Question

For staffing purposes, a retail store manager would like to standardize the number of checkout lanes to keep open on a particular shift. She believes that if the standard deviation of the hourly customer arrival rates is 10 customers or less, then a fixed number of checkout lanes can be staffed without excessive customer waiting time or excessive clerk idle time. However, before determining how many checkout lanes (and thus clerks) to use, she must verify that the standard deviation of the arrival rates does not exceed 10. Accordingly, a sample of 25 hourly customer arrival rates was compiled for that shift over the past week. picture Click here for the Excel Data File a. Select the hypotheses to test whether the standard deviation of the customer arrival rates exceeds 10. Ho: 2 s 100; HA: 02 > 100 Ho: o2 = 100; HA: 02 - 100 Ho: 02 2 100; HA: 02 < 100 b. Calculate the value of the test statistic. Assume that customer arrival rates are normally distributed. (Round intermediate calculations to at least 4 decimal places and final answer to 3 decimal places.) Test statistic c. Find the p-value. p-value 20.10 0.05 s p-value < 0.10 0.025 s p-value < 0.05 0.01 s p-value < 0.025 p-value < 0.01 d-1. At a = 0.05, what is your conclusion? HO. At the 5% significance conclude level, we the standard deviation of the customer arrival rates exceeds 10/hour. d-2. Would your conclusion change at the 1% significance level? Yes 0 No

Hourly Arrival Rates
125
120
101
118
97
120
113
116
116
121
125
122
127
121
124
120
119
94
124
108
133
91
121
111
119

Answer 1

Ans:- From given information: a) Test whether the standard deviabion of the customer Ornival rate exceeds_10 standard deviabion = 6 = Jvariance - 62 = 8=100 Hoi 6?<loo j H4:6?> 100 1 b) Test statistic y =(n-1) s² - Sample Variance 22 Cxi-x)? 12 6-1 X = EX; 6.24 2906 = 25 x= Ex = 29046 = 116.24 Moto X (X; - 832 = 2672.56 = 111.3567 S? - 2672.56 25-1

get = (1-12 52 ch-i) s = (25-1) x 111.35 100 = 26.724. c) p-value at x 26724 and degree of freedom is D-1 = 25-1 -24. D-value = 0.31673 o p-value > 0.10 d) At x=0.05 P-value = 0.3167 > =0.05 so.we. do not rejed Ho at 5% level of significance Do not reject Ho Or Accept Ho. At 57 significance level, we do not conclude that the standard deviation of the customer arrive. rates exceeds 10 hours