Question

6. Researchers took random samples of right-handed men and women and determined the numbers whose feet were the same size (L-R), the left foot larger than the right (L > R) and the number where the left foot was smaller than the right (L < R) Do the following data indicate that gender has an effect on the development of foot asymmetry? Men 2 10 28 Women 55 18 14 (a) State appropriate null and alternate hypotheses. (b) Compute the value of an appropriate test statistic (c) State your conclusions. (d) What is the approximate p-value? C1] L2] 01] x3(2) = 4.61, χる5(2) = 5.99, χ2 25(2) = 7.38, χ2 1 (2) = 9.21, χ2 05(2) = 10.59; x3 (3) = 6.25, 2 5(3) = 7.81, χ225(3) = 9.35,墻1(3) = 11.34, χ·005(3)-12.84; x3 (4) = 7.78, χ2 5(4) = 9.49, χ2 25(4) = 11.14, xh(4) = 13.27, χ205(4) = 14.86.

Answer 1

(a)

H0: Null Hypothesis: Gender has no effect on the development of foot asymmetry.

HA: Alternative Hypothesis: Gender has effect on the development of foot asymmetry.

(b)

Observed frequencies:

	L>R	L=R	L<R	Total
Men	2	10	28	40
Women	55	18	14	87
Total	57	28	42	127

Expected frequencies:

	L>R	L=R	L<R	Total
Men	57X40/127=17.95	8.82	13.23	40
Women	39.05	19.18	28.77	87
Total	57	28	42	127

The Test statistic ( $\chi ^{2}$ ) is calculated as follows:

O	E	(O - E)2/E
2	17.95	14.18
10	8.82	0.16
28	13.23	16.50
55	39.05	6.52
18	19.18	0.07
14	28.77	7.58
Total = $\chi ^{2}$ =		45.00

Test statistic = $\chi ^{2}$ = 45.00

(c)

Take $\alpha$ = 0.05

ndf = (r - 1) X (c - 1)

= (3 - 1) X (2 - 1) = 2

From Table, critical value of $\chi ^{2}$ = 5.99

Sinc the calculated value of $\chi ^{2}$ = 45.00 is greater than critical value of $\chi ^{2}$ = 5.99, the difference is significant. Reject null hypothesis.

Conclusion:

The data support the claim that Gender has effect on the development of foot asymmetry.

(d)

Test statistic = $\chi ^{2}$ = 45.00

ndf = 2

By Technology, p - value = 0.00001