Question

them: good grades, athletic ability, or popularity. Children were classified by sex and their response to the survey question. With the significance level of 0.05, test he nui hypothesis that a child's answer to the survey question is independent of heir sex 4. In a study the children were asked which of the following was most important to 50 91 60 30 Boys Girls 130

Answer 1

4)

	Good grades	Athletic	Popularity	Total
Boys	117	60	50	227
Girls	130	30	91	251
Total	247	90	141	478

Claim: A child answers to the survey question is independent of their sex.

The null and alternative hypothesis is

H0: A child answers to the survey question is independent of their sex.

H1: A child answers to the survey question is not independent of their sex.

Level of significance = 0.05

Test statistic is

$\chi ^{2}=\sum \frac{(O-E)^{2}}{E}$

O: Observed frequency
E: Expected frequency.
E = ( Row total*Column total) / Grand total

O	E	(O-E)	(O-E)^2	(O-E)^2/E
117	117.2992	-0.29916	0.089499	0.000763
60	42.74059	17.25941	297.8874	6.969661
50	66.96025	-16.9603	287.6501	4.295834
130	129.7008	0.299163	0.089499	0.00069
30	47.25941	-17.2594	297.8874	6.303239
91	74.03975	16.96025	287.6501	3.885077
			Total	21.455

21.455

Degrees of freedom = ( Number of rows - 1 ) * ( Number of column - 1) = ( 2 - 1) * (3 - 1) = 1 * 2 = 2

Critical value = 5.991

( From chi-square table)

Test statistic > Critical value we reject null hypothesis.

Conclusion:

A child answers to the survey question is not independent of their sex.

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