Question

1. [10 points] Using the data shown in the following table, test at 0.05 level of significance whether a person's ability in mathematics is independent of his or her ability in statistics. LowM and HighM denote the numbers of low and high ability individuals in mathematics. For statistics, these are LowS and HighS LowM HighM LowS 67 HighS 14 35 29

Answer 1

Claim: A person's ability in mathematics is independent of his or her ability in statistics.

The null and alternative hypothesis is

H0: A person's ability in mathematics is independent of his or her ability in statistics.

H1: A person's ability in mathematics is not independent of his or her ability in statistics.

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

	LowM	HighM	Total
LowS	67	29	96
HighS	14	35	49
Total	81	64	145

O	E	(O-E)	(O-E)^2	(O-E)^2/E
67	53.62759	13.37241	178.8215	3.334505
29	42.37241	-13.3724	178.8215	4.220233
14	27.37241	-13.3724	178.8215	6.532908
35	21.62759	13.37241	178.8215	8.268211
			Total	22.36

$\chi ^{2}=\sum \frac{(O-E)^{2}}{E} =22.36$

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

Critical value = 3.841

Test statistic > critical value we reject null hypothesis.

Conclusion:

A person's ability in mathematics is not independent of his or her ability in statistics.