mean or is it within chance fluctuation? We will run the test using the five-step approach. 

• Step 1.  Set up hypotheses and determine level of significance

H 0 : μ = 3,302 H 1 : μ < 3,302           α =0.05

The research hypothesis is that expenditures have decreased, and therefore a lower-tailed test is used.

This is a lower tailed test, using a Z statistic and a 5% level of significance.   Reject H 0 if Z < -1.645.

•   Step 4. Compute the test statistic.  

We do not reject H 0 because -1.26 > -1.645. We do not have statistically significant evidence at α=0.05 to show that the mean expenditures on health care and prescription drugs are lower in 2005 than the mean of $3,302 reported in 2002.  

Recall that when we fail to reject H 0 in a test of hypothesis that either the null hypothesis is true (here the mean expenditures in 2005 are the same as those in 2002 and equal to $3,302) or we committed a Type II error (i.e., we failed to reject H 0 when in fact it is false). In summarizing this test, we conclude that we do not have sufficient evidence to reject H 0 . We do not conclude that H 0 is true, because there may be a moderate to high probability that we committed a Type II error. It is possible that the sample size is not large enough to detect a difference in mean expenditures.      

The NCHS reported that the mean total cholesterol level in 2002 for all adults was 203. Total cholesterol levels in participants who attended the seventh examination of the Offspring in the Framingham Heart Study are summarized as follows: n=3,310, x̄ =200.3, and s=36.8. Is there statistical evidence of a difference in mean cholesterol levels in the Framingham Offspring?

Here we want to assess whether the sample mean of 200.3 in the Framingham sample is statistically significantly different from 203 (i.e., beyond what we would expect by chance). We will run the test using the five-step approach.

H 0 : μ= 203 H 1 : μ≠ 203                       α=0.05

The research hypothesis is that cholesterol levels are different in the Framingham Offspring, and therefore a two-tailed test is used.

•   Step 3. Set up decision rule.  

This is a two-tailed test, using a Z statistic and a 5% level of significance. Reject H 0 if Z < -1.960 or is Z > 1.960.

We reject H 0 because -4.22 ≤ -1. .960. We have statistically significant evidence at α=0.05 to show that the mean total cholesterol level in the Framingham Offspring is different from the national average of 203 reported in 2002.   Because we reject H 0 , we also approximate a p-value. Using the two-sided significance levels, p < 0.0001.  

Statistical Significance versus Clinical (Practical) Significance

This example raises an important concept of statistical versus clinical or practical significance. From a statistical standpoint, the total cholesterol levels in the Framingham sample are highly statistically significantly different from the national average with p < 0.0001 (i.e., there is less than a 0.01% chance that we are incorrectly rejecting the null hypothesis). However, the sample mean in the Framingham Offspring study is 200.3, less than 3 units different from the national mean of 203. The reason that the data are so highly statistically significant is due to the very large sample size. It is always important to assess both statistical and clinical significance of data. This is particularly relevant when the sample size is large. Is a 3 unit difference in total cholesterol a meaningful difference?  

Consider again the NCHS-reported mean total cholesterol level in 2002 for all adults of 203. Suppose a new drug is proposed to lower total cholesterol. A study is designed to evaluate the efficacy of the drug in lowering cholesterol.   Fifteen patients are enrolled in the study and asked to take the new drug for 6 weeks. At the end of 6 weeks, each patient's total cholesterol level is measured and the sample statistics are as follows:   n=15, x̄ =195.9 and s=28.7. Is there statistical evidence of a reduction in mean total cholesterol in patients after using the new drug for 6 weeks? We will run the test using the five-step approach. 

H 0 : μ= 203 H 1 : μ< 203                   α=0.05

•  Step 2. Select the appropriate test statistic.  

Because the sample size is small (n<30) the appropriate test statistic is

This is a lower tailed test, using a t statistic and a 5% level of significance. In order to determine the critical value of t, we need degrees of freedom, df, defined as df=n-1. In this example df=15-1=14. The critical value for a lower tailed test with df=14 and a =0.05 is -2.145 and the decision rule is as follows:   Reject H 0 if t < -2.145.

We do not reject H 0 because -0.96 > -2.145. We do not have statistically significant evidence at α=0.05 to show that the mean total cholesterol level is lower than the national mean in patients taking the new drug for 6 weeks. Again, because we failed to reject the null hypothesis we make a weaker concluding statement allowing for the possibility that we may have committed a Type II error (i.e., failed to reject H 0 when in fact the drug is efficacious).

This example raises an important issue in terms of study design. In this example we assume in the null hypothesis that the mean cholesterol level is 203. This is taken to be the mean cholesterol level in patients without treatment. Is this an appropriate comparator? Alternative and potentially more efficient study designs to evaluate the effect of the new drug could involve two treatment groups, where one group receives the new drug and the other does not, or we could measure each patient's baseline or pre-treatment cholesterol level and then assess changes from baseline to 6 weeks post-treatment. These designs are also discussed here.

Video - Comparing a Sample Mean to Known Population Mean (8:20)

Link to transcript of the video

Tests with One Sample, Dichotomous Outcome

Hypothesis testing applications with a dichotomous outcome variable in a single population are also performed according to the five-step procedure. Similar to tests for means, a key component is setting up the null and research hypotheses. The objective is to compare the proportion of successes in a single population to a known proportion (p 0 ). That known proportion is generally derived from another study or report and is sometimes called a historical control. It is important in setting up the hypotheses in a one sample test that the proportion specified in the null hypothesis is a fair and reasonable comparator.    

In one sample tests for a dichotomous outcome, we set up our hypotheses against an appropriate comparator. We select a sample and compute descriptive statistics on the sample data. Specifically, we compute the sample size (n) and the sample proportion which is computed by taking the ratio of the number of successes to the sample size,

We then determine the appropriate test statistic (Step 2) for the hypothesis test. The formula for the test statistic is given below.

Test Statistic for Testing H 0 : p = p 0

if min(np 0 , n(1-p 0 )) > 5

The formula above is appropriate for large samples, defined when the smaller of np 0 and n(1-p 0 ) is at least 5. This is similar, but not identical, to the condition required for appropriate use of the confidence interval formula for a population proportion, i.e.,

Here we use the proportion specified in the null hypothesis as the true proportion of successes rather than the sample proportion. If we fail to satisfy the condition, then alternative procedures, called exact methods must be used to test the hypothesis about the population proportion.

Example:  

The NCHS report indicated that in 2002 the prevalence of cigarette smoking among American adults was 21.1%.  Data on prevalent smoking in n=3,536 participants who attended the seventh examination of the Offspring in the Framingham Heart Study indicated that 482/3,536 = 13.6% of the respondents were currently smoking at the time of the exam. Suppose we want to assess whether the prevalence of smoking is lower in the Framingham Offspring sample given the focus on cardiovascular health in that community. Is there evidence of a statistically lower prevalence of smoking in the Framingham Offspring study as compared to the prevalence among all Americans?

H 0 : p = 0.211 H 1 : p < 0.211                     α=0.05

We must first check that the sample size is adequate.   Specifically, we need to check min(np 0 , n(1-p 0 )) = min( 3,536(0.211), 3,536(1-0.211))=min(746, 2790)=746. The sample size is more than adequate so the following formula can be used:

This is a lower tailed test, using a Z statistic and a 5% level of significance. Reject H 0 if Z < -1.645.

We reject H 0 because -10.93 < -1.645. We have statistically significant evidence at α=0.05 to show that the prevalence of smoking in the Framingham Offspring is lower than the prevalence nationally (21.1%). Here, p < 0.0001.  

The NCHS report indicated that in 2002, 75% of children aged 2 to 17 saw a dentist in the past year. An investigator wants to assess whether use of dental services is similar in children living in the city of Boston. A sample of 125 children aged 2 to 17 living in Boston are surveyed and 64 reported seeing a dentist over the past 12 months. Is there a significant difference in use of dental services between children living in Boston and the national data?

Calculate this on your own before checking the answer.

Video - Hypothesis Test for One Sample and a Dichotomous Outcome (3:55)

Tests with Two Independent Samples, Continuous Outcome

There are many applications where it is of interest to compare two independent groups with respect to their mean scores on a continuous outcome. Here we compare means between groups, but rather than generating an estimate of the difference, we will test whether the observed difference (increase, decrease or difference) is statistically significant or not. Remember, that hypothesis testing gives an assessment of statistical significance, whereas estimation gives an estimate of effect and both are important.

Here we discuss the comparison of means when the two comparison groups are independent or physically separate. The two groups might be determined by a particular attribute (e.g., sex, diagnosis of cardiovascular disease) or might be set up by the investigator (e.g., participants assigned to receive an experimental treatment or placebo). The first step in the analysis involves computing descriptive statistics on each of the two samples. Specifically, we compute the sample size, mean and standard deviation in each sample and we denote these summary statistics as follows:

for sample 1:

for sample 2:

The designation of sample 1 and sample 2 is arbitrary. In a clinical trial setting the convention is to call the treatment group 1 and the control group 2. However, when comparing men and women, for example, either group can be 1 or 2.  

In the two independent samples application with a continuous outcome, the parameter of interest in the test of hypothesis is the difference in population means, μ 1 -μ 2 . The null hypothesis is always that there is no difference between groups with respect to means, i.e.,

The null hypothesis can also be written as follows: H 0 : μ 1 = μ 2 . In the research hypothesis, an investigator can hypothesize that the first mean is larger than the second (H 1 : μ 1 > μ 2 ), that the first mean is smaller than the second (H 1 : μ 1 < μ 2 ), or that the means are different (H 1 : μ 1 ≠ μ 2 ). The three different alternatives represent upper-, lower-, and two-tailed tests, respectively. The following test statistics are used to test these hypotheses.

Test Statistics for Testing H 0 : μ 1 = μ 2

• if n 1 > 30 and n 2 > 30
• if n 1 < 30 or n 2 < 30

NOTE: The formulas above assume equal variability in the two populations (i.e., the population variances are equal, or s 1 2 = s 2 2 ). This means that the outcome is equally variable in each of the comparison populations. For analysis, we have samples from each of the comparison populations. If the sample variances are similar, then the assumption about variability in the populations is probably reasonable. As a guideline, if the ratio of the sample variances, s 1 2 /s 2 2 is between 0.5 and 2 (i.e., if one variance is no more than double the other), then the formulas above are appropriate. If the ratio of the sample variances is greater than 2 or less than 0.5 then alternative formulas must be used to account for the heterogeneity in variances.    

The test statistics include Sp, which is the pooled estimate of the common standard deviation (again assuming that the variances in the populations are similar) computed as the weighted average of the standard deviations in the samples as follows:

Because we are assuming equal variances between groups, we pool the information on variability (sample variances) to generate an estimate of the variability in the population. Note: Because Sp is a weighted average of the standard deviations in the sample, Sp will always be in between s 1 and s 2 .)

Data measured on n=3,539 participants who attended the seventh examination of the Offspring in the Framingham Heart Study are shown below.  

Suppose we now wish to assess whether there is a statistically significant difference in mean systolic blood pressures between men and women using a 5% level of significance.  

H 0 : μ 1 = μ 2

H 1 : μ 1 ≠ μ 2                       α=0.05

Because both samples are large ( > 30), we can use the Z test statistic as opposed to t. Note that statistical computing packages use t throughout. Before implementing the formula, we first check whether the assumption of equality of population variances is reasonable. The guideline suggests investigating the ratio of the sample variances, s 1 2 /s 2 2 . Suppose we call the men group 1 and the women group 2. Again, this is arbitrary; it only needs to be noted when interpreting the results. The ratio of the sample variances is 17.5 2 /20.1 2 = 0.76, which falls between 0.5 and 2 suggesting that the assumption of equality of population variances is reasonable. The appropriate test statistic is

We now substitute the sample data into the formula for the test statistic identified in Step 2. Before substituting, we will first compute Sp, the pooled estimate of the common standard deviation.

Notice that the pooled estimate of the common standard deviation, Sp, falls in between the standard deviations in the comparison groups (i.e., 17.5 and 20.1). Sp is slightly closer in value to the standard deviation in the women (20.1) as there were slightly more women in the sample.   Recall, Sp is a weight average of the standard deviations in the comparison groups, weighted by the respective sample sizes.  

Now the test statistic:

We reject H 0 because 2.66 > 1.960. We have statistically significant evidence at α=0.05 to show that there is a difference in mean systolic blood pressures between men and women. The p-value is p < 0.010.  

Here again we find that there is a statistically significant difference in mean systolic blood pressures between men and women at p < 0.010. Notice that there is a very small difference in the sample means (128.2-126.5 = 1.7 units), but this difference is beyond what would be expected by chance. Is this a clinically meaningful difference? The large sample size in this example is driving the statistical significance. A 95% confidence interval for the difference in mean systolic blood pressures is: 1.7 + 1.26 or (0.44, 2.96). The confidence interval provides an assessment of the magnitude of the difference between means whereas the test of hypothesis and p-value provide an assessment of the statistical significance of the difference.  

Above we performed a study to evaluate a new drug designed to lower total cholesterol. The study involved one sample of patients, each patient took the new drug for 6 weeks and had their cholesterol measured. As a means of evaluating the efficacy of the new drug, the mean total cholesterol following 6 weeks of treatment was compared to the NCHS-reported mean total cholesterol level in 2002 for all adults of 203. At the end of the example, we discussed the appropriateness of the fixed comparator as well as an alternative study design to evaluate the effect of the new drug involving two treatment groups, where one group receives the new drug and the other does not. Here, we revisit the example with a concurrent or parallel control group, which is very typical in randomized controlled trials or clinical trials (refer to the EP713 module on Clinical Trials).  

A new drug is proposed to lower total cholesterol. A randomized controlled trial is designed to evaluate the efficacy of the medication in lowering cholesterol. Thirty participants are enrolled in the trial and are randomly assigned to receive either the new drug or a placebo. The participants do not know which treatment they are assigned. Each participant is asked to take the assigned treatment for 6 weeks. At the end of 6 weeks, each patient's total cholesterol level is measured and the sample statistics are as follows.

Is there statistical evidence of a reduction in mean total cholesterol in patients taking the new drug for 6 weeks as compared to participants taking placebo? We will run the test using the five-step approach.

H 0 : μ 1 = μ 2 H 1 : μ 1 < μ 2                         α=0.05

Because both samples are small (< 30), we use the t test statistic. Before implementing the formula, we first check whether the assumption of equality of population variances is reasonable. The ratio of the sample variances, s 1 2 /s 2 2 =28.7 2 /30.3 2 = 0.90, which falls between 0.5 and 2, suggesting that the assumption of equality of population variances is reasonable. The appropriate test statistic is:

This is a lower-tailed test, using a t statistic and a 5% level of significance. The appropriate critical value can be found in the t Table (in More Resources to the right). In order to determine the critical value of t we need degrees of freedom, df, defined as df=n 1 +n 2 -2 = 15+15-2=28. The critical value for a lower tailed test with df=28 and α=0.05 is -1.701 and the decision rule is: Reject H 0 if t < -1.701.

Now the test statistic,

We reject H 0 because -2.92 < -1.701. We have statistically significant evidence at α=0.05 to show that the mean total cholesterol level is lower in patients taking the new drug for 6 weeks as compared to patients taking placebo, p < 0.005.

The clinical trial in this example finds a statistically significant reduction in total cholesterol, whereas in the previous example where we had a historical control (as opposed to a parallel control group) we did not demonstrate efficacy of the new drug. Notice that the mean total cholesterol level in patients taking placebo is 217.4 which is very different from the mean cholesterol reported among all Americans in 2002 of 203 and used as the comparator in the prior example. The historical control value may not have been the most appropriate comparator as cholesterol levels have been increasing over time. In the next section, we present another design that can be used to assess the efficacy of the new drug.

Video - Comparison of Two Independent Samples With a Continuous Outcome (8:02)

Tests with Matched Samples, Continuous Outcome

In the previous section we compared two groups with respect to their mean scores on a continuous outcome. An alternative study design is to compare matched or paired samples. The two comparison groups are said to be dependent, and the data can arise from a single sample of participants where each participant is measured twice (possibly before and after an intervention) or from two samples that are matched on specific characteristics (e.g., siblings). When the samples are dependent, we focus on difference scores in each participant or between members of a pair and the test of hypothesis is based on the mean difference, μ d . The null hypothesis again reflects "no difference" and is stated as H 0 : μ d =0 . Note that there are some instances where it is of interest to test whether there is a difference of a particular magnitude (e.g., μ d =5) but in most instances the null hypothesis reflects no difference (i.e., μ d =0).  

The appropriate formula for the test of hypothesis depends on the sample size. The formulas are shown below and are identical to those we presented for estimating the mean of a single sample presented (e.g., when comparing against an external or historical control), except here we focus on difference scores.

Test Statistics for Testing H 0 : μ d =0

A new drug is proposed to lower total cholesterol and a study is designed to evaluate the efficacy of the drug in lowering cholesterol. Fifteen patients agree to participate in the study and each is asked to take the new drug for 6 weeks. However, before starting the treatment, each patient's total cholesterol level is measured. The initial measurement is a pre-treatment or baseline value. After taking the drug for 6 weeks, each patient's total cholesterol level is measured again and the data are shown below. The rightmost column contains difference scores for each patient, computed by subtracting the 6 week cholesterol level from the baseline level. The differences represent the reduction in total cholesterol over 4 weeks. (The differences could have been computed by subtracting the baseline total cholesterol level from the level measured at 6 weeks. The way in which the differences are computed does not affect the outcome of the analysis only the interpretation.)

Because the differences are computed by subtracting the cholesterols measured at 6 weeks from the baseline values, positive differences indicate reductions and negative differences indicate increases (e.g., participant 12 increases by 2 units over 6 weeks). The goal here is to test whether there is a statistically significant reduction in cholesterol. Because of the way in which we computed the differences, we want to look for an increase in the mean difference (i.e., a positive reduction). In order to conduct the test, we need to summarize the differences. In this sample, we have

The calculations are shown below.  

Is there statistical evidence of a reduction in mean total cholesterol in patients after using the new medication for 6 weeks? We will run the test using the five-step approach.

H 0 : μ d = 0 H 1 : μ d > 0                 α=0.05

NOTE: If we had computed differences by subtracting the baseline level from the level measured at 6 weeks then negative differences would have reflected reductions and the research hypothesis would have been H 1 : μ d < 0. 

• Step 2 . Select the appropriate test statistic.

This is an upper-tailed test, using a t statistic and a 5% level of significance. The appropriate critical value can be found in the t Table at the right, with df=15-1=14. The critical value for an upper-tailed test with df=14 and α=0.05 is 2.145 and the decision rule is Reject H 0 if t > 2.145.

We now substitute the sample data into the formula for the test statistic identified in Step 2.

We reject H 0 because 4.61 > 2.145. We have statistically significant evidence at α=0.05 to show that there is a reduction in cholesterol levels over 6 weeks.  

Here we illustrate the use of a matched design to test the efficacy of a new drug to lower total cholesterol. We also considered a parallel design (randomized clinical trial) and a study using a historical comparator. It is extremely important to design studies that are best suited to detect a meaningful difference when one exists. There are often several alternatives and investigators work with biostatisticians to determine the best design for each application. It is worth noting that the matched design used here can be problematic in that observed differences may only reflect a "placebo" effect. All participants took the assigned medication, but is the observed reduction attributable to the medication or a result of these participation in a study.

Video - Hypothesis Testing With a Matched Sample and a Continuous Outcome (3:11)

Tests with Two Independent Samples, Dichotomous Outcome

There are several approaches that can be used to test hypotheses concerning two independent proportions. Here we present one approach - the chi-square test of independence is an alternative, equivalent, and perhaps more popular approach to the same analysis. Hypothesis testing with the chi-square test is addressed in the third module in this series: BS704_HypothesisTesting-ChiSquare.

In tests of hypothesis comparing proportions between two independent groups, one test is performed and results can be interpreted to apply to a risk difference, relative risk or odds ratio. As a reminder, the risk difference is computed by taking the difference in proportions between comparison groups, the risk ratio is computed by taking the ratio of proportions, and the odds ratio is computed by taking the ratio of the odds of success in the comparison groups. Because the null values for the risk difference, the risk ratio and the odds ratio are different, the hypotheses in tests of hypothesis look slightly different depending on which measure is used. When performing tests of hypothesis for the risk difference, relative risk or odds ratio, the convention is to label the exposed or treated group 1 and the unexposed or control group 2.      

For example, suppose a study is designed to assess whether there is a significant difference in proportions in two independent comparison groups. The test of interest is as follows:

H 0 : p 1 = p 2 versus H 1 : p 1 ≠ p 2 .  

The following are the hypothesis for testing for a difference in proportions using the risk difference, the risk ratio and the odds ratio. First, the hypotheses above are equivalent to the following:

• For the risk difference, H 0 : p 1 - p 2 = 0 versus H 1 : p 1 - p 2 ≠ 0 which are, by definition, equal to H 0 : RD = 0 versus H 1 : RD ≠ 0.
• If an investigator wants to focus on the risk ratio, the equivalent hypotheses are H 0 : RR = 1 versus H 1 : RR ≠ 1.
• If the investigator wants to focus on the odds ratio, the equivalent hypotheses are H 0 : OR = 1 versus H 1 : OR ≠ 1.  

Suppose a test is performed to test H 0 : RD = 0 versus H 1 : RD ≠ 0 and the test rejects H 0 at α=0.05. Based on this test we can conclude that there is significant evidence, α=0.05, of a difference in proportions, significant evidence that the risk difference is not zero, significant evidence that the risk ratio and odds ratio are not one. The risk difference is analogous to the difference in means when the outcome is continuous. Here the parameter of interest is the difference in proportions in the population, RD = p 1 -p 2 and the null value for the risk difference is zero. In a test of hypothesis for the risk difference, the null hypothesis is always H 0 : RD = 0. This is equivalent to H 0 : RR = 1 and H 0 : OR = 1. In the research hypothesis, an investigator can hypothesize that the first proportion is larger than the second (H 1 : p 1 > p 2 , which is equivalent to H 1 : RD > 0, H 1 : RR > 1 and H 1 : OR > 1), that the first proportion is smaller than the second (H 1 : p 1 < p 2 , which is equivalent to H 1 : RD < 0, H 1 : RR < 1 and H 1 : OR < 1), or that the proportions are different (H 1 : p 1 ≠ p 2 , which is equivalent to H 1 : RD ≠ 0, H 1 : RR ≠ 1 and H 1 : OR ≠

1). The three different alternatives represent upper-, lower- and two-tailed tests, respectively.  

The formula for the test of hypothesis for the difference in proportions is given below.

Test Statistics for Testing H 0 : p 1 = p

The formula above is appropriate for large samples, defined as at least 5 successes (np > 5) and at least 5 failures (n(1-p > 5)) in each of the two samples. If there are fewer than 5 successes or failures in either comparison group, then alternative procedures, called exact methods must be used to estimate the difference in population proportions.

The following table summarizes data from n=3,799 participants who attended the fifth examination of the Offspring in the Framingham Heart Study. The outcome of interest is prevalent CVD and we want to test whether the prevalence of CVD is significantly higher in smokers as compared to non-smokers.

The prevalence of CVD (or proportion of participants with prevalent CVD) among non-smokers is 298/3,055 = 0.0975 and the prevalence of CVD among current smokers is 81/744 = 0.1089. Here smoking status defines the comparison groups and we will call the current smokers group 1 (exposed) and the non-smokers (unexposed) group 2. The test of hypothesis is conducted below using the five step approach.

H 0 : p 1 = p 2     H 1 : p 1 ≠ p 2                 α=0.05

• Step 2.  Select the appropriate test statistic.  

We must first check that the sample size is adequate. Specifically, we need to ensure that we have at least 5 successes and 5 failures in each comparison group. In this example, we have more than enough successes (cases of prevalent CVD) and failures (persons free of CVD) in each comparison group. The sample size is more than adequate so the following formula can be used:

Reject H 0 if Z < -1.960 or if Z > 1.960.

We now substitute the sample data into the formula for the test statistic identified in Step 2. We first compute the overall proportion of successes:

We now substitute to compute the test statistic.

• Step 5. Conclusion.

We do not reject H 0 because -1.960 < 0.927 < 1.960. We do not have statistically significant evidence at α=0.05 to show that there is a difference in prevalent CVD between smokers and non-smokers.  

A 95% confidence interval for the difference in prevalent CVD (or risk difference) between smokers and non-smokers as 0.0114 + 0.0247, or between -0.0133 and 0.0361. Because the 95% confidence interval for the risk difference includes zero we again conclude that there is no statistically significant difference in prevalent CVD between smokers and non-smokers.    

Smoking has been shown over and over to be a risk factor for cardiovascular disease. What might explain the fact that we did not observe a statistically significant difference using data from the Framingham Heart Study? HINT: Here we consider prevalent CVD, would the results have been different if we considered incident CVD?

A randomized trial is designed to evaluate the effectiveness of a newly developed pain reliever designed to reduce pain in patients following joint replacement surgery. The trial compares the new pain reliever to the pain reliever currently in use (called the standard of care). A total of 100 patients undergoing joint replacement surgery agreed to participate in the trial. Patients were randomly assigned to receive either the new pain reliever or the standard pain reliever following surgery and were blind to the treatment assignment. Before receiving the assigned treatment, patients were asked to rate their pain on a scale of 0-10 with higher scores indicative of more pain. Each patient was then given the assigned treatment and after 30 minutes was again asked to rate their pain on the same scale. The primary outcome was a reduction in pain of 3 or more scale points (defined by clinicians as a clinically meaningful reduction). The following data were observed in the trial.

We now test whether there is a statistically significant difference in the proportions of patients reporting a meaningful reduction (i.e., a reduction of 3 or more scale points) using the five step approach.  

H 0 : p 1 = p 2     H 1 : p 1 ≠ p 2              α=0.05

Here the new or experimental pain reliever is group 1 and the standard pain reliever is group 2.

We must first check that the sample size is adequate. Specifically, we need to ensure that we have at least 5 successes and 5 failures in each comparison group, i.e.,

In this example, we have min(50(0.46), 50(1-0.46), 50(0.22), 50(1-0.22)) = min(23, 27, 11, 39) = 11. The sample size is adequate so the following formula can be used

We reject H 0 because 2.526 > 1960. We have statistically significant evidence at a =0.05 to show that there is a difference in the proportions of patients on the new pain reliever reporting a meaningful reduction (i.e., a reduction of 3 or more scale points) as compared to patients on the standard pain reliever.

A 95% confidence interval for the difference in proportions of patients on the new pain reliever reporting a meaningful reduction (i.e., a reduction of 3 or more scale points) as compared to patients on the standard pain reliever is 0.24 + 0.18 or between 0.06 and 0.42. Because the 95% confidence interval does not include zero we concluded that there was a statistically significant difference in proportions which is consistent with the test of hypothesis result. 

Again, the procedures discussed here apply to applications where there are two independent comparison groups and a dichotomous outcome. There are other applications in which it is of interest to compare a dichotomous outcome in matched or paired samples. For example, in a clinical trial we might wish to test the effectiveness of a new antibiotic eye drop for the treatment of bacterial conjunctivitis. Participants use the new antibiotic eye drop in one eye and a comparator (placebo or active control treatment) in the other. The success of the treatment (yes/no) is recorded for each participant for each eye. Because the two assessments (success or failure) are paired, we cannot use the procedures discussed here. The appropriate test is called McNemar's test (sometimes called McNemar's test for dependent proportions).  

Vide0 - Hypothesis Testing With Two Independent Samples and a Dichotomous Outcome (2:55)

Here we presented hypothesis testing techniques for means and proportions in one and two sample situations. Tests of hypothesis involve several steps, including specifying the null and alternative or research hypothesis, selecting and computing an appropriate test statistic, setting up a decision rule and drawing a conclusion. There are many details to consider in hypothesis testing. The first is to determine the appropriate test. We discussed Z and t tests here for different applications. The appropriate test depends on the distribution of the outcome variable (continuous or dichotomous), the number of comparison groups (one, two) and whether the comparison groups are independent or dependent. The following table summarizes the different tests of hypothesis discussed here.

• Continuous Outcome, One Sample: H0: μ = μ0
• Continuous Outcome, Two Independent Samples: H0: μ1 = μ2
• Continuous Outcome, Two Matched Samples: H0: μd = 0
• Dichotomous Outcome, One Sample: H0: p = p 0
• Dichotomous Outcome, Two Independent Samples: H0: p1 = p2, RD=0, RR=1, OR=1

Once the type of test is determined, the details of the test must be specified. Specifically, the null and alternative hypotheses must be clearly stated. The null hypothesis always reflects the "no change" or "no difference" situation. The alternative or research hypothesis reflects the investigator's belief. The investigator might hypothesize that a parameter (e.g., a mean, proportion, difference in means or proportions) will increase, will decrease or will be different under specific conditions (sometimes the conditions are different experimental conditions and other times the conditions are simply different groups of participants). Once the hypotheses are specified, data are collected and summarized. The appropriate test is then conducted according to the five step approach. If the test leads to rejection of the null hypothesis, an approximate p-value is computed to summarize the significance of the findings. When tests of hypothesis are conducted using statistical computing packages, exact p-values are computed. Because the statistical tables in this textbook are limited, we can only approximate p-values. If the test fails to reject the null hypothesis, then a weaker concluding statement is made for the following reason.

In hypothesis testing, there are two types of errors that can be committed. A Type I error occurs when a test incorrectly rejects the null hypothesis. This is referred to as a false positive result, and the probability that this occurs is equal to the level of significance, α. The investigator chooses the level of significance in Step 1, and purposely chooses a small value such as α=0.05 to control the probability of committing a Type I error. A Type II error occurs when a test fails to reject the null hypothesis when in fact it is false. The probability that this occurs is equal to β. Unfortunately, the investigator cannot specify β at the outset because it depends on several factors including the sample size (smaller samples have higher b), the level of significance (β decreases as a increases), and the difference in the parameter under the null and alternative hypothesis.    

We noted in several examples in this chapter, the relationship between confidence intervals and tests of hypothesis. The approaches are different, yet related. It is possible to draw a conclusion about statistical significance by examining a confidence interval. For example, if a 95% confidence interval does not contain the null value (e.g., zero when analyzing a mean difference or risk difference, one when analyzing relative risks or odds ratios), then one can conclude that a two-sided test of hypothesis would reject the null at α=0.05. It is important to note that the correspondence between a confidence interval and test of hypothesis relates to a two-sided test and that the confidence level corresponds to a specific level of significance (e.g., 95% to α=0.05, 90% to α=0.10 and so on). The exact significance of the test, the p-value, can only be determined using the hypothesis testing approach and the p-value provides an assessment of the strength of the evidence and not an estimate of the effect.

Answers to Selected Problems

Dental services problem - bottom of page 5.

• Step 1: Set up hypotheses and determine the level of significance.

α=0.05

• Step 2: Select the appropriate test statistic.

First, determine whether the sample size is adequate.

Therefore the sample size is adequate, and we can use the following formula:

• Step 3: Set up the decision rule.

Reject H0 if Z is less than or equal to -1.96 or if Z is greater than or equal to 1.96.

• Step 4: Compute the test statistic
• Step 5: Conclusion.

We reject the null hypothesis because -6.15<-1.96. Therefore there is a statistically significant difference in the proportion of children in Boston using dental services compated to the national proportion.

Statistics Made Easy

Introduction to Hypothesis Testing

A statistical hypothesis is an assumption about a population parameter .

For example, we may assume that the mean height of a male in the U.S. is 70 inches.

The assumption about the height is the statistical hypothesis and the true mean height of a male in the U.S. is the population parameter .

A hypothesis test is a formal statistical test we use to reject or fail to reject a statistical hypothesis.

The Two Types of Statistical Hypotheses

To test whether a statistical hypothesis about a population parameter is true, we obtain a random sample from the population and perform a hypothesis test on the sample data.

There are two types of statistical hypotheses:

The null hypothesis , denoted as H 0 , is the hypothesis that the sample data occurs purely from chance.

The alternative hypothesis , denoted as H 1 or H a , is the hypothesis that the sample data is influenced by some non-random cause.

Hypothesis Tests

A hypothesis test consists of five steps:

1. State the hypotheses. 

State the null and alternative hypotheses. These two hypotheses need to be mutually exclusive, so if one is true then the other must be false.

2. Determine a significance level to use for the hypothesis.

Decide on a significance level. Common choices are .01, .05, and .1. 

3. Find the test statistic.

Find the test statistic and the corresponding p-value. Often we are analyzing a population mean or proportion and the general formula to find the test statistic is: (sample statistic – population parameter) / (standard deviation of statistic)

4. Reject or fail to reject the null hypothesis.

Using the test statistic or the p-value, determine if you can reject or fail to reject the null hypothesis based on the significance level.

The p-value  tells us the strength of evidence in support of a null hypothesis. If the p-value is less than the significance level, we reject the null hypothesis.

5. Interpret the results. 

Interpret the results of the hypothesis test in the context of the question being asked. 

The Two Types of Decision Errors

There are two types of decision errors that one can make when doing a hypothesis test:

Type I error: You reject the null hypothesis when it is actually true. The probability of committing a Type I error is equal to the significance level, often called  alpha , and denoted as α.

Type II error: You fail to reject the null hypothesis when it is actually false. The probability of committing a Type II error is called the Power of the test or  Beta , denoted as β.

One-Tailed and Two-Tailed Tests

A statistical hypothesis can be one-tailed or two-tailed.

A one-tailed hypothesis involves making a “greater than” or “less than ” statement.

For example, suppose we assume the mean height of a male in the U.S. is greater than or equal to 70 inches. The null hypothesis would be H0: µ ≥ 70 inches and the alternative hypothesis would be Ha: µ < 70 inches.

A two-tailed hypothesis involves making an “equal to” or “not equal to” statement.

For example, suppose we assume the mean height of a male in the U.S. is equal to 70 inches. The null hypothesis would be H0: µ = 70 inches and the alternative hypothesis would be Ha: µ ≠ 70 inches.

Note: The “equal” sign is always included in the null hypothesis, whether it is =, ≥, or ≤.

Related:   What is a Directional Hypothesis?

Types of Hypothesis Tests

There are many different types of hypothesis tests you can perform depending on the type of data you’re working with and the goal of your analysis.

The following tutorials provide an explanation of the most common types of hypothesis tests:

Introduction to the One Sample t-test Introduction to the Two Sample t-test Introduction to the Paired Samples t-test Introduction to the One Proportion Z-Test Introduction to the Two Proportion Z-Test

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Module 10: Inference for Means

Hypothesis Test for a Population Mean (1 of 5)

Learning outcomes.

• Recognize when to use a hypothesis test or a confidence interval to draw a conclusion about a population mean.
• Under appropriate conditions, conduct a hypothesis test about a population mean. State a conclusion in context.

Introduction

In Inference for Means , our focus is on inference when the variable is quantitative, so the parameters and statistics are means. In “Estimating a Population Mean,” we learned how to use a sample mean to calculate a confidence interval. The confidence interval estimates a population mean. In “Hypothesis Test for a Population Mean,” we learn to use a sample mean to test a hypothesis about a population mean.

We did hypothesis tests in earlier modules. In Inference for One Proportion , each claim involved a single population proportion. In Inference for Two Proportions , the claim was a statement about a treatment effect or a difference in population proportions. In “Hypothesis Test for a Population Mean,” the claims are statements about a population mean. But we will see that the steps and the logic of the hypothesis test are the same. Before we get into the details, let’s practice identifying research questions and studies that involve a population mean.

Cell Phone Data

Cell phones and cell phone plans can be very expensive, so consumers must think carefully when choosing a cell phone and service. This decision is as much about choosing the right cellular company as it is about choosing the right phone. Many people use the data/Internet capabilities of a phone as much as, if not more than, they use voice capability. The data service of a cell company is therefore an important factor in this decision. In the following example, a student named Melanie from Los Angeles applies what she learned in her statistics class to help her make a decision about buying a data plan for her smartphone.

Melanie read an advertisement from the Cell Phone Giants (CPG, for short, and yes, we’re using a fictitious company name) that she thinks is too good to be true. The CPG ad states that customers in Los Angeles get average data download speeds of 4 Mbps. With this speed, the ad claims, it takes, on average, only 12 seconds to download a typical 3-minute song from iTunes.

Only 12 seconds on average to download a 3-minute song from iTunes! Melanie has her doubts about this claim, so she gathers data to test it. She asks a friend who uses the CPG plan to download a song, and it takes 13 seconds to download a 3-minute song using the CPG network. Melanie decides to gather more evidence. She uses her friend’s phone and times the download of the same 3-minute song from various locations in Los Angeles. She gets a mean download time of 13.5 seconds for her sample of downloads.

What can Melanie conclude? Her sample has a mean download time that is greater than 12 seconds. Isn’t this evidence that the CPG claim is wrong? Why is a hypothesis test necessary? Isn’t the conclusion clear?

Let’s review the reason Melanie needs to do a hypothesis test before she can reach a conclusion.

Why should Melanie do a hypothesis test?

Melanie’s data (with a mean of 13.5 seconds) suggest that the average download time overall is greater than the 12 seconds claimed by the manufacturer. But wait. We know that samples will vary. If the CPG claim is correct, we don’t expect all samples to have a mean download time exactly equal to 12 seconds. There will be variability in the sample means. But if the overall average download time is 12 seconds, how much variability in sample means do we expect to see? We need to determine if the difference Melanie observed can be explained by chance.

We have to judge Melanie’s data against random samples that come from a population with a mean of 12. For this reason, we must do a simulation or use a mathematical model to examine the sampling distribution of sample means. Based on the sampling distribution, we ask, Is it likely that the samples will have mean download times that are greater than 13.5 seconds if the overall mean is 12 seconds? This probability (the P-value) determines whether Melanie’s data provides convincing evidence against the CPG claim.

Now let’s do the hypothesis test.

Step 1: Determine the hypotheses.

As always, hypotheses come from the research question. The null hypothesis is a hypothesis that the population mean equals a specific value. The alternative hypothesis reflects our claim. The alternative hypothesis says the population mean is “greater than” or “less than” or “not equal to” the value we assume is true in the null hypothesis.

Melanie’s hypotheses:

• H 0 : It takes 12 seconds on average to download Melanie’s song from iTunes with the CPG network in Los Angeles.
• H a : It takes more than 12 seconds on average to download Melanie’s song from iTunes using the CPG network in Los Angeles.

We can write the hypotheses in terms of µ. When we do so, we should always define µ. Here μ = the average number of seconds it takes to download Melanie’s song on the CPG network in Los Angeles.

• H 0 : μ = 12
• H a : μ > 12

Step 2: Collect the data.

To conduct a hypothesis test, Melanie knows she has to use a t-model of the sampling distribution. She thinks ahead to the conditions required, which helps her collect a useful sample.

Recall the conditions for use of a t-model.

• There is no reason to think the download times are normally distributed (they might be, but this isn’t something Melanie could know for sure). So the sample has to be large (more than 30).
• The sample has to be random. Melanie decides to use one phone but randomly selects days, times, and locations in Los Angeles.

Melanie collects a random sample of 45 downloads by using her friend’s phone to download her song from iTunes according to the randomly selected days, times, and locations.

Melanie’s sample of size 45 downloads has an average download time of 13.5 seconds. The standard deviation for the sample is 3.2 seconds. Now Melanie needs to determine how unlikely this data is if CPG’s claim is actually true.

Step 3: Assess the evidence.

Assuming the average download time for Melanie’s song is really 12 seconds, what is the probability that 45 random downloads of this song will have a mean of 13.5 seconds or more?

This is a question about sampling variability. Melanie must determine the standard error. She knows the standard error of random sample means is [latex]\sigma/\sqrt{n}[/latex]. Since she has no way of knowing the population standard deviation, σ, Melanie uses the sample standard deviation, s = 3.2, as an approximation. Therefore, Melanie approximates the standard error of all sample means ( n = 45) to be

[latex]s/\sqrt{n} = 3.2/\sqrt{45}=0.48[/latex]

Now she can assess how far away her sample is from the claimed mean in terms of standard errors. That is, she can compute the t-score of her sample mean.

T = [latex]\frac{statistic-parameter}{standarderror} = \frac{\bar{x}-\mu}{s/\sqrt{n}}=\frac{13.5-12}{0.48}[/latex]=3.14

The sample mean for Melanie’s random sample is approximately 3.14 standard errors above the overall mean of 12. We know from previous experience that a sample mean this far above µ is very unlikely. With a t-score this large, the P-value is very small. We use a simulation of the t-model for 44 degrees of freedom to verify this.

We want the probability that the sample mean is greater than 13.5. This corresponds to the probability that T is greater than 3.14. The P-value is 0.0015.

Step 4: State a conclusion.

Here the logic is the same as for other hypothesis tests. We use the P-value to make a decision. The P-value helps us determine if the difference we see between the data and the hypothesized value of µ is statistically significant or due to chance. One of two outcomes can occur:

• One possibility is that results similar to the actual sample are extremely unlikely. This means the data does not fit with results from random samples selected from the population described by the null hypothesis. In this case, it is unlikely that the data came from this population. The probability as measured by the P-value is small, so we view this as strong evidence against the null hypothesis. We reject the null hypothesis in favor of the alternative hypothesis.
• The other possibility is that results similar to the actual sample are fairly likely (not unusual). This means the data fits with typical results from random samples selected from the population described by the null hypothesis. The probability as measured by the P-value is large. In this case, we do not have evidence against the null hypothesis, so we cannot reject it in favor of the alternative hypothesis.

Melanie’s data is very unlikely if µ = 12. The probability is essentially zero (P-value = 0.0015). This means we will rarely see sample means greater than 13.5 if µ = 12. So we reject the null and accept the alternative hypothesis. In other words, this sample provides strong evidence that CPG has overstated the speed of its data download capability.

The following activities give you an opportunity to practice parts of the hypothesis testing process for a population mean. Later you will have the opportunity to practice the hypothesis test from start to finish.

For the following scenarios, give the null and alternative hypotheses and state in words what µ represents in your hypotheses. A good definition of µ describes both the variable and the population.

In the previous example, Melanie did not state a significance level for her test. If she had, the logic is the same as we used for hypothesis tests in Modules 8 and 9. To come to a conclusion about H 0 , we compare the P-value to the significance level α.

• If P ≤ α, we reject H 0 . We conclude there is significant evidence in favor of H a .
• If P > α, we fail to reject H 0 . We conclude the sample does not provide significant evidence in favor of H a .

Use this simulation when needed to answer questions below.

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Module 10: Inference for Means

Hypothesis test for a population mean (2 of 5), learning outcomes.

• Under appropriate conditions, conduct a hypothesis test about a population mean. State a conclusion in context.

More on Checking Conditions for a T-Test

In practice, you will often see the use of a t-test with small samples. Technically, we can use the t-test with small samples only if we know the variable has a normal distribution in the population. But this is hard to verify. In addition, no variable has a perfect normal distribution. So what does the requirement that the “variable be normally distributed in the population” really mean?

We call a confidence interval or a hypothesis test robust if the confidence level or P-value does not change very much when the conditions for use of the procedure are not met.

T-procedures are robust when the variable is not normally distributed in the population, as long as the distribution is not heavily skewed. But how can we determine if the distribution of the variable in the population is heavily skewed? In this introductory course, we examine the distribution of the variable in the sample and make an educated guess about what is going on in the population.

Now we investigate this question: Can we tell from a sample whether the variable is normally distributed in the population?

Variable Skewed in the Population

Let’s start with a skewed distribution in the population. Can we tell that this distribution is not normal by looking at random samples?

The following figure shows the monthly payment on first home mortgages for 5,000 people, as reported in the 2000 U.S. Census. Think of this as data from the population of a small town. From this population, we randomly selected 20 people. We did this three times. Notice that for each random sample, the shape of the distribution of the monthly payments in the sample is skewed to the right, just like the distribution in the population. In 2 of the 3 samples, we also see outliers, just as we see in the population. So by looking at the sample, we can get a pretty good sense that the variable is not normally distributed in the population.

In this example, the sample size is less than 30. We can use the t-test only if the variable is normally distributed in the population. The shape of the distribution in any one of these samples suggests that the variable has a skewed distribution in the population, so we would not conduct a t-test with any of these samples.

Variable Normal in the Population

Now we look at a variable that has an approximately normal distribution in the population. Can we tell that this distribution is approximately normal by looking at random samples?

The following graphs show the heights (in centimeters) of 5,000 women. Think of this as data from the population of a small town. From this population, we randomly selected 20 women. We did this three times. Notice that for each random sample, the shape of the distribution of the heights in the sample is not skewed, and there are no outliers. By looking at the sample, we can get a pretty good sense that the variable is not skewed in the population, which suggests that the variable may be somewhat normally distributed in the population.

In this example, the sample size is less than 30. We can only use the t-test if the variable is normally distributed in the population. The shape of the distribution in any one of these samples indicates that the variable does not have a skewed distribution in the population, suggesting that the distribution in the population is somewhat normal. Since the t-procedures are robust, we would conduct a t-test with any of these samples.

What’s the Main Point?

We previously stated the conditions for use of the t-procedures as follows:

(1) If the variable is normally distributed in the population, you can always use the t-procedures.

(2) If the variable is not normally distributed in the population (or you can’t determine this factor), the sample size must be greater than 30 for safe use of the t-procedures.

We are now loosening these conditions somewhat because the t-procedures are robust.

(3) If the sample is small ( n ≤ 30), plot the data. If the distribution in the sample is not heavily skewed and does not have outliers, then we assume the variable is somewhat normally distributed in the population, so we use t-procedures.

If we use a t-procedure for a small sample ( n ≤ 30), it is good practice to include a disclaimer with the conclusion. We might say something like, “On the basis of the sample, we are assuming that the variable is distributed without strong skew or extreme outliers in the population. The conclusion from this test is valid only if this assumption is true.”

Each histogram in the following questions represents a random sample. We do not know if the variable has a normal distribution in the population, but we want to run a t-test to test a claim about the population mean. For each histogram, choose the option that best describes how to proceed with the hypothesis test.

Recall that the sample mean and standard deviation are not resistant to outliers. An outlier in the data can make the mean and standard deviation poor measures of center and spread. So why can we use data from large samples even if the data has an outlier? Well, if the sample is large enough, the distribution of sample means will still be approximately normal. And the t-model will be a good fit when we estimate the standard error of the sample means using the sample standard deviation. This is the important point. The P-value and confidence level come from a model of the sampling distribution, not from a model of the population’s distribution.

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8.3: Hypothesis Testing of Single Mean

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Learning Objectives

• To learn how to apply the five-step test procedure for test of hypotheses concerning a population mean when the sample size is small.

In the previous section hypotheses testing for population means was described in the case of large samples. The statistical validity of the tests was insured by the Central Limit Theorem, with essentially no assumptions on the distribution of the population. When sample sizes are small, as is often the case in practice, the Central Limit Theorem does not apply. One must then impose stricter assumptions on the population to give statistical validity to the test procedure. One common assumption is that the population from which the sample is taken has a normal probability distribution to begin with. Under such circumstances, if the population standard deviation is known, then the test statistic

\[\frac{(\bar{x}-\mu _0)}{\sigma /\sqrt{n}} \nonumber \]

still has the standard normal distribution, as in the previous two sections. If \(\sigma\) is unknown and is approximated by the sample standard deviation \(s\), then the resulting test statistic

\[\dfrac{(\bar{x}-\mu _0)}{s/\sqrt{n}} \nonumber \]

follows Student’s \(t\)-distribution with \(n-1\) degrees of freedom.

Standardized Test Statistics for Small Sample Hypothesis Tests Concerning a Single Population Mean

 If \(\sigma\) is known: \[Z=\frac{\bar{x}-\mu _0}{\sigma /\sqrt{n}} \nonumber \]

If \(\sigma\) is unknown: \[T=\frac{\bar{x}-\mu _0}{s /\sqrt{n}} \nonumber \]

• The first test statistic (\(\sigma\) known) has the standard normal distribution.
• The second test statistic (\(\sigma\) unknown) has Student’s \(t\)-distribution with \(n-1\) degrees of freedom.
• The population must be normally distributed.

The distribution of the second standardized test statistic (the one containing \(s\)) and the corresponding rejection region for each form of the alternative hypothesis (left-tailed, right-tailed, or two-tailed), is shown in Figure \(\PageIndex{1}\). This is just like Figure 8.2.1 except that now the critical values are from the \(t\)-distribution. Figure 8.2.1 still applies to the first standardized test statistic (the one containing (\(\sigma\)) since it follows the standard normal distribution.

The \(p\)-value of a test of hypotheses for which the test statistic has Student’s \(t\)-distribution can be computed using statistical software, but it is impractical to do so using tables, since that would require \(30\) tables analogous to Figure 7.1.5, one for each degree of freedom from \(1\) to \(30\). Figure 7.1.6 can be used to approximate the \(p\)-value of such a test, and this is typically adequate for making a decision using the \(p\)-value approach to hypothesis testing, although not always. For this reason the tests in the two examples in this section will be made following the critical value approach to hypothesis testing summarized at the end of Section 8.1, but after each one we will show how the \(p\)-value approach could have been used.

Example \(\PageIndex{1}\)

The price of a popular tennis racket at a national chain store is \(\$179\). Portia bought five of the same racket at an online auction site for the following prices:

\[155\; 179\; 175\; 175\; 161 \nonumber \]

Assuming that the auction prices of rackets are normally distributed, determine whether there is sufficient evidence in the sample, at the \(5\%\) level of significance, to conclude that the average price of the racket is less than \(\$179\) if purchased at an online auction.

• Step 1 . The assertion for which evidence must be provided is that the average online price \(\mu\) is less than the average price in retail stores, so the hypothesis test is \[H_0: \mu =179\\ \text{vs}\\ H_a: \mu <179\; @\; \alpha =0.05 \nonumber \]
• Step 2 . The sample is small and the population standard deviation is unknown. Thus the test statistic is \[T=\frac{\bar{x}-\mu _0}{s /\sqrt{n}} \nonumber \] and has the Student \(t\)-distribution with \(n-1=5-1=4\) degrees of freedom.
• Step 3 . From the data we compute \(\bar{x}=169\) and \(s=10.39\). Inserting these values into the formula for the test statistic gives \[T=\frac{\bar{x}-\mu _0}{s /\sqrt{n}}=\frac{169-179}{10.39/\sqrt{5}}=-2.152 \nonumber \]
• Step 4 . Since the symbol in \(H_a\) is “\(<\)” this is a left-tailed test, so there is a single critical value, \(-t_\alpha =-t_{0.05}[df=4]\). Reading from the row labeled \(df=4\) in Figure 7.1.6 its value is \(-2.132\). The rejection region is \((-\infty ,-2.132]\).
• Step 5 . As shown in Figure \(\PageIndex{2}\) the test statistic falls in the rejection region. The decision is to reject \(H_0\). In the context of the problem our conclusion is:

The data provide sufficient evidence, at the \(5\%\) level of significance, to conclude that the average price of such rackets purchased at online auctions is less than \(\$179\).

To perform the test in Example \(\PageIndex{1}\) using the \(p\)-value approach, look in the row in Figure 7.1.6 with the heading \(df=4\) and search for the two \(t\)-values that bracket the unsigned value \(2.152\) of the test statistic. They are \(2.132\) and \(2.776\), in the columns with headings \(t_{0.050}\) and \(t_{0.025}\). They cut off right tails of area \(0.050\) and \(0.025\), so because \(2.152\) is between them it must cut off a tail of area between \(0.050\) and \(0.025\). By symmetry \(-2.152\) cuts off a left tail of area between \(0.050\) and \(0.025\), hence the \(p\)-value corresponding to \(t=-2.152\) is between \(0.025\) and \(0.05\). Although its precise value is unknown, it must be less than \(\alpha =0.05\), so the decision is to reject \(H_0\).

Example \(\PageIndex{2}\)

A small component in an electronic device has two small holes where another tiny part is fitted. In the manufacturing process the average distance between the two holes must be tightly controlled at \(0.02\) mm, else many units would be defective and wasted. Many times throughout the day quality control engineers take a small sample of the components from the production line, measure the distance between the two holes, and make adjustments if needed. Suppose at one time four units are taken and the distances are measured as

Determine, at the \(1\%\) level of significance, if there is sufficient evidence in the sample to conclude that an adjustment is needed. Assume the distances of interest are normally distributed.

• Step 1 . The assumption is that the process is under control unless there is strong evidence to the contrary. Since a deviation of the average distance to either side is undesirable, the relevant test is \[H_0: \mu =0.02\\ \text{vs}\\ H_a: \mu \neq 0.02\; @\; \alpha =0.01 \nonumber \] where \(\mu\) denotes the mean distance between the holes.
• Step 2 . The sample is small and the population standard deviation is unknown. Thus the test statistic is \[T=\frac{\bar{x}-\mu _0}{s /\sqrt{n}} \nonumber \] and has the Student \(t\)-distribution with \(n-1=4-1=3\) degrees of freedom.
• Step 3 . From the data we compute \(\bar{x}=0.02075\) and \(s=0.00171\). Inserting these values into the formula for the test statistic gives \[T=\frac{\bar{x}-\mu _0}{s /\sqrt{n}}=\frac{0.02075-0.02}{0.00171\sqrt{4}}=0.877 \nonumber \]
• Step 4 . Since the symbol in \(H_a\) is “\(\neq\)” this is a two-tailed test, so there are two critical values, \(\pm t_{\alpha/2} =-t_{0.005}[df=3]\). Reading from the row in Figure 7.1.6 labeled \(df=3\) their values are \(\pm 5.841\). The rejection region is \((-\infty ,-5.841]\cup [5.841,\infty )\).
• Step 5 . As shown in Figure \(\PageIndex{3}\) the test statistic does not fall in the rejection region. The decision is not to reject \(H_0\). In the context of the problem our conclusion is:

The data do not provide sufficient evidence, at the \(1\%\) level of significance, to conclude that the mean distance between the holes in the component differs from \(0.02\) mm.

To perform the test in "Example \(\PageIndex{2}\)" using the \(p\)-value approach, look in the row in Figure 7.1.6 with the heading \(df=3\) and search for the two \(t\)-values that bracket the value \(0.877\) of the test statistic. Actually \(0.877\) is smaller than the smallest number in the row, which is \(0.978\), in the column with heading \(t_{0.200}\). The value \(0.978\) cuts off a right tail of area \(0.200\), so because \(0.877\) is to its left it must cut off a tail of area greater than \(0.200\). Thus the \(p\)-value, which is the double of the area cut off (since the test is two-tailed), is greater than \(0.400\). Although its precise value is unknown, it must be greater than \(\alpha =0.01\), so the decision is not to reject \(H_0\).

Key Takeaway

• There are two formulas for the test statistic in testing hypotheses about a population mean with small samples. One test statistic follows the standard normal distribution, the other Student’s \(t\)-distribution.
• The population standard deviation is used if it is known, otherwise the sample standard deviation is used.
• Either five-step procedure, critical value or \(p\)-value approach, is used with either test statistic.