When only two treatments are involved, ANOVA and the Student's t-test (Chapter 11) result in the same conclusions. Also, for computed test statistics, f² = F. To demonstrate this relationship, use the following example. Fourteen randomly selected students enrolled in a history course were divided into two groups, one consisting of six students who took the course in the normal lecture format. The other group of eight students took the course as a distance course format. At the end of the course, each group was examined with a 50-item test. The following is a list of the number correct for each of the two groups. Traditional Lecture Distance 36 45 31 35 38 45 31 35 31 49 40 33 42 42 Click here for the Excel Data File a-1. Complete the ANOVA table. (Round your SS, MS, and F values to 2 decimal places and p-value and F crit to 4 decimal places.) Source of Variation Treatement Error Total SS df MS a-2. Use a α = 0.05 level of significance, find or compute the critical value of F. (Round your answer to 2 decimal places.) The critical value of F is b. Using the t-test, compute t. (Negative amount should be indicated by a minus sign. Round your answer to 3 decimal places.) t is c. There is any difference in the mean test scores. (Reject or Do Not Reject) F p-value F crit Ho. There is (a statistically significant difference or no statistically significant difference) in the mean scoures between lecture and distance-based formats.
When only two treatments are involved, ANOVA and the Student's t-test (Chapter 11) result in the same conclusions. Also, for computed test statistics, f² = F. To demonstrate this relationship, use the following example. Fourteen randomly selected students enrolled in a history course were divided into two groups, one consisting of six students who took the course in the normal lecture format. The other group of eight students took the course as a distance course format. At the end of the course, each group was examined with a 50-item test. The following is a list of the number correct for each of the two groups. Traditional Lecture Distance 36 45 31 35 38 45 31 35 31 49 40 33 42 42 Click here for the Excel Data File a-1. Complete the ANOVA table. (Round your SS, MS, and F values to 2 decimal places and p-value and F crit to 4 decimal places.) Source of Variation Treatement Error Total SS df MS a-2. Use a α = 0.05 level of significance, find or compute the critical value of F. (Round your answer to 2 decimal places.) The critical value of F is b. Using the t-test, compute t. (Negative amount should be indicated by a minus sign. Round your answer to 3 decimal places.) t is c. There is any difference in the mean test scores. (Reject or Do Not Reject) F p-value F crit Ho. There is (a statistically significant difference or no statistically significant difference) in the mean scoures between lecture and distance-based formats.
Glencoe Algebra 1, Student Edition, 9780079039897, 0079039898, 2018
18th Edition
ISBN:9780079039897
Author:Carter
Publisher:Carter
Chapter4: Equations Of Linear Functions
Section4.5: Correlation And Causation
Problem 11PPS
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Transcribed Image Text:When only two treatments are involved, ANOVA and the Student's t-test (Chapter 11) result in the same conclusions. Also, for computed test statistics, f² = F. To demonstrate this relationship, use the following example. Fourteen randomly selected
students enrolled in a history course were divided into two groups, one consisting of six students who took the course in the normal lecture format. The other group of eight students took the course as a distance course format. At the end of the
course, each group was examined with a 50-item test. The following is a list of the number correct for each of the two groups.
Traditional Lecture
Distance
36
45
31
35
38
45
31
35
31
49
40
33
42
42
Click here for the Excel Data File
a-1. Complete the ANOVA table. (Round your SS, MS, and F values to 2 decimal places and p-value and F crit to 4 decimal places.)
Source of Variation
Treatement
Error
Total
SS
df
MS
a-2. Use a α = 0.05 level of significance, find or compute the critical value of F. (Round your answer to 2 decimal places.)
The critical value of F is
b. Using the t-test, compute t. (Negative amount should be indicated by a minus sign. Round your answer to 3 decimal places.)
t is
c. There is any difference in the mean test scores.
(Reject or Do Not Reject)
F
p-value
F crit
Ho. There is (a statistically significant difference or no statistically significant difference)
in the mean scoures between lecture and distance-based formats.
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