Question

we considered the mean waiting time at the drive-through of a fast-food restaurant. In addition to concern about the amount of time cars spend in the drive-through, the manager is also worried about the variability in wait times. Prior to the new drive-through system, the standard deviation of wait times was 22.1 seconds. Use the data in the table below to decide whether there is evidence to suggest the standard deviation wait-time is less than 22.1 seconds. Recall that in Homework 1-3, we verified this data could have come from a population that is normally distributed. Use the α = 0.05 level of significance. 102.5 67.4 58.0 75.9 65.1 68.3 80.4 95.5 87.3 70.9 71.

On a separate sheet of paper, write down the hypotheses (H₀ and H_a) to be tested.

Conditions:
The conditions for the χ² ("chi-square") test for standard deviations  (are / are not) satisfied for this data.

Rejection Region:
To test the given hypotheses, we will use a (left / right / two) -tailed test.  
The appropriate critical value(s) for this test is/are .  (Report your answer exactly as it appears in Table VII. For two-tailed tests, report both critical values in the answer blank separated by only a single space.)

On a separate sheet of paper, sketch the rejection region(s) for this test. You will need this sketch in Question 6.

Refer to the data given in Question 5. Recall that in Homework 1-3, we found the sample standard deviation of wait times for the given data to be 13.613 seconds.

The test statistic for this test is χ²₀=.  (Calculate this value in a single step in your calculator, and report your answer rounded to 3 decimal places.)

Label the test statistic in your sketch from Question 5. Use this sketch to conclude the hypothesis test.

We  (reject / fail to reject) H₀.
The given data  (does / does not) provide significant evidence that the standard deviation of wait times under the new method is less than 22.1 seconds.

Answer 1

From given information , Sample size = n = 11

We have to test whether there is evidence to suggest the standard deviation wait-time is less than 22.1 seconds.

i.e In notation

Hypothesis :

( Claim )

Left tailed test.

The data come from a population that is normally distribute. So, The conditions for the χ² ("chi-square") test for standard deviations are satisfied for this data.

Rejection region :

Significance level = = 0.05

df = n - 1 = 11-1 = 10

Critical value for this left tailed test is,

Using Excel function , =CHISQ.INV( , df ) , This function returns left tailed inverse of chi-square distribution.

=CHISQ.INV(0.05,10) = 3.94  

Critical value = 3.94  

Rejection region : { : < 3.94 }

Test statistic :

Where ,  

S is sample standard deviation ( Given , s = 13.613 )

hypothesized standard deviation = 22.1

So , Chi-square test statistic is ,

Decision about null hypothesis:

It is observed that test statistic fall in rejection region.

i.e test statistic( 3.794 ) is less than chi-square critical value (3.94 )

So , Reject null hypothesis.

Conclusion :

The given data does provide significant evidence that the standard deviation of wait times under the new method is less than 22.1 seconds.