hypothesis testing z statistic

Z-Test for Statistical Hypothesis Testing Explained

The Z-test is a statistical hypothesis test used to determine where the distribution of the test statistic we are measuring, like the mean , is part of the normal distribution .

There are multiple types of Z-tests, however, we’ll focus on the easiest and most well known one, the one sample mean test. This is used to determine if the difference between the mean of a sample and the mean of a population is statistically significant.

What Is a Z-Test?

A Z-test is a type of statistical hypothesis test where the test-statistic follows a normal distribution.  

The name Z-test comes from the Z-score of the normal distribution. This is a measure of how many standard deviations away a raw score or sample statistics is from the populations’ mean.

Z-tests are the most common statistical tests conducted in fields such as healthcare and data science . Therefore, it’s an essential concept to understand.

Requirements for a Z-Test

In order to conduct a Z-test, your statistics need to meet a few requirements, including:

• A Sample size that’s greater than 30. This is because we want to ensure our sample mean comes from a distribution that is normal. As stated by the c entral limit theorem , any distribution can be approximated as normally distributed if it contains more than 30 data points.
• The standard deviation and mean of the population is known .
• The sample data is collected/acquired randomly .

More on Data Science:   What Is Bootstrapping Statistics?

Z-Test Steps

There are four steps to complete a Z-test. Let’s examine each one.

4 Steps to a Z-Test

• State the null hypothesis.
• State the alternate hypothesis.
• Choose your critical value.
• Calculate your Z-test statistics. 

1. State the Null Hypothesis

The first step in a Z-test is to state the null hypothesis, H_0 . This what you believe to be true from the population, which could be the mean of the population, μ_0 :

2. State the Alternate Hypothesis

Next, state the alternate hypothesis, H_1 . This is what you observe from your sample. If the sample mean is different from the population’s mean, then we say the mean is not equal to μ_0:

3. Choose Your Critical Value

Then, choose your critical value, α , which determines whether you accept or reject the null hypothesis. Typically for a Z-test we would use a statistical significance of 5 percent which is z = +/- 1.96 standard deviations from the population’s mean in the normal distribution:

This critical value is based on confidence intervals.

4. Calculate Your Z-Test Statistic

Compute the Z-test Statistic using the sample mean, μ_1 , the population mean, μ_0 , the number of data points in the sample, n and the population’s standard deviation, σ :

If the test statistic is greater (or lower depending on the test we are conducting) than the critical value, then the alternate hypothesis is true because the sample’s mean is statistically significant enough from the population mean.

Another way to think about this is if the sample mean is so far away from the population mean, the alternate hypothesis has to be true or the sample is a complete anomaly.

More on Data Science: Basic Probability Theory and Statistics Terms to Know

Z-Test Example

Let’s go through an example to fully understand the one-sample mean Z-test.

A school says that its pupils are, on average, smarter than other schools. It takes a sample of 50 students whose average IQ measures to be 110. The population, or the rest of the schools, has an average IQ of 100 and standard deviation of 20. Is the school’s claim correct?

The null and alternate hypotheses are:

Where we are saying that our sample, the school, has a higher mean IQ than the population mean.

Now, this is what’s called a right-sided, one-tailed test as our sample mean is greater than the population’s mean. So, choosing a critical value of 5 percent, which equals a Z-score of 1.96 , we can only reject the null hypothesis if our Z-test statistic is greater than 1.96.

If the school claimed its students’ IQs were an average of 90, then we would use a left-tailed test, as shown in the figure above. We would then only reject the null hypothesis if our Z-test statistic is less than -1.96.

Computing our Z-test statistic, we see:

Therefore, we have sufficient evidence to reject the null hypothesis, and the school’s claim is right.

Hope you enjoyed this article on Z-tests. In this post, we only addressed the most simple case, the one-sample mean test. However, there are other types of tests, but they all follow the same process just with some small nuances.  

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10 Chapter 10: Hypothesis Testing with Z

Setting up the hypotheses.

When setting up the hypotheses with z, the parameter is associated with a sample mean (in the previous chapter examples the parameters for the null used 0). Using z is an occasion in which the null hypothesis is a value other than 0. For example, if we are working with mothers in the U.S. whose children are at risk of low birth weight, we can use 7.47 pounds, the average birth weight in the US, as our null value and test for differences against that. For now, we will focus on testing a value of a single mean against what we expect from the population.

Using birthweight as an example, our null hypothesis takes the form: H 0 : μ = 7.47 Notice that we are testing the value for μ, the population parameter, NOT the sample statistic ̅X (or M). We are referring to the data right now in raw form (we have not standardized it using z yet). Again, using inferential statistics, we are interested in understanding the population, drawing from our sample observations. For the research question, we have a mean value from the sample to use, we have specific data is – it is observed and used as a comparison for a set point.

As mentioned earlier, the alternative hypothesis is simply the reverse of the null hypothesis, and there are three options, depending on where we expect the difference to lie. We will set the criteria for rejecting the null hypothesis based on the directionality (greater than, less than, or not equal to) of the alternative.

If we expect our obtained sample mean to be above or below the null hypothesis value (knowing which direction), we set a directional hypothesis. O ur alternative hypothesis takes the form based on the research question itself. In our example with birthweight, this could be presented as H A : μ > 7.47 or H A : μ < 7.47. 

Note that we should only use a directional hypothesis if we have a good reason, based on prior observations or research, to suspect a particular direction. When we do not know the direction, such as when we are entering a new area of research, we use a non-directional alternative hypothesis. In our birthweight example, this could be set as H A : μ ≠ 7.47

In working with data for this course we will need to set a critical value of the test statistic for alpha (α) for use of test statistic tables in the back of the book. This is determining the critical rejection region that has a set critical value based on α.

Determining Critical Value from α

We set alpha (α) before collecting data in order to determine whether or not we should reject the null hypothesis. We set this value beforehand to avoid biasing ourselves by viewing our results and then determining what criteria we should use.

When a research hypothesis predicts an effect but does not predict a direction for the effect, it is called a non-directional hypothesis . To test the significance of a non-directional hypothesis, we have to consider the possibility that the sample could be extreme at either tail of the comparison distribution. We call this a two-tailed test .

Figure 1. showing a 2-tail test for non-directional hypothesis for z for area C is the critical rejection region.

When a research hypothesis predicts a direction for the effect, it is called a directional hypothesis . To test the significance of a directional hypothesis, we have to consider the possibility that the sample could be extreme at one-tail of the comparison distribution. We call this a one-tailed test .

Figure 2. showing a 1-tail test for a directional hypothesis (predicting an increase) for z for area C is the critical rejection region.

Determining Cutoff Scores with Two-Tailed Tests

Typically we specify an α level before analyzing the data. If the data analysis results in a probability value below the α level, then the null hypothesis is rejected; if it is not, then the null hypothesis is not rejected. In other words, if our data produce values that meet or exceed this threshold, then we have sufficient evidence to reject the null hypothesis ; if not, we fail to reject the null (we never “accept” the null). According to this perspective, if a result is significant, then it does not matter how significant it is. Moreover, if it is not significant, then it does not matter how close to being significant it is. Therefore, if the 0.05 level is being used, then probability values of 0.049 and 0.001 are treated identically. Similarly, probability values of 0.06 and 0.34 are treated identically. Note we will discuss ways to address effect size (which is related to this challenge of NHST).

When setting the probability value, there is a special complication in a two-tailed test. We have to divide the significance percentage between the two tails. For example, with a 5% significance level, we reject the null hypothesis only if the sample is so extreme that it is in either the top 2.5% or the bottom 2.5% of the comparison distribution. This keeps the overall level of significance at a total of 5%. A one-tailed test does have such an extreme value but with a one-tailed test only one side of the distribution is considered.

Figure 3. Critical value differences in one and two-tail tests. Photo Credit

Let’s re view th e set critical values for Z.

We discussed z-scores and probability in chapter 8.  If we revisit the z-score for 5% and 1%, we can identify the critical regions for the critical rejection areas from the unit standard normal table.

• A two-tailed test at the 5% level has a critical boundary Z score of +1.96 and -1.96
• A one-tailed test at the 5% level has a critical boundary Z score of +1.64 or -1.64
• A two-tailed test at the 1% level has a critical boundary Z score of +2.58 and -2.58
• A one-tailed test at the 1% level has a critical boundary Z score of +2.33 or -2.33.

Review: Critical values, p-values, and significance level

There are two criteria we use to assess whether our data meet the thresholds established by our chosen significance level, and they both have to do with our discussions of probability and distributions. Recall that probability refers to the likelihood of an event, given some situation or set of conditions. In hypothesis testing, that situation is the assumption that the null hypothesis value is the correct value, or that there is no effec t. The value laid out in H 0 is our condition under which we interpret our results. To reject this assumption, and thereby reject the null hypothesis, we need results that would be very unlikely if the null was true.

Now recall that values of z which fall in the tails of the standard normal distribution represent unlikely values. That is, the proportion of the area under the curve as or more extreme than z is very small as we get into the tails of the distribution. Our significance level corresponds to the area under the tail that is exactly equal to α: if we use our normal criterion of α = .05, then 5% of the area under the curve becomes what we call the rejection region (also called the critical region) of the distribution. This is illustrated in Figure 4.

Figure 4: The rejection region for a one-tailed test

The shaded rejection region takes us 5% of the area under the curve. Any result which falls in that region is sufficient evidence to reject the null hypothesis.

The rejection region is bounded by a specific z-value, as is any area under the curve. In hypothesis testing, the value corresponding to a specific rejection region is called the critical value, z crit (“z-crit”) or z* (hence the other name “critical region”). Finding the critical value works exactly the same as finding the z-score corresponding to any area under the curve like we did in Unit 1. If we go to the normal table, we will find that the z-score corresponding to 5% of the area under the curve is equal to 1.645 (z = 1.64 corresponds to 0.0405 and z = 1.65 corresponds to 0.0495, so .05 is exactly in between them) if we go to the right and -1.645 if we go to the left. The direction must be determined by your alternative hypothesis, and drawing then shading the distribution is helpful for keeping directionality straight.

Suppose, however, that we want to do a non-directional test. We need to put the critical region in both tails, but we don’t want to increase the overall size of the rejection region (for reasons we will see later). To do this, we simply split it in half so that an equal proportion of the area under the curve falls in each tail’s rejection region. For α = .05, this means 2.5% of the area is in each tail, which, based on the z-table, corresponds to critical values of z* = ±1.96. This is shown in Figure 5.

Figure 5: Two-tailed rejection region

Thus, any z-score falling outside ±1.96 (greater than 1.96 in absolute value) falls in the rejection region. When we use z-scores in this way, the obtained value of z (sometimes called z-obtained) is something known as a test statistic, which is simply an inferential statistic used to test a null hypothesis.

Calculate the test statistic: Z

Now that we understand setting up the hypothesis and determining the outcome, let’s examine hypothesis testing with z!  The next step is to carry out the study and get the actual results for our sample. Central to hypothesis test is comparison of the population and sample means. To make our calculation and determine where the sample is in the hypothesized distribution we calculate the Z for the sample data.

Make a decision

To decide whether to reject the null hypothesis, we compare our sample’s Z score to the Z score that marks our critical boundary. If our sample Z score falls inside the rejection region of the comparison distribution (is greater than the z-score critical boundary) we reject the null hypothesis.

The formula for our z- statistic has not changed:

To formally test our hypothesis, we compare our obtained z-statistic to our critical z-value. If z obt > z crit , that means it falls in the rejection region (to see why, draw a line for z = 2.5 on Figure 1 or Figure 2) and so we reject H 0 . If z obt < z crit , we fail to reject. Remember that as z gets larger, the corresponding area under the curve beyond z gets smaller. Thus, the proportion, or p-value, will be smaller than the area for α, and if the area is smaller, the probability gets smaller. Specifically, the probability of obtaining that result, or a more extreme result, under the condition that the null hypothesis is true gets smaller.

Conversely, if we fail to reject, we know that the proportion will be larger than α because the z-statistic will not be as far into the tail. This is illustrated for a one- tailed test in Figure 6.

Figure 6. Relation between α, z obt , and p

When the null hypothesis is rejected, the effect is said to be statistically significant . Do not confuse statistical significance with practical significance. A small effect can be highly significant if the sample size is large enough.

Why does the word “significant” in the phrase “statistically significant” mean something so different from other uses of the word? Interestingly, this is because the meaning of “significant” in everyday language has changed. It turns out that when the procedures for hypothesis testing were developed, something was “significant” if it signified something. Thus, finding that an effect is statistically significant signifies that the effect is real and not due to chance. Over the years, the meaning of “significant” changed, leading to the potential misinterpretation.

Review: Steps of the Hypothesis Testing Process

The process of testing hypotheses follows a simple four-step procedure. This process will be what we use for the remained of the textbook and course, and though the hypothesis and statistics we use will change, this process will not.

Step 1: State the Hypotheses

Your hypotheses are the first thing you need to lay out. Otherwise, there is nothing to test! You have to state the null hypothesis (which is what we test) and the alternative hypothesis (which is what we expect). These should be stated mathematically as they were presented above AND in words, explaining in normal English what each one means in terms of the research question.

Step 2: Find the Critical Values

Next, we formally lay out the criteria we will use to test our hypotheses. There are two pieces of information that inform our critical values: α, which determines how much of the area under the curve composes our rejection region, and the directionality of the test, which determines where the region will be.

Step 3: Compute the Test Statistic

Once we have our hypotheses and the standards we use to test them, we can collect data and calculate our test statistic, in this case z . This step is where the vast majority of differences in future chapters will arise: different tests used for different data are calculated in different ways, but the way we use and interpret them remains the same.

Step 4: Make the Decision

Finally, once we have our obtained test statistic, we can compare it to our critical value and decide whether we should reject or fail to reject the null hypothesis. When we do this, we must interpret the decision in relation to our research question, stating what we concluded, what we based our conclusion on, and the specific statistics we obtained.

Example: Movie Popcorn

Let’s see how hypothesis testing works in action by working through an example. Say that a movie theater owner likes to keep a very close eye on how much popcorn goes into each bag sold, so he knows that the average bag has 8 cups of popcorn and that this varies a little bit, about half a cup. That is, the known population mean is μ = 8.00 and the known population standard deviation is σ =0.50. The owner wants to make sure that the newest employee is filling bags correctly, so over the course of a week he randomly assesses 25 bags filled by the employee to test for a difference (n = 25). He doesn’t want bags overfilled or under filled, so he looks for differences in both directions. This scenario has all of the information we need to begin our hypothesis testing procedure.

Our manager is looking for a difference in the mean cups of popcorn bags compared to the population mean of 8. We will need both a null and an alternative hypothesis written both mathematically and in words. We’ll always start with the null hypothesis:

H 0 : There is no difference in the cups of popcorn bags from this employee H 0 : μ = 8.00

Notice that we phrase the hypothesis in terms of the population parameter μ, which in this case would be the true average cups of bags filled by the new employee.

Our assumption of no difference, the null hypothesis, is that this mean is exactly

the same as the known population mean value we want it to match, 8.00. Now let’s do the alternative:

H A : There is a difference in the cups of popcorn bags from this employee H A : μ ≠ 8.00

In this case, we don’t know if the bags will be too full or not full enough, so we do a two-tailed alternative hypothesis that there is a difference.

Our critical values are based on two things: the directionality of the test and the level of significance. We decided in step 1 that a two-tailed test is the appropriate directionality. We were given no information about the level of significance, so we assume that α = 0.05 is what we will use. As stated earlier in the chapter, the critical values for a two-tailed z-test at α = 0.05 are z* = ±1.96. This will be the criteria we use to test our hypothesis. We can now draw out our distribution so we can visualize the rejection region and make sure it makes sense

Figure 7: Rejection region for z* = ±1.96

Step 3: Calculate the Test Statistic

Now we come to our formal calculations. Let’s say that the manager collects data and finds that the average cups of this employee’s popcorn bags is ̅X = 7.75 cups. We can now plug this value, along with the values presented in the original problem, into our equation for z:

So our test statistic is z = -2.50, which we can draw onto our rejection region distribution:

Figure 8: Test statistic location

Looking at Figure 5, we can see that our obtained z-statistic falls in the rejection region. We can also directly compare it to our critical value: in terms of absolute value, -2.50 > -1.96, so we reject the null hypothesis. We can now write our conclusion:

When we write our conclusion, we write out the words to communicate what it actually means, but we also include the average sample size we calculated (the exact location doesn’t matter, just somewhere that flows naturally and makes sense) and the z-statistic and p-value. We don’t know the exact p-value, but we do know that because we rejected the null, it must be less than α.

Effect Size

When we reject the null hypothesis, we are stating that the difference we found was statistically significant, but we have mentioned several times that this tells us nothing about practical significance. To get an idea of the actual size of what we found, we can compute a new statistic called an effect size. Effect sizes give us an idea of how large, important, or meaningful a statistically significant effect is.

For mean differences like we calculated here, our effect size is Cohen’s d :

Effect sizes are incredibly useful and provide important information and clarification that overcomes some of the weakness of hypothesis testing. Whenever you find a significant result, you should always calculate an effect size

Table 1. Interpretation of Cohen’s d

Example: Office Temperature

Let’s do another example to solidify our understanding. Let’s say that the office building you work in is supposed to be kept at 74 degree Fahrenheit but is allowed

to vary by 1 degree in either direction. You suspect that, as a cost saving measure, the temperature was secretly set higher. You set up a formal way to test your hypothesis.

You start by laying out the null hypothesis:

H 0 : There is no difference in the average building temperature H 0 : μ = 74

Next you state the alternative hypothesis. You have reason to suspect a specific direction of change, so you make a one-tailed test:

H A : The average building temperature is higher than claimed H A : μ > 74

Now that you have everything set up, you spend one week collecting temperature data:

You calculate the average of these scores to be 𝑋̅ = 76.6 degrees. You use this to calculate the test statistic, using μ = 74 (the supposed average temperature), σ = 1.00 (how much the temperature should vary), and n = 5 (how many data points you collected):

z = 76.60 − 74.00 = 2.60    = 5.78

          1.00/√5            0.45

This value falls so far into the tail that it cannot even be plotted on the distribution!

Figure 7: Obtained z-statistic

You compare your obtained z-statistic, z = 5.77, to the critical value, z* = 1.645, and find that z > z*. Therefore you reject the null hypothesis, concluding: Based on 5 observations, the average temperature (𝑋̅ = 76.6 degrees) is statistically significantly higher than it is supposed to be, z = 5.77, p < .05.

d = (76.60-74.00)/ 1= 2.60

The effect size you calculate is definitely large, meaning someone has some explaining to do!

Example: Different Significance Level

First, let’s take a look at an example phrased in generic terms, rather than in the context of a specific research question, to see the individual pieces one more time. This time, however, we will use a stricter significance level, α = 0.01, to test the hypothesis.

We will use 60 as an arbitrary null hypothesis value: H 0 : The average score does not differ from the population H 0 : μ = 50

We will assume a two-tailed test: H A : The average score does differ H A : μ ≠ 50

We have seen the critical values for z-tests at α = 0.05 levels of significance several times. To find the values for α = 0.01, we will go to the standard normal table and find the z-score cutting of 0.005 (0.01 divided by 2 for a two-tailed test) of the area in the tail, which is z crit * = ±2.575. Notice that this cutoff is much higher than it was for α = 0.05. This is because we need much less of the area in the tail, so we need to go very far out to find the cutoff. As a result, this will require a much larger effect or much larger sample size in order to reject the null hypothesis.

We can now calculate our test statistic.  The average of 10 scores is M = 60.40 with a µ = 60. We will use σ = 10 as our known population standard deviation. From this information, we calculate our z-statistic as:

Our obtained z-statistic, z = 0.13, is very small. It is much less than our critical value of 2.575. Thus, this time, we fail to reject the null hypothesis. Our conclusion would look something like:

Notice two things about the end of the conclusion. First, we wrote that p is greater than instead of p is less than, like we did in the previous two examples. This is because we failed to reject the null hypothesis. We don’t know exactly what the p- value is, but we know it must be larger than the α level we used to test our hypothesis. Second, we used 0.01 instead of the usual 0.05, because this time we tested at a different level. The number you compare to the p-value should always be the significance level you test at. Because we did not detect a statistically significant effect, we do not need to calculate an effect size. Note: some statisticians will suggest to always calculate effects size as a possibility of Type II error. Although insignificant, calculating d = (60.4-60)/10 = .04 which suggests no effect (and not a possibility of Type II error).

Review Considerations in Hypothesis Testing

Errors in hypothesis testing.

Keep in mind that rejecting the null hypothesis is not an all-or-nothing decision. The Type I error rate is affected by the α level: the lower the α level the lower the Type I error rate. It might seem that α is the probability of a Type I error. However, this is not correct. Instead, α is the probability of a Type I error given that the null hypothesis is true. If the null hypothesis is false, then it is impossible to make a Type I error. The second type of error that can be made in significance testing is failing to reject a false null hypothesis. This kind of error is called a Type II error.

Statistical Power

The statistical power of a research design is the probability of rejecting the null hypothesis given the sample size and expected relationship strength. Statistical power is the complement of the probability of committing a Type II error. Clearly, researchers should be interested in the power of their research designs if they want to avoid making Type II errors. In particular, they should make sure their research design has adequate power before collecting data. A common guideline is that a power of .80 is adequate. This means that there is an 80% chance of rejecting the null hypothesis for the expected relationship strength.

Given that statistical power depends primarily on relationship strength and sample size, there are essentially two steps you can take to increase statistical power: increase the strength of the relationship or increase the sample size. Increasing the strength of the relationship can sometimes be accomplished by using a stronger manipulation or by more carefully controlling extraneous variables to reduce the amount of noise in the data (e.g., by using a within-subjects design rather than a between-subjects design). The usual strategy, however, is to increase the sample size. For any expected relationship strength, there will always be some sample large enough to achieve adequate power.

Inferential statistics uses data from a sample of individuals to reach conclusions about the whole population. The degree to which our inferences are valid depends upon how we selected the sample (sampling technique) and the characteristics (parameters) of population data. Statistical analyses assume that sample(s) and population(s) meet certain conditions called statistical assumptions.

It is easy to check assumptions when using statistical software and it is important as a researcher to check for violations; if violations of statistical assumptions are not appropriately addressed then results may be interpreted incorrectly.

Learning Objectives

Having read the chapter, students should be able to:

• Conduct a hypothesis test using a z-score statistics, locating critical region, and make a statistical decision including.
• Explain the purpose of measuring effect size and power, and be able to compute Cohen’s d.

Exercises – Ch. 10

• List the main steps for hypothesis testing with the z-statistic. When and why do you calculate an effect size?
• z = 1.99, two-tailed test at α = 0.05
• z = 1.99, two-tailed test at α = 0.01
• z = 1.99, one-tailed test at α = 0.05
• You are part of a trivia team and have tracked your team’s performance since you started playing, so you know that your scores are normally distributed with μ = 78 and σ = 12. Recently, a new person joined the team, and you think the scores have gotten better. Use hypothesis testing to see if the average score has improved based on the following 8 weeks’ worth of score data: 82, 74, 62, 68, 79, 94, 90, 81, 80.
• A study examines self-esteem and depression in teenagers.  A sample of 25 teens with a low self-esteem are given the Beck Depression Inventory.  The average score for the group is 20.9.  For the general population, the average score is 18.3 with σ = 12.  Use a two-tail test with α = 0.05 to examine whether teenagers with low self-esteem show significant differences in depression.
• You get hired as a server at a local restaurant, and the manager tells you that servers’ tips are $42 on average but vary about $12 (μ = 42, σ = 12). You decide to track your tips to see if you make a different amount, but because this is your first job as a server, you don’t know if you will make more or less in tips. After working 16 shifts, you find that your average nightly amount is $44.50 from tips. Test for a difference between this value and the population mean at the α = 0.05 level of significance.

Answers to Odd- Numbered Exercises – Ch. 10

1. List hypotheses. Determine critical region. Calculate z.  Compare z to critical region. Draw Conclusion.  We calculate an effect size when we find a statistically significant result to see if our result is practically meaningful or important

5. Step 1: H 0 : μ = 42 “My average tips does not differ from other servers”, H A : μ ≠ 42 “My average tips do differ from others”

Introduction to Statistics for Psychology Copyright © 2021 by Alisa Beyer is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License , except where otherwise noted.

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Z test is a statistical test that is conducted on data that approximately follows a normal distribution. The z test can be performed on one sample, two samples, or on proportions for hypothesis testing. It checks if the means of two large samples are different or not when the population variance is known.

A z test can further be classified into left-tailed, right-tailed, and two-tailed hypothesis tests depending upon the parameters of the data. In this article, we will learn more about the z test, its formula, the z test statistic, and how to perform the test for different types of data using examples.

What is Z Test?

A z test is a test that is used to check if the means of two populations are different or not provided the data follows a normal distribution. For this purpose, the null hypothesis and the alternative hypothesis must be set up and the value of the z test statistic must be calculated. The decision criterion is based on the z critical value.

Z Test Definition

A z test is conducted on a population that follows a normal distribution with independent data points and has a sample size that is greater than or equal to 30. It is used to check whether the means of two populations are equal to each other when the population variance is known. The null hypothesis of a z test can be rejected if the z test statistic is statistically significant when compared with the critical value.

Z Test Formula

The z test formula compares the z statistic with the z critical value to test whether there is a difference in the means of two populations. In hypothesis testing , the z critical value divides the distribution graph into the acceptance and the rejection regions. If the test statistic falls in the rejection region then the null hypothesis can be rejected otherwise it cannot be rejected. The z test formula to set up the required hypothesis tests for a one sample and a two-sample z test are given below.

One-Sample Z Test

A one-sample z test is used to check if there is a difference between the sample mean and the population mean when the population standard deviation is known. The formula for the z test statistic is given as follows:

z = \(\frac{\overline{x}-\mu}{\frac{\sigma}{\sqrt{n}}}\). \(\overline{x}\) is the sample mean, \(\mu\) is the population mean, \(\sigma\) is the population standard deviation and n is the sample size.

The algorithm to set a one sample z test based on the z test statistic is given as follows:

Left Tailed Test:

Null Hypothesis: \(H_{0}\) : \(\mu = \mu_{0}\)

Alternate Hypothesis: \(H_{1}\) : \(\mu < \mu_{0}\)

Decision Criteria: If the z statistic < z critical value then reject the null hypothesis.

Right Tailed Test:

Alternate Hypothesis: \(H_{1}\) : \(\mu > \mu_{0}\)

Decision Criteria: If the z statistic > z critical value then reject the null hypothesis.

Two Tailed Test:

Alternate Hypothesis: \(H_{1}\) : \(\mu \neq \mu_{0}\)

Two Sample Z Test

A two sample z test is used to check if there is a difference between the means of two samples. The z test statistic formula is given as follows:

z = \(\frac{(\overline{x_{1}}-\overline{x_{2}})-(\mu_{1}-\mu_{2})}{\sqrt{\frac{\sigma_{1}^{2}}{n_{1}}+\frac{\sigma_{2}^{2}}{n_{2}}}}\). \(\overline{x_{1}}\), \(\mu_{1}\), \(\sigma_{1}^{2}\) are the sample mean, population mean and population variance respectively for the first sample. \(\overline{x_{2}}\), \(\mu_{2}\), \(\sigma_{2}^{2}\) are the sample mean, population mean and population variance respectively for the second sample.

The two-sample z test can be set up in the same way as the one-sample test. However, this test will be used to compare the means of the two samples. For example, the null hypothesis is given as \(H_{0}\) : \(\mu_{1} = \mu_{2}\).

Z Test for Proportions

A z test for proportions is used to check the difference in proportions. A z test can either be used for one proportion or two proportions. The formulas are given as follows.

One Proportion Z Test

A one proportion z test is used when there are two groups and compares the value of an observed proportion to a theoretical one. The z test statistic for a one proportion z test is given as follows:

z = \(\frac{p-p_{0}}{\sqrt{\frac{p_{0}(1-p_{0})}{n}}}\). Here, p is the observed value of the proportion, \(p_{0}\) is the theoretical proportion value and n is the sample size.

The null hypothesis is that the two proportions are the same while the alternative hypothesis is that they are not the same.

Two Proportion Z Test

A two proportion z test is conducted on two proportions to check if they are the same or not. The test statistic formula is given as follows:

z =\(\frac{p_{1}-p_{2}-0}{\sqrt{p(1-p)\left ( \frac{1}{n_{1}} +\frac{1}{n_{2}}\right )}}\)

where p = \(\frac{x_{1}+x_{2}}{n_{1}+n_{2}}\)

\(p_{1}\) is the proportion of sample 1 with sample size \(n_{1}\) and \(x_{1}\) number of trials.

\(p_{2}\) is the proportion of sample 2 with sample size \(n_{2}\) and \(x_{2}\) number of trials.

How to Calculate Z Test Statistic?

The most important step in calculating the z test statistic is to interpret the problem correctly. It is necessary to determine which tailed test needs to be conducted and what type of test does the z statistic belong to. Suppose a teacher claims that his section's students will score higher than his colleague's section. The mean score is 22.1 for 60 students belonging to his section with a standard deviation of 4.8. For his colleague's section, the mean score is 18.8 for 40 students and the standard deviation is 8.1. Test his claim at \(\alpha\) = 0.05. The steps to calculate the z test statistic are as follows:

• Identify the type of test. In this example, the means of two populations have to be compared in one direction thus, the test is a right-tailed two-sample z test.
• Set up the hypotheses. \(H_{0}\): \(\mu_{1} = \mu_{2}\), \(H_{1}\): \(\mu_{1} > \mu_{2}\).
• Find the critical value at the given alpha level using the z table. The critical value is 1.645.
• Determine the z test statistic using the appropriate formula. This is given by z = \(\frac{(\overline{x_{1}}-\overline{x_{2}})-(\mu_{1}-\mu_{2})}{\sqrt{\frac{\sigma_{1}^{2}}{n_{1}}+\frac{\sigma_{2}^{2}}{n_{2}}}}\). Substitute values in this equation. \(\overline{x_{1}}\) = 22.1, \(\sigma_{1}\) = 4.8, \(n_{1}\) = 60, \(\overline{x_{2}}\) = 18.8, \(\sigma_{2}\) = 8.1, \(n_{2}\) = 40 and \(\mu_{1} - \mu_{2} = 0\). Thus, z = 2.32
• Compare the critical value and test statistic to arrive at a conclusion. As 2.32 > 1.645 thus, the null hypothesis can be rejected. It can be concluded that there is enough evidence to support the teacher's claim that the scores of students are better in his class.

Z Test vs T-Test

Both z test and t-test are univariate tests used on the means of two datasets. The differences between both tests are outlined in the table given below:

Related Articles:

• Probability and Statistics
• Data Handling
• Summary Statistics

Important Notes on Z Test

• Z test is a statistical test that is conducted on normally distributed data to check if there is a difference in means of two data sets.
• The sample size should be greater than 30 and the population variance must be known to perform a z test.
• The one-sample z test checks if there is a difference in the sample and population mean,
• The two sample z test checks if the means of two different groups are equal.

Examples on Z Test

Example 1: A teacher claims that the mean score of students in his class is greater than 82 with a standard deviation of 20. If a sample of 81 students was selected with a mean score of 90 then check if there is enough evidence to support this claim at a 0.05 significance level.

Solution: As the sample size is 81 and population standard deviation is known, this is an example of a right-tailed one-sample z test.

\(H_{0}\) : \(\mu = 82\)

\(H_{1}\) : \(\mu > 82\)

From the z table the critical value at \(\alpha\) = 1.645

z = \(\frac{\overline{x}-\mu}{\frac{\sigma}{\sqrt{n}}}\)

\(\overline{x}\) = 90, \(\mu\) = 82, n = 81, \(\sigma\) = 20

As 3.6 > 1.645 thus, the null hypothesis is rejected and it is concluded that there is enough evidence to support the teacher's claim.

Answer: Reject the null hypothesis

Example 2: An online medicine shop claims that the mean delivery time for medicines is less than 120 minutes with a standard deviation of 30 minutes. Is there enough evidence to support this claim at a 0.05 significance level if 49 orders were examined with a mean of 100 minutes?

Solution: As the sample size is 49 and population standard deviation is known, this is an example of a left-tailed one-sample z test.

\(H_{0}\) : \(\mu = 120\)

\(H_{1}\) : \(\mu < 120\)

From the z table the critical value at \(\alpha\) = -1.645. A negative sign is used as this is a left tailed test.

\(\overline{x}\) = 100, \(\mu\) = 120, n = 49, \(\sigma\) = 30

As -4.66 < -1.645 thus, the null hypothesis is rejected and it is concluded that there is enough evidence to support the medicine shop's claim.

Example 3: A company wants to improve the quality of products by reducing defects and monitoring the efficiency of assembly lines. In assembly line A, there were 18 defects reported out of 200 samples while in line B, 25 defects out of 600 samples were noted. Is there a difference in the procedures at a 0.05 alpha level?

Solution: This is an example of a two-tailed two proportion z test.

\(H_{0}\): The two proportions are the same.

\(H_{1}\): The two proportions are not the same.

As this is a two-tailed test the alpha level needs to be divided by 2 to get 0.025.

Using this, the critical value from the z table is 1.96.

\(n_{1}\) = 200, \(n_{2}\) = 600

\(p_{1}\) = 18 / 200 = 0.09

\(p_{2}\) = 25 / 600 = 0.0416

p = (18 + 25) / (200 + 600) = 0.0537

z =\(\frac{p_{1}-p_{2}-0}{\sqrt{p(1-p)\left ( \frac{1}{n_{1}} +\frac{1}{n_{2}}\right )}}\) = 2.62

As 2.62 > 1.96 thus, the null hypothesis is rejected and it is concluded that there is a significant difference between the two lines.

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FAQs on Z Test

What is a z test in statistics.

A z test in statistics is conducted on data that is normally distributed to test if the means of two datasets are equal. It can be performed when the sample size is greater than 30 and the population variance is known.

What is a One-Sample Z Test?

A one-sample z test is used when the population standard deviation is known, to compare the sample mean and the population mean. The z test statistic is given by the formula \(\frac{\overline{x}-\mu}{\frac{\sigma}{\sqrt{n}}}\).

What is the Two-Sample Z Test Formula?

The two sample z test is used when the means of two populations have to be compared. The z test formula is given as \(\frac{(\overline{x_{1}}-\overline{x_{2}})-(\mu_{1}-\mu_{2})}{\sqrt{\frac{\sigma_{1}^{2}}{n_{1}}+\frac{\sigma_{2}^{2}}{n_{2}}}}\).

What is a One Proportion Z test?

A one proportion z test is used to check if the value of the observed proportion is different from the value of the theoretical proportion. The z statistic is given by \(\frac{p-p_{0}}{\sqrt{\frac{p_{0}(1-p_{0})}{n}}}\).

What is a Two Proportion Z Test?

When the proportions of two samples have to be compared then the two proportion z test is used. The formula is given by \(\frac{p_{1}-p_{2}-0}{\sqrt{p(1-p)\left ( \frac{1}{n_{1}} +\frac{1}{n_{2}}\right )}}\).

How Do You Find the Z Test?

The steps to perform the z test are as follows:

• Set up the null and alternative hypotheses.
• Find the critical value using the alpha level and z table.
• Calculate the z statistic.
• Compare the critical value and the test statistic to decide whether to reject or not to reject the null hypothesis.

What is the Difference Between the Z Test and the T-Test?

A z test is used on large samples n ≥ 30 and normally distributed data while a t-test is used on small samples (n < 30) following a student t distribution . Both tests are used to check if the means of two datasets are the same.

Introduction to Statistics and Data Analysis

6 hypothesis testing: the z-test.

We’ve all had the experience of standing at a crosswalk waiting staring at a pedestrian traffic light showing the little red man. You’re waiting for the little green man so you can cross. After a little while you’re still waiting and there aren’t any cars around. You might think ‘this light is really taking a long time’, but you continue waiting. Minutes pass and there’s still no little green man. At some point you come to the conclusion that the light is broken and you’ll never see that little green man. You cross on the little red man when it’s clear.

You may not have known this but you just conducted a hypothesis test. When you arrived at the crosswalk, you assumed that the light was functioning properly, although you will always entertain the possibility that it’s broken. In terms of hypothesis testing, your ‘null hypothesis’ is that the light is working and your ‘alternative hypothesis’ is that it’s broken. As time passes, it seems less and less likely that light is working properly. Eventually, the probability of the light working given how long you’ve been waiting becomes so low that you reject the null hypothesis in favor of the alternative hypothesis.

This sort of reasoning is the backbone of hypothesis testing and inferential statistics. It’s also the point in the course where we turn the corner from descriptive statistics to inferential statistics. Rather than describing our data in terms of means and plots, we will now start using our data to make inferences, or generalizations, about the population that our samples are drawn from. In this course we’ll focus on standard hypothesis testing where we set up a null hypothesis and determine the probability of our observed data under the assumption that the null hypothesis is true (the much maligned p-value). If this probability is small enough, then we conclude that our data suggests that the null hypothesis is false, so we reject it.

In this chapter, we’ll introduce hypothesis testing with examples from a ‘z-test’, when we’re comparing a single mean to what we’d expect from a population with known mean and standard deviation. In this case, we can convert our observed mean into a z-score for the standard normal distribution. Hence the name z-test.

It’s time to introduce the hypothesis test flow chart . It’s pretty self explanatory, even if you’re not familiar with all of these hypothesis tests. The z-test is (1) based on means, (2) with only one mean, and (3) where we know \(\sigma\) , the standard deviation of the population. Here’s how to find the z-test in the flow chart:

6.1 Women’s height example

Let’s work with the example from the end of the last chapter where we started with the fact that the heights of US women has a mean of 63 and a standard deviation of 2.5 inches. We calculated that the average height of the 122 women in Psych 315 is 64.7 inches. We then used the central limit theorem and calculated the probability of a random sample 122 heights from this population having a mean of 64.7 or greater is 2.4868996^{-14}. This is a very, very small number.

Here’s how we do it using R:

Let’s think of our sample as a random sample of UW psychology students, which is a reasonable assumption since all psychology students have to take a statistics class. What does this sample say about the psychology students that are women at UW compared to the US population? It could be that these psychology students at UW have the same mean and standard deviation as the US population, but our sample just happens to have an unusual number of tall women, but we calculated that the probability of this happening is really low. Instead, it makes more sense that the population that we’re drawing from has a mean that’s greater than the US population mean. Notice that we’re making a conclusion about the whole population of women psychology students based on our one sample.

Using the terminology of hypothesis testing, we first assumed the null hypothesis that UW women psych students have the same mean (and standard deviation) as the US population. The null hypothesis is written as:

\[ H_{0}: \mu = 63 \] In this example, our alternative hypothesis is that the mean of our population is larger than the mean of null hypothesis population. We write this as:

\[ H_{A}: \mu > 63 \]

Next, after obtaining a random sample and calculate the mean, we calculate the probability of drawing a mean this large (or larger) from the null hypothesis distribution.

If this probability is low enough, we reject the null hypothesis in favor of the alternative hypothesis. When our probability allows us to reject the null hypothesis, we say that our observed results are ‘statistically significant’.

In statistics terms, we never say we ‘accept that alternative hypothesis’ as true. All we can say is that we don’t think the null hypothesis is true. I know it’s subtle, but in science can never prove that a hypothesis is true or not. There’s always the possibility that we just happened to grab an unusual sample from the null hypothesis distribution.

6.2 The hated p<.05

The probability that we obtain our observed mean or greater given that the null hypothesis is true is called the p-value. How improbable is improbable enough to reject the null hypothesis? The p-value for our example above on women’s heights is astronomically low, so it’s clear that we should reject \(H_{0}\) .

The p-value that’s on the border of rejection is called the alpha ( \(\alpha\) ) value. We reject \(H_{0}\) when our p-value is less than \(\alpha\) .

You probably know that the most common value of alpha is \(\alpha = .05\) .

The first publication of this value dates back to Sir Ronald Fisher, in his seminal 1925 book Statistical Methods for Research Workers where he states:

“It is convenient to take this point as a limit in judging whether a deviation is considered significant or not. Deviations exceeding twice the standard deviation are thus formally regarded as significant.” (p. 47)

If you read the chapter on the normal distribution, then you should know that 95% of the area under the normal distribution lies within \(\pm\) two standard deviations of the mean. So the probability of obtaining a sample that exceeds two standard deviations from the mean (in either direction) is .05.

6.3 IQ example

Let’s do an example using IQ scores. IQ scores are normalized to have a mean of 100 and a standard deviation of 15 points. Because they’re normalized, they are a rare example of a population which has a known mean and standard deviation. In the next chapter we’ll discuss the t-test, which is used in the more common situation when we don’t know the population standard deviation.

Suppose you have the suspicion that graduate students have higher IQ’s than the general population. You have enough time to go and measure the IQ’s of 25 randomly sampled grad students and obtain a mean of 105. Is this difference between our this observed mean and 100 statistically significant using an alpha value of \(\alpha = 0.05\) ?

Here the null hypothesis is:

\[ H_{0}: \mu = 100\]

And the alternative hypothesis is:

\[ H_{A}: \mu > 100 \]

We know that the parameters for the null hypothesis are:

\[ \mu = 100 \] and \[ \sigma = 15 \]

From this, we can calculate the probability of observing our mean of 105 or higher using the central limit theorem and what we know about the normal distribution:

\[ \sigma_{\bar{x}} = \frac{\sigma_{x}}{\sqrt{n}} = \frac{15}{\sqrt{25}} = 3 \] From this, we can calculate the probability of our observed mean using R’s ‘pnorm’ function. Here’s how to do the whole thing in R.

Since our p-value of 0.0478 is (just barely) less than our chosen value of \(\alpha = 0.05\) as our criterion, we reject \(H_{0}\) for this (contrived) example and conclude that our observed mean of 105 is significantly greater than 100, so our study suggests that the average graduate student has a higher IQ than the overall population.

You should feel uncomfortable making such a hard, binary decision for such a borderline case. After all, if we had chosen our second favorite value of alpha, \(\alpha = .01\) , we would have failed to reject \(H_{0}\) . This discomfort is a primary reason why statisticians are moving away from this discrete decision making process. Later on we’ll discuss where things are going, including reporting effect sizes, and using confidence intervals.

6.4 Alpha values vs. critical values

Using R’s qnorm function, we can find the z-score for which only 5% of the area lies above:

So the probability of a randomly sampled z-score exceeding 1.644854 is less than 5%. It follows that if we convert our observed mean into z-score values, we will reject \(H_{0}\) if and only if our z-score is greater than 1.644854. This value is called the ‘critical value’ because it lies on the boundary between rejecting and failing to reject \(H_{0}\) .

In our last example, the z-score for our observed mean is:

\[ z = \frac{X-\mu}{\frac{\sigma}{\sqrt{n}}} = \frac{105 - 100}{3} = 1.67 \] Our z-score is just barely greater than the critical value of 1.644854, which makes sense because our p-value is just barely less than 0.05.

Sometimes you’ll see textbooks will compare critical values to observed scores for the decision making process in hypothesis testing. This dates back to days were computers were less available and we had to rely on tables instead. There wasn’t enough space in a book to hold complete tables which prohibited the ability to look up a p-value for any observed value. Instead only critical values for specific values of alpha were included. If you look at really old papers, you’ll see statistics reported as \(p<.05\) or \(p<.01\) instead of actual p-values for this reason.

It may help to visualize the relationship between p-values, alpha values and critical values like this:

The red shaded region is the upper 5% of the standard normal distribution which starts at the critical value of z=1.644854. This is sometimes called the ‘rejection region’. The blue vertical line is drawn at our observed value of z=1.67. You can see that the red line falls just inside the rejection region, so we Reject \(H_{0}\) !

6.5 One vs. two-tailed tests

Recall that our alternative hypothesis was to reject if our mean IQ was significantly greater than the null hypothesis mean: \(H_{A}: \mu > 100\) . This implies that the situation where \(\mu < 100\) is never even in consideration, which is weird. In science, we’re trying to understand the true state of the world. Although we have a hunch that grad student IQ’s are higher than average, there is always the possibility that they are lower than average. If our sample came up with an IQ well below 100, we’d simply fail to reject \(H_{0}\) and move on. This feels like throwing out important information.

The test we just ran is called a ‘one-tailed’ test because we only reject \(H_{0}\) if our results fall in one of the two tails of the population distribution.

Instead, it might make more sense to reject \(H_{0}\) if we get either an unusually large or small score. This means we need two critical values - one above and one below zero. At first thought you might think we just duplicate our critical value from a one-tailed test to the other side. But will double the area of the rejection region. That’s not a good thing because if \(H_{0}\) is true, there’s actually a \(2\alpha\) probability that we’ll draw a score in the rejection region.

Instead, we divide the area into two tails, each containing an area of \(\frac{\alpha}{2}\) . For \(\alpha\) = 0.05, we can find the critical value of z with qnorm:

So with a two-tailed test at \(\alpha = 0.05\) we reject \(H_{0}\) if our observed z-score is either above z = 1.96 or less than -1.96. This is that value around 2 that Sir Ronald Fischer was talking about!

Here’s what the critical regions and observed value of z looks like for our example with a two-tailed test:

You can see that splitting the area of \(\alpha = 0.05\) into two halves increased the critical value in the positive direction from 1.64 to 1.96, making it harder to reject \(H_{0}\) . For our example, this changes our decision: our observed value of z = 1.67 no longer falls into the rejection region, so now we fail to reject \(H_{0}\) .

If we now fail to reject \(H_{0}\) , what about the p-value? Remember, for a one-tailed test, p = \(\alpha\) if our observed z-score lands right on the critical value of z. The same is true for a two-tailed test. But the z-score moved so that the area above that score is \(\frac{\alpha}{2}\) . So for a two-tailed test, in order to have a p-value of \(\alpha\) when our z-score lands right on the critical value, we need to double p-value hat we’d get for a one-tailed test.

For our example, the p-value for the one tailed test was \(p=0.0478\) . So if we use a two-tailed test, our p-value is \((2)(0.0478) = 0.0956\) . This value is greater than \(\alpha\) = 0.05, which makes sense because we just showed above that we fail to reject \(H_{0}\) with a two tailed test.

Which is the right test, one-tailed or two-tailed? Ideally, as scientists, we should be agnostic about the results of our experiment. But in reality, we all know that the results are more interesting if they are statistically significant. So you can imagine that for this example, given a choice between one and two-tailed, we’d choose a one-tailed test so that we can reject \(H_{0}\) .

There are two problems with this. First, we should never adjust our choice of hypothesis test after we observe the data. That would be an example of ‘p-hacking’, a topic we’ll discuss later. Second, most statisticians these days strongly recommend against one-tailed tests. The only reason for a one-tailed test is if there is no logical or physical possibility for a population mean to fall below the null hypothesis mean.

A Z-test is a type of statistical hypothesis test used to test the mean of a normally distributed test statistic. It tests whether there is a significant difference between an observed population mean and the population mean under the null hypothesis, H 0 .

A Z-test can only be used when the population variance is known (or can be estimated with a high degree of accuracy), or if the sample size of the experiment is large (typically n>30). Also, the test statistic must exhibit a normal distribution; if it exhibits a distribution that is clearly not normal, the Z-test is not applicable. In many cases, population parameters may not be known, or it may not be possible to estimate them accurately. In such cases, or in cases where the sample size is small, a Student's t-test is more appropriate.

How to conduct a Z-test

The procedure for conducting a Z-test is similar to that of other statistical hypothesis tests, and is generally as follows:

• State the null (H 0 ) and alternative hypotheses (H a ).
• Select a significance level, α.
• Calculate the Z-score.
• Determine the critical value(s) of Z or the p-value.
• Compare the Z-score of the observed value to the critical value of Z (or compare the p-value to α) to determine if the null hypothesis should be rejected in favor of the alternative hypothesis, or if the null hypothesis should not be rejected.

H 0 and H a

The null hypothesis is typically a statement of no difference. For example, assume that the average score received on the SAT by high schoolers in a given state was a 1200 with a known standard deviation. If the average score of students in a given high school is a 1230, we may use a Z-test to determine whether this result is better, statistically, than the state average. The null hypothesis in this case would be that the average score of students in the high school is not better than the state average, or H 0 : μ ≤ μ 0 , or μ ≤ 1200.

The alternative hypothesis is a statement of difference from the null hypothesis. It can take one of three forms:

• Given H 0 : μ ≤ μ 0 , H a : μ > μ 0
• Given H 0 : μ ≥ μ 0 , H a : μ 0
• Given H 0 : μ = μ 0 , H a : μ ≠ μ 0

In this example, it is believed that a score of 1230 is statistically significant, and that students in this high school performed better than the state average. Therefore, the alternative hypothesis takes on the first form in the list, H a : μ > μ 0 , or μ > 1200.

Significance level

The significance level, α, is the probability of a study rejecting the null hypothesis when the null hypothesis is true. Commonly used significance levels include 0.01, 0.05, and 0.10. A significance level of 0.05, or 5%, means that there is a 5% chance of concluding that a difference exists (thus rejecting H 0 ) when there is no actual difference. The lower the significance level, the more evidence required before the null hypothesis can be rejected. The significance level is compared to the p-value: if a p-value is less than the significance level, the null hypothesis is rejected in favor of the alternative hypothesis.

Calculating a Z-score is a necessary part of conducting a Z-test. A Z-score indicates the number of standard deviations that an observed value is from the mean in a standard normal distribution. For example, an observed value with a Z-score of 1.2 indicates that the observed value is 1.2 standard deviations from the mean. If the population mean and standard deviation are known, the Z-score is calculated using the following formula:

where μ is the mean of the population, σ is the standard deviation of the population, and x is the observed value. In many cases the population mean and standard deviation are not known. In such cases, these population parameters can be estimated using a sample mean and sample standard deviation, and the Z-score can be computed as follows:

where x is the sample mean, s is the sample standard deviation, and x is the observed value.

Critical value and p-value

Once a Z-score has been calculated, there are two methods for drawing conclusions about the test statistic: using the critical value(s), or using a p-value. To form a conclusion for a hypothesis test using a critical value, the Z-score of the observed value is compared to the critical value(s) of the selected significance level; to use a p-value, the p-value of the observed value is compared to the significance level.

Critical value

A critical value is a value that indicates the critical region(s) (or rejection region) of the standard normal distribution, where a critical region is the area of the distribution in which a value must lie in order to reject the null hypothesis.

The critical value is dependent on the significance level as well as whether a one-tailed or two-tailed test is being conducted. A one-tailed test is used when we want to know if a value is significantly larger or smaller than the Z-score. There is only one critical region in a one-tailed Z-test. It is either a left-tailed test (or lower-tailed) or right-tailed test (or upper-tailed) based on the position of the critical region, as shown in the figure below.

The critical regions are shown in pink. If a test statistic lies within the pink region, the null hypothesis is rejected in favor of the alternative hypothesis. Otherwise, the null hypothesis is not rejected.

If a test value lies in either of the critical regions shown in pink, the null hypothesis is rejected in favor of the alternative hypothesis; if it lies within the green region, the null hypothesis is not rejected.

After selecting the significance level and type of test, the critical Z value can be determined using a Z table by finding the Z value that corresponds to the selected significance level. For example, for a one-tailed test and a significance level of 0.05, find the probability closest to 0.05 and read the Z value that results in this probability; the Z value for α = 0.05 for a one-tailed Z-test is -1.96 for a left-tailed Z-test and 1.96 for a right-tailed Z-test. For a two-tailed Z-test, divide α by 2, then determine the corresponding Z-value. For α = 0.05, each tail will comprise an area of 0.025 in the standard normal distribution, which corresponds to Z-values of -1.645 and 1.645. Thus, the critical regions are Z 1.645. The critical values for common significance levels are shown in the table below:

The p-value indicates the probability of obtaining test results that are at least as extreme as the observed results, assuming that the null hypothesis is true. It tells us how likely it is for an outcome to occur solely based on chance. For example, a p-value of 0.05 means that there is a 5% chance that an outcome occurred solely by chance. The smaller the p-value, the less likely it is for an outcome to occur solely by chance, and the more evidence there is to reject the null hypothesis.

Like critical values, a p-value can be determined using a Z table. For a left-tailed Z-test, the p-value is the area under the standard normal distribution to the left of the Z-score of the observed value; for a right-tailed Z-test, it is the area to the right of the Z-score; for a two-tailed Z-test, it is the sum of the area to the left and right of the Z-score. If the p-value is less than or equal to the significance level, the null hypothesis is rejected in favor of the alternative hypothesis. Otherwise, the null hypothesis is not rejected.

It is important to note that the p-value is not the probability that the null hypothesis is true. It is the probability that the data could deviate from the null hypothesis as much, or more than it did. The calculation of the p-value assumes that the null hypothesis is true, so it is not a measure of whether or not the null hypothesis is correct. Rather, it is a measure of how well the data fits the null hypothesis. Also, the p-value (or critical value) may provide evidence that the null hypothesis should be rejected in favor of the alternative hypothesis at the chosen level of significance . This does not mean that the alternative hypothesis is being accepted, because it is possible that the null hypothesis would not be rejected at a different significance level. Similarly, if the p-value is greater than the significance level, this does not mean that the null hypothesis is being accepted, just that the null hypothesis is not rejected.

Finally, p-values and critical values only indicate statistical significance, and may not necessarily indicate that the study's findings are significant within their context. For example, if a new medicine and a placebo are tested on different populations, and the medicine is found to have a statistically significant effect, it may not necessarily mean that there is clinical significance. It is possible for a finding to be both statistically and clinically significant, or only one or the other. For large sample sizes, it is possible for results to indicate statistical significance even when the effect is actually small and unimportant. Conversely, a small sample may not exhibit statistical significance even when the effect is large and potentially important. Thus, it is important to fully understand the scope of a study, as well as the statistical methods used, in order to effectively interpret the results and draw accurate, unbiased conclusions.

The average score on a national mathematics exam taken by high school seniors is an 82 with a standard deviation of 8. A sample of 1000 seniors achieved an average score of 68. Perform a Z-test to determine whether there is a statistically significant difference between the national average and that of the sample of seniors at a significance level of 0.05.

We want to determine whether there is any difference, so the null hypothesis is that there is no difference, or

H 0 : μ = 82

and the alternative hypothesis is:

H a : μ ≠ 82

Thus, a two-tailed Z-test should be conducted since differences on either side of the distribution must be accounted for.

The selected significance level is:

α = 0.05

This value must be greater than the p-value in order to conclude that the difference in scores is statistically significant.

Since the population standard deviation and mean are known, the Z-score can be computed as:

Based on the selected significance level and the use of a two-tailed Z-test, the critical values are Z = ± 1.96. Since the Z-score of the observed value lies between both tails (rather than within one of them), we fail to reject the null hypothesis, as depicted in the figure below.

Thus, we conclude that the difference between the observed mean and the population mean is not statistically significant for a significance level of 0.05.

However, had we selected a significance level of 0.10, the critical values would be Z = ±1.645, and Z = -1.75 would lie within the left tail of the distribution. In this case, we would reject the null hypothesis in favor of the alternative hypothesis, and conclude that the observed value is statistically significant for a significance level of 0.10.

The above discussion involved hypothesis testing for one sample, where an observed value was compared to the expected population parameter. In certain cases, scientists may want to compare the means of two samples. In such cases, a two-sample Z-test is used instead.

Two-sample Z-test

A two-sample Z-test is conducted using the same procedures described above for a one-sample Z-test, with the exception that the Z-score is computed using the following formula:

where μ 1 and μ 2 are the means of the two respective populations, x 1 and x 2 are the sample means, and n 1 and n 2 are the sample sizes.

Researchers want to test whether a certain drug has any effect on the scores received by patients who are administered the drug prior to performing a physical stress test. The researchers place patients into 2 groups: 500 are placed into the experimental group and are administered the drug; 300 are placed into the control group and are administered a placebo. Both groups then perform the physical stress test, the results of which are as follows:

Determine whether or not there is a statistically significant difference between the two groups at a significance level of 0.05.

The null hypothesis is that there is no difference, so:

H 0 : μ 1 = μ 2

Also, since it is assumed that the null hypothesis is true, μ 1 - μ 2 = 0.

The alternative hypothesis is that there is a difference, so:

H a : μ 1 ≠ μ 2

The selected significance level is 0.05, and we conduct a two-tailed test since we are looking for any observable difference.

The Z-score is then calculated as follows:

Using a Z table (or a p-value calculator), the p-value for a two-tailed Z-test for a Z-score of 2.604 is 0.009214. Since the p-value is less than the selected significance level, we reject the null hypothesis in favor of the alternative hypothesis, and conclude that the drug has a statistically significant effect on the performance of the patients. Since the Z-score lies in the right tail, we may conclude that patients who received the drug scored significantly better than those who received the placebo. If the Z-score were to lie in left tail, we would conclude the opposite: that patients who received the drug performed significantly worse.

We could also have used the critical values Z = ±1.96 for a significance level of 0.05 to reach the same conclusion, since 2.604 lies within the critical region denoted by the right tail of the distribution, as shown in the figure below.

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13.1: The one-sample z-test

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• Page ID 4021

• Danielle Navarro
• University of New South Wales

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In this section I’ll describe one of the most useless tests in all of statistics: the z-test . Seriously – this test is almost never used in real life. Its only real purpose is that, when teaching statistics, it’s a very convenient stepping stone along the way towards the t-test, which is probably the most (over)used tool in all statistics.

inference problem that the test addresses

To introduce the idea behind the z-test, let’s use a simple example. A friend of mine, Dr Zeppo, grades his introductory statistics class on a curve. Let’s suppose that the average grade in his class is 67.5, and the standard deviation is 9.5. Of his many hundreds of students, it turns out that 20 of them also take psychology classes. Out of curiosity, I find myself wondering: do the psychology students tend to get the same grades as everyone else (i.e., mean 67.5) or do they tend to score higher or lower? He emails me the zeppo.Rdata file, which I use to pull up the grades of those students,

and calculate the mean:

Hm. It might be that the psychology students are scoring a bit higher than normal: that sample mean of \(\bar{X}\) = 72.3 is a fair bit higher than the hypothesised population mean of μ=67.5, but on the other hand, a sample size of N=20 isn’t all that big. Maybe it’s pure chance.

To answer the question, it helps to be able to write down what it is that I think I know. Firstly, I know that the sample mean is \(\bar{X}\) =72.3. If I’m willing to assume that the psychology students have the same standard deviation as the rest of the class then I can say that the population standard deviation is σ=9.5. I’ll also assume that since Dr Zeppo is grading to a curve, the psychology student grades are normally distributed.

Next, it helps to be clear about what I want to learn from the data. In this case, my research hypothesis relates to the population mean μ for the psychology student grades, which is unknown. Specifically, I want to know if μ=67.5 or not. Given that this is what I know, can we devise a hypothesis test to solve our problem? The data, along with the hypothesised distribution from which they are thought to arise, are shown in Figure 13.1. Not entirely obvious what the right answer is, is it? For this, we are going to need some statistics.

Constructing the hypothesis test

The first step in constructing a hypothesis test is to be clear about what the null and alternative hypotheses are. This isn’t too hard to do. Our null hypothesis, H 0 , is that the true population mean μ for psychology student grades is 67.5%; and our alternative hypothesis is that the population mean isn’t 67.5%. If we write this in mathematical notation, these hypotheses become,

H 0 :μ=67.5

H 1 :μ≠67.5

though to be honest this notation doesn’t add much to our understanding of the problem, it’s just a compact way of writing down what we’re trying to learn from the data. The null hypotheses H 0 and the alternative hypothesis H 1 for our test are both illustrated in Figure 13.2. In addition to providing us with these hypotheses, the scenario outlined above provides us with a fair amount of background knowledge that might be useful. Specifically, there are two special pieces of information that we can add:

1 The psychology grades are normally distributed. 1 The true standard deviation of these scores σ is known to be 9.5.

For the moment, we’ll act as if these are absolutely trustworthy facts. In real life, this kind of absolutely trustworthy background knowledge doesn’t exist, and so if we want to rely on these facts we’ll just have make the assumption that these things are true. However, since these assumptions may or may not be warranted, we might need to check them. For now though, we’ll keep things simple.

The next step is to figure out what we would be a good choice for a diagnostic test statistic; something that would help us discriminate between H 0 and H 1 . Given that the hypotheses all refer to the population mean μ, you’d feel pretty confident that the sample mean \(\bar{X}\) would be a pretty useful place to start. What we could do, is look at the difference between the sample mean \(\bar{X}\) and the value that the null hypothesis predicts for the population mean. In our example, that would mean we calculate \(\bar{X}\) - 67.5. More generally, if we let μ 0 refer to the value that the null hypothesis claims is our population mean, then we’d want to calculate

\(\bar{X}-\mu_{0}\)

If this quantity equals or is very close to 0, things are looking good for the null hypothesis. If this quantity is a long way away from 0, then it’s looking less likely that the null hypothesis is worth retaining. But how far away from zero should it be for us to reject H 0 ?

To figure that out, we need to be a bit more sneaky, and we’ll need to rely on those two pieces of background knowledge that I wrote down previously, namely that the raw data are normally distributed, and we know the value of the population standard deviation σ. If the null hypothesis is actually true, and the true mean is μ 0 , then these facts together mean that we know the complete population distribution of the data: a normal distribution with mean μ 0 and standard deviation σ. Adopting the notation from Section 9.5, a statistician might write this as:

X∼Normal(μ 0 ,σ 2 )

Okay, if that’s true, then what can we say about the distribution of \(\bar{X}\)? Well, as we discussed earlier (see Section 10.3.3), the sampling distribution of the mean \(\bar{X}\) is also normal, and has mean μ. But the standard deviation of this sampling distribution SE (\(\bar{X}\)), which is called the standard error of the mean , is

\(\operatorname{SE}(\bar{X})=\dfrac{\sigma}{\sqrt{N}}\)

In other words, if the null hypothesis is true then the sampling distribution of the mean can be written as follows:

\(\bar{X}\)∼Normal(μ 0 ,SE(\(\bar{X}\)))

Now comes the trick. What we can do is convert the sample mean \(\bar{X}\) into a standard score (Section 5.6). This is conventionally written as z, but for now I’m going to refer to it as \(z_{\bar{X}}\). (The reason for using this expanded notation is to help you remember that we’re calculating standardised version of a sample mean, not a standardised version of a single observation, which is what a z-score usually refers to). When we do so, the z-score for our sample mean is

\(\ z_{\bar{X}} = {{\bar{X}-\mu_{0}} \over SE(\bar{X})}\)

or, equivalently

\(\ z_{\bar{X}} = {{\bar{X}-\mu_{0}} \over \sigma/ \sqrt{N} }\)

This z-score is our test statistic. The nice thing about using this as our test statistic is that like all z-scores, it has a standard normal distribution:

\(\ z_{\bar{X}}\)∼Normal(0,1)

(again, see Section 5.6 if you’ve forgotten why this is true). In other words, regardless of what scale the original data are on, the z-statistic iteself always has the same interpretation: it’s equal to the number of standard errors that separate the observed sample mean\(\bar{X}\) from the population mean μ 0 predicted by the null hypothesis. Better yet, regardless of what the population parameters for the raw scores actually are, the 5% critical regions for z-test are always the same, as illustrated in Figures 13.4 and 13.3. And what this meant, way back in the days where people did all their statistics by hand, is that someone could publish a table like this:

which in turn meant that researchers could calculate their z-statistic by hand, and then look up the critical value in a text book. That was an incredibly handy thing to be able to do back then, but it’s kind of unnecessary these days, since it’s trivially easy to do it with software like R.

worked example using R

Now, as I mentioned earlier, the z-test is almost never used in practice. It’s so rarely used in real life that the basic installation of R doesn’t have a built in function for it. However, the test is so incredibly simple that it’s really easy to do one manually. Let’s go back to the data from Dr Zeppo’s class. Having loaded the grades data, the first thing I need to do is calculate the sample mean:

Then, I create variables corresponding to known population standard deviation (σ=9.5), and the value of the population mean that the null hypothesis specifies (μ 0 =67.5):

Let’s also create a variable for the sample size. We could count up the number of observations ourselves, and type N <- 20 at the command prompt, but counting is tedious and repetitive. Let’s get R to do the tedious repetitive bit by using the length() function, which tells us how many elements there are in a vector:

Next, let’s calculate the (true) standard error of the mean:

And finally, we calculate our z-score:

At this point, we would traditionally look up the value 2.26 in our table of critical values. Our original hypothesis was two-sided (we didn’t really have any theory about whether psych students would be better or worse at statistics than other students) so our hypothesis test is two-sided (or two-tailed) also. Looking at the little table that I showed earlier, we can see that 2.26 is bigger than the critical value of 1.96 that would be required to be significant at α=.05, but smaller than the value of 2.58 that would be required to be significant at a level of α=.01. Therefore, we can conclude that we have a significant effect, which we might write up by saying something like this:

With a mean grade of 73.2 in the sample of psychology students, and assuming a true population standard deviation of 9.5, we can conclude that the psychology students have significantly different statistics scores to the class average (z=2.26, N=20, p<.05).

However, what if want an exact p-value? Well, back in the day, the tables of critical values were huge, and so you could look up your actual z-value, and find the smallest value of α for which your data would be significant (which, as discussed earlier, is the very definition of a p-value). However, looking things up in books is tedious, and typing things into computers is awesome. So let’s do it using R instead. Now, notice that the α level of a z-test (or any other test, for that matter) defines the total area “under the curve” for the critical region, right? That is, if we set α=.05 for a two-sided test, then the critical region is set up such that the area under the curve for the critical region is .05. And, for the z-test, the critical value of 1.96 is chosen that way because the area in the lower tail (i.e., below −1.96) is exactly .025 and the area under the upper tail (i.e., above 1.96) is exactly .025. So, since our observed z-statistic is 2.26, why not calculate the area under the curve below −2.26 or above 2.26? In R we can calculate this using the pnorm() function. For the upper tail:

The lower.tail = FALSE is me telling R to calculate the area under the curve from 2.26 and upwards . If I’d told it that lower.tail = TRUE , then R would calculate the area from 2.26 and below , and it would give me an answer 0.9880771. Alternatively, to calculate the area from −2.26 and below, we get

Thus we get our p-value:

Assumptions of the z-test

As I’ve said before, all statistical tests make assumptions. Some tests make reasonable assumptions, while other tests do not. The test I’ve just described – the one sample z-test – makes three basic assumptions. These are:

• Normality . As usually described, the z-test assumes that the true population distribution is normal. 186 is often pretty reasonable, and not only that, it’s an assumption that we can check if we feel worried about it (see Section 13.9).
• Independence . The second assumption of the test is that the observations in your data set are not correlated with each other, or related to each other in some funny way. This isn’t as easy to check statistically: it relies a bit on good experimetal design. An obvious (and stupid) example of something that violates this assumption is a data set where you “copy” the same observation over and over again in your data file: so you end up with a massive “sample size”, consisting of only one genuine observation. More realistically, you have to ask yourself if it’s really plausible to imagine that each observation is a completely random sample from the population that you’re interested in. In practice, this assumption is never met; but we try our best to design studies that minimise the problems of correlated data.
• Known standard deviation . The third assumption of the z-test is that the true standard deviation of the population is known to the researcher. This is just stupid. In no real world data analysis problem do you know the standard deviation σ of some population, but are completely ignorant about the mean μ. In other words, this assumption is always wrong.

In view of the stupidity of assuming that σ is known, let’s see if we can live without it. This takes us out of the dreary domain of the z-test, and into the magical kingdom of the t-test, with unicorns and fairies and leprechauns, and um…

Statistics Made Easy

T-Score vs. Z-Score: When to Use Each

Two terms that often confuse students in statistics classes are t-scores and z-scores .

Both are used extensively when performing hypothesis tests or constructing confidence intervals , but they’re slightly different.

Here’s the formula for each:

• t-score = ( x – μ) / (s/√ n )
• x : Sample mean
• μ : Population mean
• s : Sample standard deviation
• n : Sample size
• z-score = ( x – μ) / σ
• x : Raw data value
• σ : Population standard deviation

This flow chart shows when you should use each, depending on your data:

The following examples show how to calculate a t-score and z-score in practice.

Example 1: Calculating a T-Score

Suppose a restaurant makes burgers that claim to have a mean weight of μ = 0.25 pounds.

Suppose we take a random sample of n = 20 burgers and find that the sample mean weight is x = 0.22 pounds with a standard deviation of s = 0.05 pounds. Perform a hypothesis test to determine if the true mean weight of all burgers produced by this restaurant is equal to 0.25 pounds.

For this example, we would use a t-score to perform the hypothesis test because neither of the following two conditions are met.

• The population standard deviation (σ) is known. (σ is not provided in this example)
• The sample size is greater than 30. (n = 20 in this example)

Thus, we would calculate the t-score as:

• t-score = (.22 – .25) / (.05 / √ 20 )
• t- score = -2.68

According to the T Score to P Value Calculator , the p-value that corresponds to this t-score is 0.01481 .

Since this p-value is less than .05, we have sufficient evidence to say that the mean weight of burgers produced at this restaurant is not equal to 0.25 pounds.

Example 2: Calculating a Z-Score

Suppose a company manufactures batteries that are known to have a lifespan that follows a normal distribution with a mean of μ = 20 hours and a standard deviation of σ = 5 hours.

Suppose we take a random sample of n = 50 batteries and find that the sample mean is x = 21 hours. Perform a hypothesis test to determine if the true mean lifespan of all batteries manufactured by this company is equal to 20 hours.

For this example, we would use a z-score to perform the hypothesis test because the following two conditions are met:

• The population standard deviation (σ) is known. (σ is equal to 5 in this example)
• The sample size is greater than 30. (n = 50 in this example)

Thus, we would calculate the z-score as:

• z-score = (21 – 20) / 5
• z- score = 0.2

According to the Z Score to P Value Calculator , the p-value that corresponds to this z-score is 0.84184 .

Since this p-value is not less than .05, we don’t have sufficient evidence to say that the mean lifespan of all batteries manufactured by this company is different than 20 hours.

Additional Resources

The following tutorials offer more information on both t-scores and z-scores:

Normal Distribution vs. t-Distribution: What’s the Difference? How to Read the t-Distribution Table How to Read the Z Table

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Hey there. My name is Zach Bobbitt. I have a Masters 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.

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Z-tests for Hypothesis testing: Formula & Examples

Z-tests are statistical hypothesis testing techniques that are used to determine whether the null hypothesis relating to comparing sample means or proportions with that of population at a given significance level can be rejected or otherwise based on the z-statistics or z-score. As a data scientist , you must get a good understanding of the z-tests and its applications to test the hypothesis for your statistical models. In this blog post, we will discuss an overview of different types of z-tests and related concepts with the help of examples. You may want to check my post on hypothesis testing titled – Hypothesis testing explained with examples

Table of Contents

What are Z-tests & Z-statistics?

Z-tests can be defined as statistical hypothesis testing techniques that are used to quantify the hypothesis testing related to claim made about the population parameters such as mean and proportion. Z-test uses the sample data to test the hypothesis about the population parameters (mean or proportion). There are different types of Z-tests which are used to estimate the population mean or proportion, or, perform hypotheses testing related to samples’ means or proportions.

Different types of Z-tests 

There are following different types of Z-tests which are used to perform different types of hypothesis testing.  

• One-sample Z-test for means
• Two-sample Z-test for means
• One sample Z-test for proportion
• Two sample Z-test for proportions

Four variables are involved in the Z-test for performing hypothesis testing for different scenarios. They are as follows:

• An independent variable that is called the “sample” and assumed to be normally distributed;
• A dependent variable that is known as the test statistic (Z) and calculated based on sample data
• Different types of Z-test that can be used for performing hypothesis testing
• A significance level or “alpha” is usually set at 0.05 but can take the values such as 0.01, 0.05, 0.1

When to use Z-test – Explained with examples

The following are different scenarios when Z-test can be used:

• Compare the sample or a single group with that of the population with respect to the parameter, mean. This is called as one-sample Z-test for means. For example, whether the student of a particular school has been scoring marks in Mathematics which is statistically significant than the other schools. This can also be thought of as a hypothesis test to check whether the sample belongs to the population or otherwise.
• Compare two groups with respect to the population parameter, mean. This is called as two-samples Z-test for means. For example, you want to compare class X students from different schools and determine if students of one school are better than others based on their score of Mathematics.
• Compare hypothesized proportion of the population to that of population theoritical proportion. For example, whether the unemployment rate of a given state is different than the well-established rate for the ccountry
• Compare the proportion of one population with the proportion of othe rproportion. For example, whether the efficacy rate of vaccination in two different population are statistically significant or otherwise.

Z-test Interview Questions 

Here is a list of a few interview questions you may expect in your data scientists interview:

• What is Z-test?
• What is Z-statistics or Z-score?
• When to use Z-test vs other tests such as T-test or Chi-square test?
• What is Z-distribution?
• What is the difference between Z-distribution and T-distribution?
• What is sampling distribution?
• What are different types of Z-tests?
• Explain different types of Z-tests with the help of real-world examples?
• What’s the difference two samples Z-test for means and two-samples Z-test for proportions? Explain with one example each.
• As data scientists, give some scenarios when you would like to use Z-test when building machine learning models?

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