Question

Suppose a cosmetics company wants to test the effectiveness of three different sales training programs. Steve, an analyst in the company's HR department, randomly assigns 18 new employees, 6 each, to the training programs. After the training program is complete, each employee takes a product knowledge and sales skills assessment test, where scores are on a scale from 0 to 100.

Steve records the assessment test results for each employee and analyzes the sample data. The distribution of test scores for each sample is approximately normal without any outliers. As population variances are unknown, Steve assumes equal variances based on the sample data because side-by-side boxplots of the samples indicate relatively equal sizes of the variances.

Steve intends to conduct a one-way ANOVA ?- test at a significance level of ?=0.05 to test if the mean assessment test scores for the three training programs are all equal. The mean square between groups, MSG, is 156.7222 and the mean square within groups, also referred to as the mean square error, MSE, is 35.0889. Steve computes the ?- statistic, which is ?=4.4379 .

QUESTION 1: Select the statement that accurately evaluates whether or not the requirements of a one-way ANOVA ?-F-test have been met in Steve's experiment. If the requirements have not been met, do not proceed to answer subsequent questions.

A) The requirements have not been met because the population variances are unknown.

B) The requirements have not been met because the sample sizes are not properly selected and are not large enough.

C) The requirements have been met because the samples are properly selected and are approximately normally distributed with equal variances.

D) The requirements have been met because the sampling distributions are approximately normal and the dependent variable is categorical.

E) The requirements have been met because the samples are properly selected, are normally distributed, and of equal sizes.

QUESTION 2: If the test requIrements have been met, compute the degrees of freedom for the means square between groups, MSG, and the degress of freedom for the mean square error, MSE.

df1 =

df2 =

QUESTION 3: Determine the critical value of ?F for Steve's hypothesis test at the significance level of ?=0.05α=0.05 using software or a
F‑distribution table . Provide the result with precision to two decimal places.

F statistic =

QUESTION 4: Select the accurate statement regarding Steve's hypothesis test decision and conclusion.

A) Steve should reject the null hypothesis. There is sufficient evidence to reject the null hypothesis because the ?-F-statistic is greater than the critical value. Therefore, Steve should conclude that one or more of the mean test scores are different.

B) Steve should reject the null hypothesis. There is sufficient evidence to reject the null hypothesis because the ?-F-statistic is greater than the critical value. Therefore, Steve should conclude that each pair of mean test scores differs.

C) Steve should reject the null hypothesis. There is sufficient evidence to reject the null hypothesis because the ?-F-statistic is less than the critical value. Therefore, Steve should conclude that at least one of the mean test scores is different from the others.

D) Steve should fail to reject the null hypothesis. There is insufficient evidence to reject the null hypothesis because the ?-F-statistic is greater than the critical value. Therefore, Steve should conclude that the mean test scores for each training program are equal.

E) Steve should fail to reject the null hypothesis. There is insufficient evidence to reject the null hypothesis because the ?-F-statistic is less than the critical value. Therefore, Steve should conclude that the mean test scores among the training programs are the same.

Answer 1

Q1) As per assumptions of ANOVA the option c) is correct.

The requirements have been met because the samples are properly selected and are approximately normally distributed with equal variances.

The samples need not be of equal size. We need randomly assigned samples which are approximately normal with equal variances all which have been met.

Q2) degrees of freedom for mean square between groups(MSG) = k-1 (where k= number of treatments / groups) in our case we have 3 different sales training programmes so K=3.

hence D.F for MSG= k-1= 3-1 = 2

similarly, D.F for mean square error(MSE) = N-k (where N is total number of observations) so in our case there were total 18 participants so N=18.

hence D.F for MSE = N-k = 18- 3 = 15

Q3) critical value of F at 0.05 level

F statistic = 3.68 ( two decimal places)

Q4) now we can clearly see that steves calculated F was 4.4379 and critical value is 3.68 so F value is greater than critical value hence there is sufficient evidence to reject the null hypothesis.

So option A) seems best.

Steve should reject the null hypothesis. There is sufficient evidence to reject the null hypothesis because the ?-F-statistic is greater than the critical value. Therefore, Steve should conclude that one or more of the mean test scores are different.

option B cannot be your answer because we cannot say from this result that each pair is different we can only say that one more are different.

Thanks and please upvote.