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UNIT 5 — MILESTONE 5
8 questions were answered correctly.
16 questions were answered incorrectly.
1
Mike tabulated the following values for heights in inches of seven of his friends: 65, 71, 74, 61, 66, 70, and 72. The
sample standard deviation is 4.577.
Select the 95% confidence interval for Mike's set of data.
59.95 to 76.91
59.95 to 72.67
64.19 to 76.91
64.19 to 72.67
RATIONALE
In order to get the 95% CI , we first need to find the critical t-score. Using a t-table, we need to find (n-1) degrees
of freedom, or (7-1) = 6 df and the corresponding CI.
Using the 95% CI in the bottom row and 6 df on the far left column, we get a t-critical score of 2.447.
We also need to calculate the mean:
https://capella.sophia.org/spcc/introduction-to-statistics-2/milestone_take_feedbacks/5465629 1/21
,8/1/2020 Sophia :: Welcome
So we use the formula to find the confidence interval:
The lower bound is:
68.43 - 4.23 = 64.20
The upper bound is:
68.43 +4.23 = 72.66
*note that when rounding you can get values that might be slightly different but the values should be very close to
what is calculated here.
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.
East Campus West Campus
Passed 39 45
Failed 61 55
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 East Campus and passed?
39 students
42 students
48.3 students
50.5 students
RATIONALE
https://capella.sophia.org/spcc/introduction-to-statistics-2/milestone_take_feedbacks/5465629 2/21
, 8/1/2020 Sophia :: Welcome
In order to get the expected counts we can note the formula is:
CONCEPT
Chi-Square Test for Homogeneity
3
Brad recorded the number of visitors at the local science museum during the week:
Day Visitors
Tuesday 18
Wednesday 24
Thursday 28
Friday 30
He expected to see 25 visitors each day. To answer whether the number of visitors follows a uniform distribution, a
chi-square test for goodness of fit should be performed. (alpha = 0.10)
What is the chi-squared test statistic? Answers are rounded to the nearest hundredth.
3.36
1.40
1.12
2.54
RATIONALE
Using the chi-square formula we can note the test statistic is
https://capella.sophia.org/spcc/introduction-to-statistics-2/milestone_take_feedbacks/5465629 3/21
Score 8/24
Sorry, you didn't pass this Milestone. Get in touch with a Learning Coach for help at
.
UNIT 5 — MILESTONE 5
8 questions were answered correctly.
16 questions were answered incorrectly.
1
Mike tabulated the following values for heights in inches of seven of his friends: 65, 71, 74, 61, 66, 70, and 72. The
sample standard deviation is 4.577.
Select the 95% confidence interval for Mike's set of data.
59.95 to 76.91
59.95 to 72.67
64.19 to 76.91
64.19 to 72.67
RATIONALE
In order to get the 95% CI , we first need to find the critical t-score. Using a t-table, we need to find (n-1) degrees
of freedom, or (7-1) = 6 df and the corresponding CI.
Using the 95% CI in the bottom row and 6 df on the far left column, we get a t-critical score of 2.447.
We also need to calculate the mean:
https://capella.sophia.org/spcc/introduction-to-statistics-2/milestone_take_feedbacks/5465629 1/21
,8/1/2020 Sophia :: Welcome
So we use the formula to find the confidence interval:
The lower bound is:
68.43 - 4.23 = 64.20
The upper bound is:
68.43 +4.23 = 72.66
*note that when rounding you can get values that might be slightly different but the values should be very close to
what is calculated here.
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.
East Campus West Campus
Passed 39 45
Failed 61 55
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 East Campus and passed?
39 students
42 students
48.3 students
50.5 students
RATIONALE
https://capella.sophia.org/spcc/introduction-to-statistics-2/milestone_take_feedbacks/5465629 2/21
, 8/1/2020 Sophia :: Welcome
In order to get the expected counts we can note the formula is:
CONCEPT
Chi-Square Test for Homogeneity
3
Brad recorded the number of visitors at the local science museum during the week:
Day Visitors
Tuesday 18
Wednesday 24
Thursday 28
Friday 30
He expected to see 25 visitors each day. To answer whether the number of visitors follows a uniform distribution, a
chi-square test for goodness of fit should be performed. (alpha = 0.10)
What is the chi-squared test statistic? Answers are rounded to the nearest hundredth.
3.36
1.40
1.12
2.54
RATIONALE
Using the chi-square formula we can note the test statistic is
https://capella.sophia.org/spcc/introduction-to-statistics-2/milestone_take_feedbacks/5465629 3/21
