Question

	Biological Offspring
Parental Handedness (father × mother)	Right-Handed	Left-Handed
Right × Right	294	37
Right × Left	29	6
Left × Right	28	6

(a)	Find the values in cells (2,2) and (3,2) of the expected table.

(b) Can a chi-square analysis be performed on the above table?

(A) No, because the population is not normal. (B) Yes, because only some of the expected frequencies are less than 5. (C) No, because at least one of the observed frequencies is less than 5. (D) Yes, because at least one of the expected frequencies is less than 5. (E) No, because at least one of the expected frequencies is less than 5. (F) Yes, because at least one of the observed frequencies is less than 5. (G) Yes, because all of the observed frequencies are at least 5.

(c)

Combine the last two rows in the above table to create a new 2 × 2 table. (The meaning of the 2nd row in this new table would be "at least one of the parents is left-handed".) If we use the resulting 2 × 2 table to test the hypothesis that the handedness of the biological offspring is independent of the handedness of the parents using the 5% significance level, what is the value of the test statistic?

(d)	Find the critical value for the test in (c).

(e)

What is the conclusion? (A) We conclude that the factors are independent since the answer in (c) is less than or equal to the answer in (d) (B) Do not reject the hypothesis of independence since the answer in (c) is greater than or equal to the answer in (d) (C) Reject the hypothesis of independence since the answer in (c) is less than or equal to the answer in (d) (D) Reject the hypothesis of independence since the answer in (c) is greater than the answer in (d) (E) Do not reject the hypothesis of independence since the answer in (c) is less than or equal to the answer in (d) (F) We conclude that the factors are independent since the answer in (c) is greater than the answer in (d) (G) We conclude that the factors are dependent since the answer in (c) is less than or equal to the answer in (d)

Answer 1

Ei=row total*column total/grand total

from above

a)

values in cells (2,2) and (3,2) are 4.2875 and 4.1650

b)

E) No, because at least one of the expected frequencies is less than 5.

c)

Applying chi square test of independence:

Expected	Ei=row total*column total/grand total	RH	LH	Total
	R*R	290.4525	40.5475	331.00
	R*L	60.5475	8.4525	69.00
	total	351.00	49.00	400.00
chi square    χ2	=(Oi-Ei)2/Ei	RH	LH	Total
	R*R	0.043	0.310	0.3537
	R*L	0.208	1.489	1.6967
	total	0.2512	1.7993	2.050
test statistic X2 =	2.0504

d)

for 1 df and 0.05 level , critical value       χ2=

3.841

e)

(E) Do not reject the hypothesis of independence since
the answer in (c) is less than or equal to the answer in (d)