Using the t-distribution, as we have the standard deviation for the sample, it is found that there is a significant difference between the wait times for the two populations.

<h3>What are the hypothesis tested?</h3>

At the null hypothesis, we test if there is no difference, that is:

$H_0: \mu_A - \mu_B = 0$

At the alternative hypothesis, it is tested if there is difference, that is:

$H_1: \mu_A - \mu_B = 0$

<h3>What are the mean and the standard error of the distribution of differences?</h3>

For each sample, we have that:

$\mu_A = 73, s_A = \frac{2}{\sqrt{100}} = 0.2$

$\mu_B = 74, s_B = \frac{4}{\sqrt{100}} = 0.4$

For the distribution of differences, we have that:

$\overline{x} = \mu_A - \mu_B = 73 - 74 = -1$

$s = \sqrt{s_A^2 + s_B^2} = \sqrt{0.2^2 + 0.4^2} = 0.447$

<h3>What is the test statistic?</h3>

It is given by:

$t = \frac{\overline{x} - \mu}{s}$

In which $\mu = 0$ is the value tested at the null hypothesis.

Hence:

$t = \frac{\overline{x} - \mu}{s}$

$t = \frac{-1 - 0}{0.447}$

$t = -2.24$

<h3>What is the p-value and the decision?</h3>

Considering a one-tailed test, as stated in the exercise, with 100 - 1 = 99 df, using a t-distribution calculator, the p-value is of 0.014.

Since the p-value is less than the significance level of 0.05, it is found that there is a significant difference between the wait times for the two populations.

More can be learned about the t-distribution at brainly.com/question/16313918

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