19 questions were answered correctly.
5 questions were answered incorrectly.
1
Adam tabulated the values for the average speeds on each day of his road trip as 60.5, 63.2, 54.7, 51.6,
72.3, 70.7, 67.2, and 65.4 mph. The sample standard deviation is 7.309.
Select the 98% confidence interval for Adam’s set of data.
46.94 to 79.46
55.45 to 70.95
46.94 to 71.33
55.45 to 79.46
RATIONALE
In order to get the 98% CI , we first need to find the critical
t-score. Using a t-table, we need to find (n-1) degrees of freedom, or (8-1) = 7 df and the corresponding
CI.
Using the 98% CI in the bottom row and 7 df on the far left column, we get a t-critical score of 2.998.
We also need to calculate the mean:
So we use the formula to find the confidence interval:
The lower bound is:
63.2-7.75 = 55.45
The upper bound is:
63.2+7.75 = 70.95
CONCEPT
,Confidence Intervals Using the T-Distribution
2
A table represents the number of students who passed or failed an aptitude test at two different
campuses.
South Campus North Campus
Passed 42 31
Failed 58 69
In order to determine if there is a significant difference between campuses and pass rate, the chi-square
test for association and independence should be performed.
What is the expected frequency of South Campus and passed?
50 students
36.5 students
42 students
43.7 students
RATIONALE
In order to get the expected counts we can note the formula is:
CONCEPT
Chi-Square Test for Homogeneity
3
Sukie interviewed 125 employees at her company and discovered that 21 of them planned to take an
extended vacation next year.
, What is the 95% confidence interval for this population proportion? Answer choices are rounded to
the hundredths place.
0.10 to 0.23
0.16 to 0.17
0.11 to 0.16
0.11 to 0.21
RATIONALE
In order to get the CI we want to use the following form.
First, we must determine the corresponding z*score for 95% Confidence Interval. Remember, this means
that we have 5% for the tails, meaning 5%, or 0.025, for each tail. Using a z-table, we can find the upper
z-score by finding (1 - 0.025) or 0.975 in the table.
This corresponding z-score is at 1.96.
We can know
So putting it all together:
The lower bound is:
0.168-0.065 =0.103 or 0.10
The upper bound is:
0.168+0.065 =0.233 or 0.23
5 questions were answered incorrectly.
1
Adam tabulated the values for the average speeds on each day of his road trip as 60.5, 63.2, 54.7, 51.6,
72.3, 70.7, 67.2, and 65.4 mph. The sample standard deviation is 7.309.
Select the 98% confidence interval for Adam’s set of data.
46.94 to 79.46
55.45 to 70.95
46.94 to 71.33
55.45 to 79.46
RATIONALE
In order to get the 98% CI , we first need to find the critical
t-score. Using a t-table, we need to find (n-1) degrees of freedom, or (8-1) = 7 df and the corresponding
CI.
Using the 98% CI in the bottom row and 7 df on the far left column, we get a t-critical score of 2.998.
We also need to calculate the mean:
So we use the formula to find the confidence interval:
The lower bound is:
63.2-7.75 = 55.45
The upper bound is:
63.2+7.75 = 70.95
CONCEPT
,Confidence Intervals Using the T-Distribution
2
A table represents the number of students who passed or failed an aptitude test at two different
campuses.
South Campus North Campus
Passed 42 31
Failed 58 69
In order to determine if there is a significant difference between campuses and pass rate, the chi-square
test for association and independence should be performed.
What is the expected frequency of South Campus and passed?
50 students
36.5 students
42 students
43.7 students
RATIONALE
In order to get the expected counts we can note the formula is:
CONCEPT
Chi-Square Test for Homogeneity
3
Sukie interviewed 125 employees at her company and discovered that 21 of them planned to take an
extended vacation next year.
, What is the 95% confidence interval for this population proportion? Answer choices are rounded to
the hundredths place.
0.10 to 0.23
0.16 to 0.17
0.11 to 0.16
0.11 to 0.21
RATIONALE
In order to get the CI we want to use the following form.
First, we must determine the corresponding z*score for 95% Confidence Interval. Remember, this means
that we have 5% for the tails, meaning 5%, or 0.025, for each tail. Using a z-table, we can find the upper
z-score by finding (1 - 0.025) or 0.975 in the table.
This corresponding z-score is at 1.96.
We can know
So putting it all together:
The lower bound is:
0.168-0.065 =0.103 or 0.10
The upper bound is:
0.168+0.065 =0.233 or 0.23
