Question

ie: 3. A person studying fathers' involvement in their children's education interviews a simple random scheduled parent-teacher sample of fathers of school-age children. One question concerns regularly conferences. Here is a two-way table of the results: Attended Attended Attended Total none some 132 10 Fathers in two-parent families Fathers in single-parent families Non-resident fathers 109 , - 38 13 15 82 298 580 Total 135 147 (a) Use appropriate graphical techniques to illustrate the relationship between these two variables. (b) Based on your graphical analysis, discuss the relationship between family type and how often fathers atended parent-teacher conferences. Support your conclusions with the appropriate statistical test.

Answer 1

a) Let's make a stacked bar graph to check the relationship between both the variables. First, each data point is divided by its row total and converted the data points into percentages. Hence, each stack bar has a total percentage of 100. The data and the graph is:

	Attended all	Attended some	Attended none	Total
Father in two-parent families	25%	39%	11%	444
Fathers in single-parent families	30%	26%	5%	38
Non-resident fathers	46%	34%	84%	98
	135	147	298	580

Chart Area Stacked Bar Graph 1.2 0.8 0.6 0.4 0.2 Two-parent Single Parent Non-Resident Attended All Attended some Attended None

There seems to be good relationship between both the variables.

b) From the graph, it looks that fathers who doesn't live with the family mostly have attended none of the meetings. For a single parent family, it is logical that they will be the most involved.  

Let's check the association with a chi-square test

Ho: There is no association between the variables

Ha: There is a significant association between the variables

The data table is as below on how the statistic is calculated:

Observed (Oi)	Expected (Ei)	(Oi-Ei)^2	(Oi-Ei)^2/Ei
109	103.34	32.036	0.310
132	112.53	379.081	3.369
203	228.12	631.014	2.766
15	8.84	37.946	4.292
10	9.63	0.137	0.014
13	19.52	42.510	2.178
11	22.80	139.240	6.107
5	24.84	393.626	15.846
82	50.35	1001.723	19.895
		SUM	54.778

Expected value is calculated by: Corresponding row total*Corresponding column total/580

Chi-square statistic = 54.778

df = 4, Hence p-value<0.001

Hence, as the p-value is so small, we can reject the null hypothesis and conclude that there is an association between both the variables.