Question

1. Waiting time for checkout line at two stores of a supermarket chain were measured for a random sample of customers at each store. The chain wants to use this data to test the research (alternative) hypothesis that the mean waiting time for checkout at Store 1 is lower than that of Store 2. (12 points)

Store 1 (in Seconds)	Store 2 (in Seconds)
470	375
394	319
167	266
293	324
187	244
115	178
195	279
400	289
228	342
315	212
195
188

1. What are the null and alternative hypothesis for this research?
2. What are the sample mean waiting time for checkout line and the sample standard deviation for the two stores?
3. Compute the test statistic t used to test the hypothesis. Note that the population standard deviations are not known and therefore you cannot use the formula in Section 10.1. Use the one in Section 10.2 instead.
4. Compute the degree of freedom for the test statistic t.
5. Can the chain conclude that the mean waiting time for checkout at Store 1 is lower than that of Store 2? Use the critical-value approach and α = 0.05 to conduct the hypothesis test.
6. Construct a 95% confidence interval for the difference of mean waiting time for checkout line at the two stores.
   

Answer 1

a)

  

b)  = 262.25

s1 = 110.934

= 282.8

s2 = 60.415

C) The test statistic is

= -0.551

d)

  

= 17

e) At = 0.05, the critical value is t_{0.05, 17} = -1.740

Since the test statistic value is not less than the critical value, so we should not reject the null hypothesis.

At 0.05 significance level, there is not sufficient evidence to conclude that the mean waiting time for checkout at store 1 is lower than that of store 2.

f) At 95% confidence level, the critical value is t^* = 2.110

The 95% confidence interval is

() +/- t^* * sqrt(s1^2/n1 + s2^2/n2)

= (262.25 - 282.8) +/- 2.11 * sqrt((110.934)^2/12 + (60.415)^2/10)

= -20.55 +/- 78.68

= -99.23, 58.13