To test a population mean, conduct the 5-step procedure previously discussed in this link: click here. To apply the third step, three test statistics are available.

Firstly, if the population standard deviation is not known but the population is normally distributed, the statistic t = \frac{\bar{x} - {\mu}_0}{s/\sqrt{n}} may be used. Here, s denotes the sample standard deviation and n is the sample size.

The test statistic to use if the population standard deviation is known is z = \frac{\bar{x} - {\mu}_0}{\sigma / \sqrt{n}}. To apply it, the population has to be normally distributed. But, if the population is not normally distributed, the test statistic may be applied if the sample size is not less than 30, that is, n ≥ 30.

If the population standard deviation is not known and the population is not normally distributed, use the statistic z = \frac{\bar{x} - {\mu}_0}{s / \sqrt{n}} provided that the sample size is not less than 30, i.e. n ≥ 30.

 

Consider the three similar sample problems below.

 

Sample Problem #1
The waiting time for customers at MacBurger Restaurants follows a normal distribution. At the Warren Road MacBurger, the quality-assurance department sampled 15 customers and found that the mean and standard deviation of the waiting time were 2.75 minutes and 1 minute, respectively. At the .05 significance level, can we conclude that the mean waiting time is less than 3 minutes?

 

Sample Problem #2
The waiting time for customers at MacBurger Restaurants follows a normal distribution with a standard deviation of 1 minute. At the Warren Road MacBurger, the quality-assurance department sampled 15 customers and found that the mean of the waiting time was 2.75 minutes. At the .05 significance level, can we conclude that the mean waiting time is less than 3 minutes?

 

Sample Problem #3
The quality-assurance department of Warren Road MacBurger is investigating whether the mean waiting time is less than 3 minutes. The department sampled 60 customers and found that the mean and standard deviation of the waiting time were 2.75 minutes and 1 minute, respectively. At the .05 significance level, can we conclude that the mean waiting time is less than 3 minutes?

 

The sample problems above call for three different test statistics. In Sample Problem #1, the population standard deviation σ is not known. The standard deviation given is the sample standard deviation. How do we know that it is the sample standard deviation, not the population standard deviation? Read the second sentence of the problem, which is highlighted below.

The phrase “standard deviation” appears in a sentence describing the sampling result. Thus, it is not a population standard deviation. In this case, the population standard deviation is not known. In addition, the waiting time is normally distributed (It is stated in the first sentence of the problem). For these reasons, to solve the problem we use t = \frac{\bar{x} - {\mu}_0}{s/\sqrt{n}}.

 

In Sample Problem #2, a standard deviation is given. Is it the population or sample standard deviation? To answer this, read carefully the sentence containing the phrase “standard deviation”. It is highlighted below.

Does the sentence talk about anything related to the sampling? No, information about sampling is in the second sentence. Moreover, the highlighted sentence tells us about something in general, that is the waiting time for customers at the restaurant. Owing to this, the stated standard deviation is the population standard deviation. So, in this case, the population standard deviation is known. Moreover, the problem states that the waiting time follows a normal distribution. All these facts lead us to the use of z = \frac{\bar{x} - {\mu}_0}{\sigma / \sqrt{n}} as the proper test statistic.

 

Referring to the second sentence of Sample Problem #3 (highlighted below), the stated standard deviation is a sample standard deviation.

The second sentence talks about the sampling conducted by the department. So, all information there, are sample data, including the standard deviation. Hence, the population standard deviation σ is not known. Together with the facts that the normality is not known and n = 60 (≥ 30), we use z = \frac{\bar{x} - {\mu}_0}{s / \sqrt{n}} as the test statistic.

 

Now, we are ready to solve the problems.

 

Solution to Sample Problem #1

Step 1: State the null and the alternate hypotheses
H0: μ ≥ 3 minutes
H1: μ < 3 minutes

Step 2: Select a level of significance
α = .05

Step 3: Select the test statistic

As discussed previously, the test statistic to use is t = \frac{\bar{x} - {\mu}_0}{s/\sqrt{n}}

Step 4: Formulate the decision rule
To do this, we have to refer to a suitable table of critical values. As in this case we use a t statistic, the proper table is The Table of t Distribution Critical Values. (To download the table, click here.) To make use of the table, we have to calculate the degree of freedom (df). The df for testing a population mean using t statistic is: df = n – 1. Since in this sample problem the sample size is n = 15, the degree of freedom is df = 15 – 1 = 14. Now, move down the column labeled .05 (that is the significance level α) until it intersects the row with 14 degrees of freedom. The value is 1.761. The rejection region in this sample problem is in the left tail, so the critical value is negative. As a consequence, the decision rule is to reject H0 if the value of t is less than -1.761. To make it clearer, the decision rule is depicted in the figure below.

The red-shaded region in the picture shows the rejection region. If the t value computed in the next step falls in the area, the null hypothesis will be rejected.

Step 5: Make a decision (Reject or Accept the Null Hypothesis)
In this problem, \bar{x} = 2.75 \: minutes, μ0 = 3 minutes, s = 1 minute, n = 15. Substitute these values for corresponding variables in t = \frac{\bar{x} - {\mu}_0}{s/\sqrt{n}}. This gives t = \frac{2.75-3}{1/\sqrt{15}} \approx -.968. Since -.968 does not fall in the rejection region, H0 is not rejected. (See the figure below.) We conclude that the mean waiting time is not less than 3 minutes.

 

Solution to Sample Problem #2
Step 1: State the null and the alternate hypotheses
H0: μ ≥ 3 minutes
H1: μ < 3 minutes

Step 2: Select a level of significance
α = .05

Step 3: Select the test statistic
As discussed previously, the test statistic to use is z = \frac{\bar{x} - {\mu}_0}{\sigma/\sqrt{n}}

Step 4: Formulate the decision rule
The null hypothesis will be rejected if the z value computed in the next step is less than to a critical value. To determine the critical value, refer to the following table.

Move down the column labeled α until the significance level .050 is found. Then, move to the right until we reach the third column (labeled zα). The value obtained is 1.64. Since the rejection region in this sample problem is in the left tail, the critical value is negative. As a consequence, the decision rule is to reject H0 if the value of z is less than -1.64. To make it clearer, the decision rule is depicted in the figure below.

The red-shaded region in the picture represents the rejection region. If the z value computed in the next step falls in the area, the null hypothesis will be rejected.

Step 5: Make a decision (Reject or Accept the Null Hypothesis)
In this problem, \bar{x} = 2.75 \: minutes, μ0 = 3 minutes, σ = 1 minute, n = 15. Substitute these values for corresponding variables in z = \frac{\bar{x} - {\mu}_0}{\sigma / \sqrt{n}}. As a result, z = \frac{2.75-3}{1/\sqrt{15}} \approx -.968. Since -.968 does not fall in the rejection region, H0 is not rejected. (See the figure below.) We conclude that the mean waiting time is not less than 3 minutes.

 

Solution to Sample Problem #3
Step 1: State the null and the alternate hypotheses
H0: μ ≥ 3 minutes
H1: μ < 3 minutes

Step 2: Select a level of significance
α = .05

Step 3: Select the test statistic

As discussed previously, the test statistic to use is z = \frac{\bar{x} - {\mu}_0}{s/\sqrt{n}}

Step 4: Formulate the decision rule
Applying the procedure described in the previous sample problem, the critical value obtained is 1.64. The rejection region in this sample problem is in the left tail, thus the critical value is negative. As a consequence, the decision rule is to reject H0 if the value of z is less than -1.64.

The red-shaded region in the figure represents the rejection region. If the z value computed in the next step falls in the area, the null hypothesis will be rejected.

Step 5: Make a decision (Reject or Accept the Null Hypothesis)
In this problem, \bar{x} = 2.75 \: minutes, μ0 = 3 minutes, s = 1 minute, n = 60. Substitute these values for corresponding variables in z = \frac{\bar{x} - {\mu}_0}{s / \sqrt{n}}. This results in z = \frac{2.75-3}{1/\sqrt{60}} \approx -1.94. As -1.94 falls in the rejection region, H0 is rejected. (See the figure below.) We conclude that the mean waiting time is significantly less than 3 minutes.

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