Question

In order to determine the possible effect of chemical treatment on women’s hair, an experiment was conducted. The results are given in the accompanying table

TREATMENT	GROWTH	NO GROWTH
CHEMICALLY TREATED	118	22
UNTREATED	120	40

1. a) State the null hypothesis and the alternative hypothesis

2. b) Calculate (and show) the expected frequency for each cell. Place each expected value in parentheses ( ) in the same cell (and to the right) as the observed value to which it corresponds

3. c) Using the formula: X^2= ∑(O-E)^2/(E)  calculate the chi-square value.

4. d) At the 5% level of significance, what is the chi-square critical value?

5. e) What decision would you make?

6. f) Draw a conclusion (in the context of the problem)?

Answer 1

a)

Hypotheses are:

H0: There is no effect of chemical treatment on women’s hair.

Ha: There is a effect of chemical treatment on women’s hair.

(b)

Following table shows the row total and column total:

	GROWTH	NO GROWTH	Total
CHEMICALLY TREATED	118	22	140
UNTREATED	120	40	160
Total	238	62	300

Expected frequencies will be calculated as follows:

Following table shows the expected frequencies:

	GROWTH	NO GROWTH	Total
CHEMICALLY TREATED	111.067	28.933	140
UNTREATED	126.933	33.067	160
Total	238	62	300

	GROWTH	NO GROWTH	Total
CHEMICALLY TREATED	118 (111.067)	22(28.933)	140
UNTREATED	120(126.933)	40(33.067)	160
Total	238	62	300

c)

Following table shows the calculations for chi square test statistics:

O	E	(O-E)^2/E
118	111.067	0.432770211
120	126.933	0.378676065
22	28.933	1.661303321
40	33.067	1.453609006
Total		3.926358604

Following is the test statistics:

d)

Degree of freedom: df =( number of rows -1)*(number of columns-1) = (2-1)*(2-1)=1

The critical value using excel function "=chiinv(0.05,1)" is 3.841.

e)

Since so we reject the null hypothesis.

f)

There is evidence to conclude that there is a effect of chemical treatment on women’s hair.