Question

A behavioral therapist records the number of children who are spanked and not spanked by their parents as a form of punishment in a sample of parents who were spanked by their parents as children. The following table shows the results of a chi-square goodness-for-fit test. If we expect to observe equal frequencies, then compute a chi-square goodness-of-fit test using a .05 level of significance and decide to retain or reject the null hypothesis. (This is the entire question as worded in the book, no other information is available).

Child Not Spanked        Child Spanked

     28                                      46

Answer 1

Null hypothesis:Ho: there is equal proportion for number of children who are spanked and not spanked by their parents

Alternate hypothesis:The proportion is different for number of children who are spanked and not spanked by their parents

degree of freedom =categories-1=

1

for 1 df and 0.05 level of signifcance critical region       χ2=

3.841

applying chi square goodness of fit test:

	category	frequency	Oi	Ei=total*p	R2i=(Oi-Ei)/√Ei	R2i=(Oi-Ei)2/Ei
16.470	Spanked	1/2	28	37.00	-1.48	2.189
27.990	Not spanked	1/2	46	37.00	1.48	2.189
	total	1.000	74	74		4.3784

as test statistic X² =4.3784 falls in rejection region ; we reject null hypothesis and conclude that The proportion is different for number of children who are spanked and not spanked by their parents