Question

The following three independent random samples are obtained from three normally distributed populations with equal variances. The dependent variable is starting hourly wage, and the groups are the types of position (internship, co-op, work study). Round answers to 4 decimal places.

Internship	Co-op	Work Study
9.25	10.5	10.75
9.5	9.75	10
10.75	11	10.5
12	10.75	11.25
10.25	10.25	9.75
10.75	9.25	10.25
10	11.5	9
9.75	9.75	10
10.25	9.5	11.25
12.75	11	10.75
10	11.5	8.75
8.25	10	10.25
9	9.5	9
10	11.75	8.75
10	9.5	8.75
11.5	13.5	9.75
10.25	11.25	10.75
11	11.75	11.25
9.75	11.25	10.75
10	11.5	11

Use Excel to conduct a single-factor ANOVA to determine if the group means are equal using α=0.05

Group means:
Internship:
Co-op:
Work Study:

Fill in the summary table for the ANOVA test:

	SS	df	MS
Between
Within
Total

From this table, obtain the necessary statistics for the ANOVA:

ANOVA summary statistics:
Test Statistic =

p-value =

Conclusion:

Answer 1

Null hypothesis

Alternative hypothesis

The means are not equal.

Source of Variation	SS	df	MS
Between Groups	4.1896	2	2.0948
Within Groups	56.3719	57	0.9890
Total	60.5615	59

Test Statistic =2.1181

p-value =0.1296

P value is =0.1296 >0.05

Therefore, we fail to reject H0 at

Conclusion:We do not have enough evidence to reject the null hypothesis at that the population means are all equal.

We can directly use here one way anova by Excel.

Step 1 ) Enter data in Excel.

Step 2) Data >>Data analysis >> One way anova >>Select data >>click on label in first row >>ok