Question

Researchers gave a family planning survey was given to the head(s) of household in a local community. In particular, one question asked the head(s) of households to indicate if they think the community has a great need, some need, or no need of family planning counseling. In addition, they were asked to indicate how many children are in the household. Researchers want to know if the number of children in the household is impacted by family planning counseling. What can they conclude with α = 0.01?

great need	some need	no need
0 0 2 3 1 2 0 1	6 2 3 1 5 4 3 5	9 4 3 2 9 4 5 4

Conduct Tukey's Post Hoc Test for the following comparisons:
1 vs. 2: difference =  ; significant:  ---Select--- Yes No
1 vs. 3: difference =  ; significant:  ---Select--- Yes No

f) Conduct Scheffe's Post Hoc Test for the following comparisons:
1 vs. 2: test statistic =  ; significant:  ---Select--- Yes No
1 vs. 3: test statistic =  ; significant:  ---Select--- Yes No

Answer 1

To perform the post hoc-test first one way ANOVA analysis is performed in excel by using following steps,

Step 1: Write the data values in excel. The screenshot is shown below,

Step 2: DATA > Data Analysis > ANOVA: single Factor > OK.  The screenshot is shown below,

Step 3: Select Input Range: All the data values column, Alpha = 0.05. The screenshot is shown below,

The result is obtained.  The ANOVA table is shown below,

ANOVA
Source of Variation	SS	df	MS	F	P-value	F crit
Between Groups	61.75	2	30.875	8.447883	0.002033	3.4668
Within Groups	76.75	21	3.654762
Total	138.5	23

Since the significance P-value for the F-statistic is less than 0.01 at 1% significance level, there is a significant difference in means.

Post-Hoc test (Tuckey HSD Test)

Now, the Tukey multiple comparisons procedure is used to test all the pairwise comparison and identify which pair is significantly different.

The Tukey method uses the formula,

Since ni = nj, the HSD value will be same for each comparison

The q value is obtained using the q distribution table for significance level = 0.05, number of groups, k = 3, degree of freedom = N - k = 24 - 3 = 21.

The HSD value is,

Decision Rule: If,

Now,

The mean value for each groups are,

Groups	Average
great need,	1.125
some need,	3.625
no need,	5

The comparison of means are,

Groups	Average		HSD
	2.5	<	3.1173	Not Significant
	3.875	>	3.1173	Significant
	1.375	<	3.1173	Not Significant

Post-Hoc test (Scheffe Test )

The scheffe method uses the formula,

Where, the data values are obtained from ANOVA table

k = 3

F critical value = 8.447883

MSE = 3.654762

Groups	Average
	2.5	<	2.5170	Not Significant
	3.875	>	2.5170	Significant
	1.375	<	2.5170	Not Significant