Question

The accompanying data are​ drive-through service times​(seconds) recorded at a fast-food restaurant during dinner times. Assuming that dinner service times at the​ restaurant's competitor have standard deviation σ = 61.4​sec, use a 0.01 significance level to test the claim that service times at the restaurant have the same variation as service times at its​ competitor's restaurant. Use the accompanying data to identify the null​ hypothesis, alternative​ hypothesis, test​ statistic, and​ P-value. Then state a conclusion about the null hypothesis. Need help with part B and beyond

Data: -44, 75, -22, -75, -42, 15, 16, 52, -5, -51, -107, -107

A) Identify null and alternative hypotheses (I already have answered, but gave just in case needed for rest). The answer I got was:

H0: σ = 611.4 minutes

Ha σ *Doesn't equal* 61.4 minutes

B) Compute the test statistic (2 decimal places for rounding).

X^2 = _____

C) Find the P-Value of the test-statistic (3 decimal places for rounding).

The P-value for the test statistic is _______.

D) State the conclusion about the null hypothesis.

There _______ ("is" or "is not") sufficient evidence to conclude that there is a difference between the waiting times in the two​restaurants, because H0 is _______ ("rejected" or "not rejected") by the hypothesis test. Thank you in advance!

Answer 1

first we need to find standard deviation of givendata

part A is correct H0 : =61.4 vs H1 : 61.4

using excel command =STDEV.S() we can find standard deviation of given data

s = 56.9113

s2 = (56.9113)2 = 3353.72

now (61.4)2 = 3769.96

B) now chi square statistics = [(n-1) *s2]/ = [(12-1)* 3353.72]/3769.96

= 36890.92/3769.96= 9.785

chi square statistics = x^2 = 9.79

C) p-value for this two tailed test is as follows

using excel command

P( X^2>9.79) is =CHISQ.DIST.RT(9.79,11) = 0.54937

P(X^2 <9.79) is =CHISQ.DIST(9.79,11,1) = 0.45063

so minimum is 0.45063 so

p-value = 2 * min[ P( X^2> 9.79) , P( X^2 <9.79)]

= 2 * 0.45063

=0.90126 =0.901

D) decision rule : if p-value < sig.level (0.01) then reject H0 otherwise do not reject H0

here p-value (0.901) is > 0.01 so we do not reject H0

There "is not" sufficient evidence to conclude that there is a difference between the waiting times in the two ​restaurants, because H0 is "not rejected" by the hypothesis test.

Hope this will help you.