Download Quiz 5 Key

BA 275
Winter 2007
Quizzes
Answer Key
Quiz #5
Answer Key
Name (please print)
12noon – 1:50pm
Section (circle one)
2:00 – 3:50pm
4:00 – 5:50pm
Question 1. A sample of 11 minivan electrical warranty repairs for “loose, not attached” wires
showed a sample mean repair cost of x  $45.66 with a sample standard deviation of s = $27.79.
Construct a 99% confidence interval for the true mean repair cost. Give your answer in the
following format: ( point estimate ) ± ( margin of error ).
(
Answer
45.66
)±(
3.169
27.79
11
)
Question 2. A company that sells educational materials, called “directed reading activities”,
wants to convince customers that its materials improve the reading ability of elementary school
pupils.
A consultant arranges for a third-grade class of 21 students to take part in these activities for an
eight-week period. A control classroom of 23 third-graders follows the same curriculum without
the activities. At the end of the eighth week, all students are given a Degree of Reading Power
(DRP) test, which measures the aspects of reading ability that these activities are designed to
improve. The summary statistics are
Group
Sample
size
Sample
mean
Sample standard
deviation
1. Class that used the activities
21
51.48
(Treatment group)
2. Class that did not use the activities
23
41.52
(Control group)
Note that the population standard deviations are unknown.
11.01
17.15
The company wants to show that its educational materials are effective, that is, the DRP result
from the treatment group (Group 1) is better than that from the control group (Group 2). A twosample t-test is conducted.
A. State the null and alternative hypotheses.
Null hypothesis H0:
treatment = control
Alternative hypothesis Ha:
treatment > control
Hsieh, P-H
1
BA 275
Winter 2007
Quizzes
Answer Key
B. Calculate and report the t statistic (or t score).
51.48  41.52  0  2.311887

t=
11.012 17.15 2

21
23
C. Based on the t statistic calculated above, which of the following statements about the p-value
of the test is correct:
a. p-value < 0.01
b. 0.02 > p-value > 0.01
c. 0.025 > p-value > 0.02
d. 0.05 > p-value > 0.025
e. p-value > 0.05
B. 0.02 > p-value > 0.01
Answer
The degrees of freedom for this test is MIN( 21 – 1, 23 – 1 ) = 20. Given that Ha: treatment >
control, this is a right-sided test. If  = 1%, then the rejection region is: Reject H0 if t > 2.528.
If  = 2%, then the rejection region is: Reject H0 if t > 2.197. The calculation in (B) shows
that the t statistic is 2.311887, which implies that H0 will be rejected if  = 2% but will not
be rejected if  = 1%. In other words, the corresponding p-value must be between 1% and
2% to obtain such result.
D. Given  = 5%, what is the conclusion of the test? (circle one)
Reject H0
Do not reject H0
Since the p-value is less than 0.02,  = 5% is greater than the p-value.
E. Based on the statistical conclusion in the previous question, does the company have evidence
to support their claim of better educational materials? (circle one)
Yes, the claim is supported.
No, the claim is not supported.
Rejecting H0 means that the mean score of the treatment group is higher than that of the
control group. Hence, the claim is supported.
F. To estimate with a 90% confidence for the mean improvement in the entire population of
third-graders, what would be the multiplier t* used in calculating the margin of error?
The multiplier t* =
1.725
Degrees of freedom = MIN( 21 – 1, 23 – 1 ) = 20. Since H0 is rejected (the treatment group has
a higher mean), we want to know “by how much”? A 90% confidence interval is:
51.48  41.52  1.725
Hsieh, P-H
11.012 17.15 2

. The multiplier t* used is 1.725.
21
23
2