Bank A and Bank B have each de
Bank A and Bank B have each developed an improved process forserving customers. The waiting period from the moment a customerenters until he or she reaches the counter needs to be shortened. Arandom sample of 10 customers is selected from each bank and theresults (in minutes) are shown in the accompanying data table
| Bank A |
| 2.18 |
| 2.38 |
| 3.02 |
| 3.02 |
| 3.11 |
| 4.57 |
| 4.29 |
| 4.37 |
| 5.28 |
| 5.83 |
| Bank B |
| 3.97 |
| 4.41 |
| 4.13 |
| 5.18 |
| 5.33 |
| 6.84 |
| 6.36 |
| 8.58 |
| 8.55 |
| 10.33 |
Assuming that the population variances from both banks areequal, is there evidence of a difference in the mean waiting timebetween the two branches? (Use
.α=0.01.)
1) Reject Ho. There is insufficient evidence that meansdiffer.
2) Do not reject Ho. There is sufficient evidence that meansdiffer
3) Do not reject Ho. There is insufficient evidence that meansdiffer
4) Reject Ho. There is sufficient evidence that meansdiffer.
B) Determine the p-value in (a) and interpret its meaning.
interpret the p-value. Choose the correct answer below.
1) It is the probability of obtaining a sample that yields a ttest statistic farther away from 0 in the negative direction thanthe computed test statistic if there is no difference in the meanwaiting time between Bank A and Bank B.
2) It is the probability of obtaining a sample that yields a ttest statistic farther away from 0 in the positive direction thanthe computed test statistic if there is no difference in the meanwaiting time between Bank A and Bank B.
3) It is the probability of obtaining a sample that yields a ttest statistic farther away from 0 in either direction than thecomputed test statistic if there is no difference in the meanwaiting time between Bank A and Bank B.
c. In addition to equal variances, what otherassumption is necessary in (a)?
1) Both sampled populations are approximately normal.
2) The sample sizes must be equal.
3) The samples are specifially chosen and not independentlysampled.
4) Both sampled populations are not approximately normal.
Answer:
Where
Substitutingthe variances and values of 
and 
inthe formula above we get 
as 3.1366. Nowcomputing the test statistic we have T=3.2359
The degrees of freedom here in this test is 
The p value for the T value at 18 degrees of freedom is0.0046<0.05 . Reject H_0.Hence there is enough evidence from thedata to believe that the two means differ
A) option 4
B) option 3 ( Since the test is two tailed, it can be eitherdirection
C) option 1 ( Both samples are assumed to be from the normalpopulation, t test is used since the population standard deviationis unknown)
"Our Prices Start at $11.99. As Our First Client, Use Coupon Code GET15 to claim 15% Discount This Month!!"
