Problem Set Ii
Chapter 12 ********** 22, 32, 40
Evan Lee Turner
#22 Random samples of five were selected from each of three populations. The sum of squares total was 100. The sum of squares due to the treatments was 40.
a. Set up the null hypothesis and the alternate hypothesis.
b. What is the decision rule? Use the .05 significance level.
c. Complete the ANOVA table. What is the value of F?
d. What is your decision regarding the null hypothesis?
SS 100
SST 40
SSE 60 k 3 n 15 null hypothesis Variance of one population equals the others alternate The variances differ
k-1 = num 2 n-k = den 12
Crit value of F 3.89 *according to appendix G Decision rule is to reject null hypothesis if the Computed value of F is greater than 6.93
Source of Variation Sum of Squares Degrees of Freedom mean square treatments 40 2 20 error 60 30 total 100 12
f= mst/mse 1.5
We reject the null hypothesis and accept the alternate hypothesis because the computed F is not greater than 3.89
#32 Evan Turner
Three assembly lines are used to produce a certain component for an airliner. To examine the production rate, a random sample of six hourly periods is chosen for each assembly line and the number of components produced during these periods for each line is recorded.
The output from a statistical software package is:
Summary
Groups Count Sum Average Variance
Line A 6 250 41.66667 0.266667
Line B 6 260 43.33333 0.666667
Line C 6 249 41.5 0.7
ANOVA
Source of Variation SS df MS F P-Value between groups 12.33333 2 6.166667 11.32653 0.001005 within groups 8.166667 15 0.544444 total 20.5 17
a. Use a .01 level of significance to test if there is a difference in the mean production of the three assembly lines.
b. Develop a 99 percent confidence
Evan Lee Turner
#22 Random samples of five were selected from each of three populations. The sum of squares total was 100. The sum of squares due to the treatments was 40.
a. Set up the null hypothesis and the alternate hypothesis.
b. What is the decision rule? Use the .05 significance level.
c. Complete the ANOVA table. What is the value of F?
d. What is your decision regarding the null hypothesis?
SS 100
SST 40
SSE 60 k 3 n 15 null hypothesis Variance of one population equals the others alternate The variances differ
k-1 = num 2 n-k = den 12
Crit value of F 3.89 *according to appendix G Decision rule is to reject null hypothesis if the Computed value of F is greater than 6.93
Source of Variation Sum of Squares Degrees of Freedom mean square treatments 40 2 20 error 60 30 total 100 12
f= mst/mse 1.5
We reject the null hypothesis and accept the alternate hypothesis because the computed F is not greater than 3.89
#32 Evan Turner
Three assembly lines are used to produce a certain component for an airliner. To examine the production rate, a random sample of six hourly periods is chosen for each assembly line and the number of components produced during these periods for each line is recorded.
The output from a statistical software package is:
Summary
Groups Count Sum Average Variance
Line A 6 250 41.66667 0.266667
Line B 6 260 43.33333 0.666667
Line C 6 249 41.5 0.7
ANOVA
Source of Variation SS df MS F P-Value between groups 12.33333 2 6.166667 11.32653 0.001005 within groups 8.166667 15 0.544444 total 20.5 17
a. Use a .01 level of significance to test if there is a difference in the mean production of the three assembly lines.
b. Develop a 99 percent confidence