Determine the truth (T/F) of each statement with regard to the one-factor analysis of variance technique.

a. The mean squares are measures of variance.

b. “There is no difference between the mean values of the random variable at the various levels of the test factor” is a possible interpretation of the null hypothesis.

c. “The factor being tested has no effect on the random variable x” is a possible interpretation of the alternative hypothesis.

d. “There is no variance among the mean values of x for each of the different factor levels” is a possible interpretation of the null hypothesis.

e. The “partitioning” of the variance occurs when SS(total) is separated into SS(factor) and SS(error).

f. We will reject the null hypothesis and conclude that the factor has an effect on the variable when the amount of variance assigned to the factor is significantly larger than the variance assigned to error.

g. In order to apply the F-test, the sample size from each factor level must be the same.

h. In order to apply the F-test, the sample standard deviation from each factor level must be the same.

i. If 20 is subtracted from every data value, then the calculated value of the F statistic is also reduced by 20.

When the calculated value of F, F, is greater than the table value for F, j. The decision will be fail to reject Ho .

k. The conclusion will be the factor being tested does have an effect on the variable.

Independent samples were collected to test the effect a factor had on a variable. The data are summarized in this ANOVA table:

SS df Factor 810 2 Error 720 8 Total 1530 10 Is there sufficient evidence to reject the null hypothesis that all levels of the test factor have the same effect on the variable?

l. The null hypothesis could be A  B

 C  D.

m. The calculated value of F is 1.125.

n. The critical value of F for   0.05 is 6.06.

o. The null hypothesis can be rejected at

  0.05.