Question

The head housekeeper at the Palisades Point Humane Habitat, a no-kill adoption organization for cats and dogs, wonders whether (a) the individual staff member or (b) the day of the week might impact the average number of the pets’ sleeping spaces or cages cleaned during one eight-hour shift. She asks an office assistant to run some statistical tests to explore these two questions, using the data in the following table. Based on the available evidence, and testing at the 10% level of significance, is there any difference between (a) the mean number of sleeping spaces or cages cleaned according to staff members and (b) the mean number of sleeping spaces or cages rooms cleaned according to days of the week? Explain, utilizing appropriate inferential analytic methods. [COMMENTS & HINTS: Use software to your advantage. Be sure to state explicitly the hypotheses (both null and research/alternative), the conclusions (technical and contextual), and the p-values both for staff and for days of the week.]

	S	M	T	W	T	F	S
Ari	34	32	43	33	37	44	33
Betty	46	38	35	42	39	41	37
Carlos	35	39	49	29	38	36	34
Dante	47	43	46	47	44	49	47
Enid	46	42	33	37	45	41	33
Fatima	44	44	46	39	41	44	37

Answer 1

SOLUTION

Anova in excel

1. Arrange the data in excel as shown below.

2. Select two-factor anova without replication from the data analysis tab.

3. Input the value as shown.

4. The output will be generated as follows.

Hypothesis 1

Ho: The average pets' sleeping spaces or cages cleaning is equal for all staff members.
H1: The average pets' sleeping spaces or cages cleaning is not equal for all staff members.

From the ANOVA output, we look at the pvalue which is 0.002(highlighted in yellow). Since the pvalue is less than 0.1, we reject the null hypothesis and conclude that the average pets' sleeping spaces or cages cleaning is not equal for all staff members.

Hypothesis 2

Ho: The average pets' sleeping spaces or cages cleaning is equal for all days of the week.
H1: The average pets' sleeping spaces or cages cleaning is not equal for all days of the week.

From the ANOVA output, we look at the pvalue which is 0.1755(highlighted in blue). Since the pvalue is greater than 0.1, we fail reject the null hypothesis and conclude that the average pets' sleeping spaces or cages cleaning is equal for all days of the week.