Dominos Pizza for years made a claim that they would deliver your pizza within 30 minutes ("30 minutes or less") of your order being placed (provided you live within a certain radius of their restaurant). A group of college students make the ultimate sacrifice and order Dominos from the same location for 20 straight days for the good of statistics. They find a mean of 23.7 minutes and a standard deviation of 2.15.

A) In constructing a 95% confidence interval for mean wait time for Dominos Pizza delivery using the data collected by the students, would your critical value be a z * or a t*? Explain your answer.

B) Staying consistent with your answer to part A, construct a 95% confidence interval for mean wait time of Dominos Pizza delivery.

C) Do you think this confidence interval is accurate for Dominos Pizza in general? Why or why not?

Question

Adriana Mcintosh

Mathematics

7 June, 12:44

Dominos Pizza for years made a claim that they would deliver your pizza within 30 minutes ("30 minutes or less") of your order being placed (provided you live within a certain radius of their restaurant). A group of college students make the ultimate sacrifice and order Dominos from the same location for 20 straight days for the good of statistics. They find a mean of 23.7 minutes and a standard deviation of 2.15.

A) In constructing a 95% confidence interval for mean wait time for Dominos Pizza delivery using the data collected by the students, would your critical value be a z * or a t*? Explain your answer.

B) Staying consistent with your answer to part A, construct a 95% confidence interval for mean wait time of Dominos Pizza delivery.

C) Do you think this confidence interval is accurate for Dominos Pizza in general? Why or why not?

+4

Answers (1)

Know the Answer?

Werner · Answer 1

a. t*

b. (22.87, 24.53)

c. yes

Step-by-step explanation:

We have the following dа ta:

mean = m = 23.7

standard deviation = sd = 2.15

x = 20

Thus:

a. The critical value would be t * since the value is m <30

b. whenever the mean is less than 30, the t distribution is used, for 95% confidence interval with degree of freedom = (x - 1) df

= 20 - 1

= 19

now, for this value we look in the table of t, we find that t * = 1.729

m + - t * (sd / x ^ 81/2))

replacing

23.7 + - 1.729 * (2.15 / 20 ^ (1/2))

23.7 + - 0.8312

then the inverval would be:

23.7 + 0.8312 = 24.5312

23.7 - 0.8312 = 22.8688

(22.87, 24.53)

c. yes, the interval is accurate since the mean of 23.7 is within this value