Question Detail:
Question:
A market researcher wants to test to determine if families of four in City A spend less per week, on average, than families of four in City B on dining out. To accomplish this, the market researcher randomly 50 families of four from City A and 60 families of four from City B and has them all keep a log of their dining out spending for one week. The sample mean for City A is $185 with a standard deviation of $55. The sample mean for City B is $207 with a standard deviation of $65. Assuming an alpha of .05, the observed z value is:
Question:
A researcher desires to estimate the difference in means of two populations. To accomplish this, she takes a random sample of 81 items from the first population. The sample yields a mean of 168 with a variance of 324. A random sample of 64 items is taken from the second population yielding a mean of 163 with a variance of 625. Compute a 94% confidence interval for the difference in population means.
Question:
A researcher is interested in testing to determine if the mean of population one is different than the mean of population two. The null hypothesis is that there is no difference in the population means (i.e. the difference is zero). The alternative hypothesis is that there is a difference. She randomly selects a sample of 7 items from population one resulting in a mean of 17.3 and a standard deviation of 3.4. She randomly selects a sample of 12 items from population two resulting in a mean of 14.7 and a standard deviation of 2.9. She is using an alpha value of .10 to conduct this test. Assume the populations are normally distributed. The observed t value is:
Question: A researcher wants to conduct a before/after study on 15 subjects to determine if a treatment results in any difference in scores. The null hypothesis is that the average difference is zero while the alternative hypothesis is that the average difference is not zero. Scores are obtained on the subjects both before and after the treatment. After subtracting the after scores from the before scores, the average difference is computed to be -2.13 with a sample standard deviation of 2.51. The observed t value for this test is:
Question
Use the following data and a=0.10 to test the stated hypotheses. Assume
x
is normally distributed in the populations and the variances of the populations are approximately equal.
Sample 1
|
Sample 2
|
n
1=20
|
n
2=20
|
x
1=120
|
x
2=113
|
s
1=23.9
|
s
2=21.6
|
Round the answer to two decimal places.
Observed
t
=
The decision is to fail to reject the null hypothesis reject the null hypothesis .
The tolerance is +/- 0.02.
Construct a 90% confidence interval to estimate
D
from the following sample information. Assume the differences are normally distributed in the population.
Client
|
Before
|
After
|
1
|
32
|
40
|
2
|
28
|
25
|
3
|
35
|
36
|
4
|
32
|
32
|
5
|
26
|
29
|
6
|
25
|
31
|
7
|
37
|
39
|
8
|
16
|
29
|
9
|
35
|
31
|
Round the answers to two decimal places.
=
D
=
Question:
Many Australians spend time worrying about paying their bills. A survey by Fleishman-Hilliard Research for MassMutual discovered that 62% of Australians with kids say that paying bills is a major concern. This proportion compares to 52% of Australians without kids. Suppose 830 Australians with kids and 930 without kids were contacted for this study. Use these data to construct a 95% confidence interval to estimate the difference in population proportions between Australians with kids and Australians without kids on this issue.
Round the answers to three decimal places.
=
µ
1
-
µ
2
=
The tolerance is +/- 0.005.
Question:
Construct a 99% confidence interval for
p
1
-
p
2
for the following.
Note: Use Table IV in Appendix C to find
z-value.
Round the answers to three decimal places.
p
1
-
p
2
Question:
A market researcher wants to test to determine if families of four in City A spend less per week, on average, than families of four in City B on dining out. To accomplish this, the market researcher randomly 50 families of four from City A and 60 families of four from City B and has them all keep a log of their dining out spending for one week. The sample mean for City A is $185 with a standard deviation of $55. The sample mean for City B is $207 with a standard deviation of $65. Assuming an alpha of .05, the observed z value is:
Question:
A researcher desires to estimate the difference in means of two populations. To accomplish this, she takes a random sample of 81 items from the first population. The sample yields a mean of 168 with a variance of 324. A random sample of 64 items is taken from the second population yielding a mean of 163 with a variance of 625. Compute a 94% confidence interval for the difference in population means.
Question:
A researcher is interested in testing to determine if the mean of population one is different than the mean of population two. The null hypothesis is that there is no difference in the population means (i.e. the difference is zero). The alternative hypothesis is that there is a difference. She randomly selects a sample of 7 items from population one resulting in a mean of 17.3 and a standard deviation of 3.4. She randomly selects a sample of 12 items from population two resulting in a mean of 14.7 and a standard deviation of 2.9. She is using an alpha value of .10 to conduct this test. Assume the populations are normally distributed. The observed t value is:
Question: A researcher wants to conduct a before/after study on 15 subjects to determine if a treatment results in any difference in scores. The null hypothesis is that the average difference is zero while the alternative hypothesis is that the average difference is not zero. Scores are obtained on the subjects both before and after the treatment. After subtracting the after scores from the before scores, the average difference is computed to be -2.13 with a sample standard deviation of 2.51. The observed t value for this test is:
Question
Use the following data and a=0.10 to test the stated hypotheses. Assume
x
is normally distributed in the populations and the variances of the populations are approximately equal.
Sample 1
|
Sample 2
|
n
1=20
|
n
2=20
|
x
1=120
|
x
2=113
|
s
1=23.9
|
s
2=21.6
|
Round the answer to two decimal places.
Observed
t
=
The decision is to fail to reject the null hypothesis reject the null hypothesis .
The tolerance is +/- 0.02.
Construct a 90% confidence interval to estimate
D
from the following sample information. Assume the differences are normally distributed in the population.
Client
|
Before
|
After
|
1
|
32
|
40
|
2
|
28
|
25
|
3
|
35
|
36
|
4
|
32
|
32
|
5
|
26
|
29
|
6
|
25
|
31
|
7
|
37
|
39
|
8
|
16
|
29
|
9
|
35
|
31
|
Round the answers to two decimal places.
=
D
=
Question:
Many Australians spend time worrying about paying their bills. A survey by Fleishman-Hilliard Research for MassMutual discovered that 62% of Australians with kids say that paying bills is a major concern. This proportion compares to 52% of Australians without kids. Suppose 830 Australians with kids and 930 without kids were contacted for this study. Use these data to construct a 95% confidence interval to estimate the difference in population proportions between Australians with kids and Australians without kids on this issue.
Round the answers to three decimal places.
=
µ
1
-
µ
2
=
The tolerance is +/- 0.005.
Question:
Construct a 99% confidence interval for
p
1
-
p
2
for the following.
Note: Use Table IV in Appendix C to find
z-value.
Round the answers to three decimal places.
p
1
-
p
2
Appendix: Tables
Appendix: Tables
|