Question

A supervisor in the Department of Rehabilitive Services is critical of the performance of one of her counselors.   The counselor is expected to arrange job training for those in need of vocational rehabilitation so that they may find employment.   Yet the counselor has managed to place just over 30% of his clients. The counselor argues that he is doing a good job and that the reason for his overall low rate of placement is that most of his clients are severely disabled, which makes them very difficult to place.  

The counselor’s case load is presented in the accompanying table.   Percentage the table appropriately, and evaluate who is correct -- the supervisor or the counselor?

Job Placement	Not Severely Disabled	Severely Disabled
Not Placed	42	100
Placed	24	40

Compute the value of the chi-square for the data in the table above. From these data, would you say that the relationship between level of disability and whether or not they are placed is weak or strong? Explain your answer....There are good examples in the book for how to do this, so don't leave any valuable points on the table! (Or, use the Excel spreadsheet in the resources.)

a) Calculate the expected frequencies, the deviations (observed - expected).   Square the deviation and divide by the expected frequencies
(in other words, calculate (O-E)²/E.   Then sum to an overall measure of discrepancy.
b) Convert the Chi-square into a p-value using degrees of freedom (d.f.) = (c-1)(r-1)
c) Is your p-value sufficiently small to reject Ho at alpha = .05?   In other words, is there enough statistical evidence for a difference in visits based on disability status?

Answer 1

using excel we have

Chi-Square Test
Observed Frequencies
	Disabled			Calculations
Job lacement	Not surely	Surely	Total		fo-fe
Not placed	42	100	142		-3.49515	3.495146
Placed	24	40	64		3.495146	-3.49515
Total	66	140	206
Expected Frequencies
	Disabled
Job lacement	Not surely	Surely	Total		(fo-fe)^2/fe
Not placed	45.49515	96.50485	142		0.268513	0.126585
Placed	20.50485	43.49515	64		0.595763	0.28086
Total	66	140	206
Data
Level of Significance	0.05
Number of Rows	2
Number of Columns	2
Degrees of Freedom	1
Results
Critical Value	3.841459
Chi-Square Test Statistic	1.271721
p-Value	0.259444
Do not reject the null hypothesis

a) the expected frequencies are given in table

b) the Chi-square into a p-value using degrees of freedom is 1
c) p value is 0.2594 ,No p-value is not sufficiently small to reject Ho at alpha = .05. there is not enough statistical evidence for a difference in visits based on disability status?