Based on the output given, answer the following question. a) Show that the standard deviation of the paired differences is 12.1879 b) Test at 5% significance level whether attending the tutorial helped to improve the students’ performance. c) Calculate a 95% confidence interval for the mean difference in the test score. Does the interval support the conclusion as in b)? Explain.
Based on the output given, answer the following question. a) Show that the standard deviation of the paired differences is 12.1879 b) Test at 5% significance level whether attending the tutorial helped to improve the students’ performance. c) Calculate a 95% confidence interval for the mean difference in the test score. Does the interval support the conclusion as in b)? Explain.
Glencoe Algebra 1, Student Edition, 9780079039897, 0079039898, 2018
18th Edition
ISBN:9780079039897
Author:Carter
Publisher:Carter
Chapter10: Statistics
Section10.1: Measures Of Center
Problem 4GP
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Topic Video
Question
Based on the output given, answer the following question.
a) Show that the standard deviation of the paired differences is 12.1879
b) Test at 5% significance level whether attending the tutorial helped to improve the
students’ performance.
c) Calculate a 95% confidence interval for the
interval support the conclusion as in b)? Explain.
![Paired T for before -
after
N
Mean
StDev
SE Mean
before
10
49.3000
12.8932
4.0772
7.6478
12.1879
10
64.4000
2.4184
3.8541
after
Difference
10
-15.1000
95% upper bound for mean difference: -8.0349
T-Test of mean difference = 0 (vs < 0): T-Value = -3.92
P-Value = 0.002](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F1e10d969-7e3b-425f-b977-60bb033302e7%2F60675529-04e3-4875-9def-7dc2970b69fb%2Fpju5093_processed.png&w=3840&q=75)
Transcribed Image Text:Paired T for before -
after
N
Mean
StDev
SE Mean
before
10
49.3000
12.8932
4.0772
7.6478
12.1879
10
64.4000
2.4184
3.8541
after
Difference
10
-15.1000
95% upper bound for mean difference: -8.0349
T-Test of mean difference = 0 (vs < 0): T-Value = -3.92
P-Value = 0.002
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