Transcript Review for Test Ch. 23, 24, 25

1)The choice between a z-test and a t-test for
a population mean depends primarily on:
A) The sample size.
B) The level of significance.
C) Whether a one- or two-tailed test is
indicated.
D) Whether the given standard deviation is
from the population or the sample.
E) A z-test should never be used.
2) One of your peers claims that boys do better in math classes than girls. Together
you run two independent simple random samples and calculate the given summary
statistics of the boys and the girls for comparable math classes. In Calculus, 15
boys had a mean percentage of 82.3 with standard deviation of 5.6 while 12 girls
had a mean percentage of 81.2 with standard deviation of 6.7. Which of the
following would be the most appropriate test for establishing whether boys do
better in math classes than girls?
A) Two sample z test for means
B) Two sample t test for means
C) Two sample z test for proportions
D) Matched pair t test
E) One sample t test for means
3) Suppose a school district has always believed that the mean height of
their eighth grade girls was 60 inches. Using the computer printout, which
of the following represents the calculation for the appropriate test?
Test of mu = 60.00 vs mu not = 60.00
Variable n
mean
stdev
SE mean
Heights 23
62.35
5.42
1.13
A)
z
62.35  0
5.42
23
62.35  60
C) t 
1.13
23
E)
62.35  0
t
5.42
23
62.35  60
B) z 
5.42
23
62.35  60
D) t 
5.42
23
The creators of a discovery-based learning system for statistics claim that students
who are taught using their method experience greater success than students taught
with a strictly lecture-based method. To test the claim , Professor Myers decided to
teach on of his introductory courses using the discovery-based method and the other
using a lecture-based course. Eighty introductory statistics students were randomly
assigned to one of the Professor Myers’ two sections. The summary statistics for his
end-of-course assessment appear below.
x
s
n
Lecture Method
80.6
5.2
40
Discovery-based Method
84.2
3.9
40
4)At a 0.01 significance level, test the claim that students who learn introductory statistics
using a discovery-based approach show higher achievement that those taught using a
lecture-based approach.
A) Reject H0, p-value < 0.01
B) Fail to reject H0, p-value < 0.01
C) Reject H0, p-value > 0.01
D) Fail to reject H0, p-value > 0.01
E) There is not enough information to test the claim
The creators of a discovery-based learning system for statistics claim that students
who are taught using their method experience greater success than students taught
with a strictly lecture-based method. To test the claim , Professor Myers decided to
teach on of his introductory courses using the discovery-based method and the other
using a lecture-based course. Eighty introductory statistics students were randomly
assigned to one of the Professor Myers’ two sections. The summary statistics for his
end-of-course assessment appear below.
x
s
n
Lecture Method
80.6
5.2
40
Discovery-based Method
84.2
3.9
40
5) Construct a 95% confidence interval for the mean difference in scores of
students taught using a lecture-based method versus a discovery-based method
A) 3.60  1.99
 5.2 
40
2

(3.9)
40
3.9 
 5.2

C) 3.60  1.99 

40 
 40
3.9 
 5.2

E) 3.60  1.96 

40 
 40
2
B) 3.60  1.96
 5.2 
2
(3.9)2

40
40
3.9 
 5.2

D) 3.60  1.96 

40 
 40
6) It is believed that using a new fertilizer will result in a
yield of 1.6 tons per acre. A botanist carries out a two-tailed
test on a field of 64 acres. Determine the P-value if the mean
yield per acre in the sample is 1.72 tons with a standard
deviation of 0.4. What is the conclusion at a level of
significance of 10%? 5%? 1%?
A) P = .0097, and so the 1.6-ton claim should be rejected at
all three of these levels.
B) P = .0193, and so the 1.6 ton claim should be rejected at the
10% and 5% levels but not at the 1% level.
C) P = .0097, and so the 1.6 ton claim should be rejected at
the 10% level but not at the 5% and 1% levels.
D) P = .0193, and so the 1.6 ton claim should be rejected at
the 1% level but not at the 10% and 5% levels.
E) P = .0097, and so there is not enough evidence to reject
the 1.6-ton claim at any of the three levels.
7) An NCAA official claims that the average 5K
time for students trying out for college crosscountry teams is 17 minutes. A Big East coach
believes the true figure is lower among Big East
universities. He picks an SRS of 30 recruits and
calculates their mean 5K time is 16 minutes and
52 seconds, with a standard deviation of 25
seconds. What is the p-value for the appropriate
test?
A) 0.04
D) 0.42
B) 0.08
E) 0.75
C) 0.37
A random sample of THS students was asked how many
credits were needed to meet the FBISD recommended
graduation requirements. The following results in the
computer output were found:
TEST MU = 24 VS MU NOT = 24
VARIABLE
N
MEAN
STDEV
CREDITS
34
22.25
1.87
8) Which of the following is the correct null and
alternative hypotheses statements for this situation?
A) H0 : µ =24 vs. Ha : µ ≠ 24
B) H0 : µ =24 vs. Ha : µ < 24
C) H0 : p =24 vs. Ha : p ≠ 24
D) H0 : µ =22.5 vs. Ha : µ ≠ 22.5
E) H0 : p =22.5 vs. Ha : p ≠ 22.5
A random sample of THS students was asked how many credits
were needed to meet the FBISD recommended graduation
requirements. The following results in the computer output were
found:
TEST MU = 24 VS MU NOT = 24
VARIABLE
N
MEAN
STDEV
CREDITS
34
22.25
1.87
9) Which of the following is the correct method of
calculating the test statistics for this situation?
A) t  22.25  24
1.87
34
D) z  22.25  0
1.87
34
B) t  22.25  24
0.32
34
E)
22.25  24
z
1.87
34
C) t  22.25  0
1.87
34
A random sample of THS students was asked how many
credits were needed to meet the FBISD recommended
graduation requirements. The following results in the
computer output were found:
TEST MU = 24 VS MU NOT = 24
VARIABLE
N
MEAN
STDEV
CREDITS
34
22.25
1.87
10) The degrees of freedom for this
situation is
A) 25
B) 30
D) 34
E) 35
C) 33
A random sample of THS students was asked how many
credits were needed to meet the FBISD recommended
graduation requirements. The following results in the
computer output were found:
TEST MU = 24 VS MU NOT = 24
VARIABLE
N
MEAN
STDEV
CREDITS
34
22.25
1.87
11) The p-value for this situation is
A) 0.00000004
C) 0.00000239
E) none of these
B) 0.00000479
D) 0.99999760
Multiple Choice – Ch 23, 24, 25 – Inferences on Means
1)D
2) B
3) D
4) A
5) A
6) B
7) A
8) A
9) A
10) C
11) B