Confidence interval in comparing service times

Question

A store manager wishes to compare the service times of the express checkout with the service times of the self-serve checkout. Suppose that independent random samples of 121 customers at express and selfserve checkouts were taken, and the service times for each customer was recorded. The mean and standard deviation of the sample of customers using the express checkout were 3.7 and 0.9 minutes, respectively. For the self-serve customers, the mean and standard deviation were 4.2 and 1.7 minutes, respectively.

a) Calculate the ratio of the maximum sample variance to the minimum sample variance. Does it appear that the population variances are equal or unequal? Explain.

b) Construct the appropriate 95% confidence interval for the difference in the mean service times for customers using the express and self-serve checkouts, and interpret your result.

My work:

For a) I'm not really sure how to get the answer of 3.57, I tried $0.9^2+1.7^2,$ which equals 3.7, but that is wrong.

For b) I tried $3.7-4.2±1.96*\sqrt{(0.81/121)+(2.89/121)}$, to which I get the confidence interval of $(-0.84,-0.16)$ The answer says it should be $-0.5 ± 0.81,$ which results in $(-1.31,0.31).$

Any help is very much appreciated!!

Answer 1

(a) The requested $ratio$ of variances is $(1.7/.9)^2 = 3.567901.$ This exceeds the quantile .975 of F(120, 120), which is 1.433 (from F tables or from software). We conclude that variances differ.

(b) A Welch (separate-variances) two-sample t interval is indicated. Minitab output:

 Sample    N   Mean  StDev  SE Mean
 1       121  3.700  0.900    0.082
 2       121   4.20   1.70     0.15

 Difference = mu (1) - mu (2)
 Estimate for difference:  -0.500
 95% CI for difference:  (-0.845, -0.155)
 T-Test of difference = 0 (vs not =): 
      T-Value = -2.86  P-Value = 0.005  DF = 182

DF must exceed 120, so it is safe to use $\pm 1.96$ for the CI. The standard error of the difference in means is $\sqrt{.9^2/121 + 1.7^2/121} = 0.175$. Thus the CI for $\mu_{exp} - \mu_{ss}$ is approximately $-.5 \pm 1.96(0.175)$ or $(-0.84, -0.16),$ consistent with the printout and with your answer.