Question

The waiting time for customers at MacBurger Restaurants follows a normal distribution with a population standard deviation of 3 minute. At the Warren Road MacBurger, the quality assurance department sampled 49 customers and found that the mean waiting time was 16.25 minutes. At the 0.05 significance level, can we conclude that the mean waiting time is less than 17 minutes? Use α = 0.05.

a. State the null hypothesis and the alternate hypothesis.

H₀: μ ≥           

H₁: μ <           

b. State whether the decision rule is true or false: Reject H₀ if z < -1.645.

(Click to select)  True  False

c. Compute the value of the test statistic. (Negative answer should be indicated by a minus sign. Round the final answer to 2 decimal places.)

Test statistic z is            .

d. What is your decision regarding H₀?

(Click to select)  Do not reject  Reject  H₀.

e. What is the p-value? (Round the final answer to 4 decimal places.)

The p-value is  .

Answer 1

The given information is:

The researcher wants to test whether the mean waiting time is less than 17 minutes.

The total number of customers that are sampled (n) = 49

The sample mean waiting time = 16.25 minutes

The population standard deviation = 3 minutes

The significance level = 0.05

Here, z test applied as is known and n > 30 in the study.

a). The null and the alternative hypothesis can be stated as:

It is a left tailed test.

b). At the significance level 0.05, the left tailed critical value obtained from the standard normal table is -1.645. Therefore, the rejection region is,

Decision rule: Reject Ho if z < -1.645.

Yes, the decision rule is true.

c). Test statistic:

Hence, the value of the z-statistic is -1.75.

d). since the calculated value (-1.75) is less than the critical value (-1.645), so the decision is to reject the null hypothesis at the significance level 0.05.

Hence, the decision is Reject Ho.

In conclusion, there is enough evidence to support that mean waiting time is less than 17 minutes.

e). Using the standard normal table, the obtained p-value is given by,

Therefore, the p-value is 0.0401.