The mean waiting time at the drive-through of a fast-food restaurant from the ti

Question

The mean waiting time at the drive-through of a fast-food restaurant from the time an order is placed to the time the order is received is 86.2 seconds. A manager devises a new drive-through system that she believes will decrease wait time. As a test, she initiates the new system at her restaurant and measures the wait time for 10 randomly selected orders. The wait times are provided in the table. Complete parts (a) and (b) below. Because the sample size is small, the manager must verify that the wait time is normally distributed and the sample does not contain any outliers. The normal probability plot and box plot are shown. Are the conditions for testing the hypothesis satisfied? Yes, the conditions are satisfied. No, the sample contains outliers. No, the sample is not normally distributed. No, the sample is not normally distributed and contains outliers. Is the new system effective? Conduct a hypothesis test using the P-value approach and a level of significance of alpha = 0.05. First determine the appropriate hypotheses. H_0: 86.2 H_1: 86.2 Find the test statistic. t_0 = (Round to two decimal places as needed.) Find the P-value. The P-value is (Round to three decimal places as needed.) Use the alpha = 0.05 level of significance. What can be concluded from the hypothesis test? The P-value is less than the level of significance so there is sufficient evidence to conclude the new system is effective. The P-value is greater than the level of significance so there is not sufficient evidence to conclude the new system is effective. The P-value is greater than the level of significance so there is sufficient evidence to conclude the new system is effective. The P-value is less than the level of significance so there is not sufficient evidence to conclude the new system is effective.

Explanation / Answer

SolutionA:

From box plot no outliers detected

from normal probability plot points ar e on line

it follows normal distribution

optionA

Solutionb:

H0:mean=86.2

H1:mean<86.2

SOlutionc:

t=-1.82

degrees of freedom

=n-1=10.1=9

alpha=0.05

The P-Value is .051

The result is not significant at p < .05.

Fail to reject Null hypothesis

Accept Null hypothesis

Conclusion:

as p> alpha there is no sufficient evidence to support the claim at 5% level of significance

OptionB

102.2 66 56.2 73 65 81.8 95.7 85.8 72.9 81.4 mean 78 std deviation 14.24008 population mean 86.2 sampl size 10 t =78-86.2/14.24008/sqrt(10) t =-8.2/14.24/3.16 t =-1.82

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