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.
What are the hypothesis tested?
At the null hypothesis, we test if there is no difference, that is:
[tex]H_0: \mu_A - \mu_B = 0[/tex]
At the alternative hypothesis, it is tested if there is difference, that is:
[tex]H_1: \mu_A - \mu_B = 0[/tex]
What are the mean and the standard error of the distribution of differences?
For each sample, we have that:
[tex]\mu_A = 73, s_A = \frac{2}{\sqrt{100}} = 0.2[/tex]
[tex]\mu_B = 74, s_B = \frac{4}{\sqrt{100}} = 0.4[/tex]
For the distribution of differences, we have that:
[tex]\overline{x} = \mu_A - \mu_B = 73 - 74 = -1[/tex]
[tex]s = \sqrt{s_A^2 + s_B^2} = \sqrt{0.2^2 + 0.4^2} = 0.447[/tex]
What is the test statistic?
It is given by:
[tex]t = \frac{\overline{x} - \mu}{s}[/tex]
In which [tex]\mu = 0[/tex] is the value tested at the null hypothesis.
Hence:
[tex]t = \frac{\overline{x} - \mu}{s}[/tex]
[tex]t = \frac{-1 - 0}{0.447}[/tex]
[tex]t = -2.24[/tex]
What is the p-value and the decision?
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 https://brainly.com/question/16313918
