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?

SS100

SST40

SSE60

k3

n 15

null hypothesisVariance of one population equals the others alternateThe variances differ

k-1 = num2

n-k = den12

Crit value of F3.89*according to appendix G

Decision rule is to reject null hypothesis if the Computed value of F is greater than 6.93

Source of VariationSum of SquaresDegrees of Freedommean square treatments40220

error6030

total10012

f= mst/mse1.5

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

GroupsCountSumAverageVariance

Line A625041.666670.266667

Line B626043.333330.666667

Line C624941.50.7

ANOVA

Source of VariationSSdfMSFP-Value

between groups12.3333326.16666711.326530.001005

within groups8.166667150.544444

total20.517

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 interval for the difference in the means between Line B and Line C.

Critical Value F6.36*according to appendix G

null hypothesis*there is no difference between production Alternate hypothesis*there is a difference between production

f=mst/mse11.32654

We must reject the null hypotheis because the computed F is greater than 6.36. There is a statistical difference between the production lines at the .01 signifigance

Confidence interval = X +- Z * (std dev/sqrt(n))

X = Sample Mean

z = (1-c)/2 where c = 99%

z = (1-.99)/2

z = 0.005

Std Deviation = sqrt(variance)1.998384

n = sample size = 60.009992

Square Root of n =2.449

Line BLine C

Sample Mean43.33341.5

Std Deviation0.8160.837

n66

Top Confidence Interval "X+"43.33541.502

Lower Confidence Interval "X-"43.33141.498

#40 Evan Lee Turner

Three supermarket chains in the Denver area each claim to have the lowest overall prices. As part of an investigative study on supermarket advertising, the Denver Daily News conducted a study. First, a random sample of nine grocery items was selected. Next, the price of each selected item was checked at each of the three chains on the same day. At the .05 significance level, is there a difference in the mean prices at the supermarkets or for the items?

itemsuper$ralphslowblaws

11.121.021.07

21.141.11.21

31.721.972.08

42.222.092.32

52.42.12.3

64.044.324.15

75.054.955.05

84.684.134.67

95.525.465.86

ANOVA

Source of VariationSum of SquaresDegrees of Freedommean square treatments4020

error600

total10012

null hypothesisthe prices do not differ between stores...