Question

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In a genetic study of chromosome structures, 132 individuals are classified according to the type of structural chromosome aberration and carriers in their parents. The following counts are obtained.

Type of Aberration	One Parent	Neither Parent	Total
Presumably Innocuous	27	20	47
Substantially Unbalanced	36	49	85
Total	63	69	132

Test the null hypothesis that type of aberration is independent of parental carrier.

Test Statistic:

According to your table, the P-value is bounded by:

<P−value<

Is there sufficient evidence to reject independence?  Choose one Yes No

Answer 1

Solution:

We have to test the null hypothesis that type of aberration is independent of parental carrier.

Hypothesis are:

H0: type of aberration is independent of parental carrier.

Vs

H1: type of aberration is NOT independent of parental carrier.

Test Statistic:

C] มี

Where

O_ij = Observed frequencies for i^th row and j^th column.

E_ij = Expected frequencies for i^th row and j^th column.

Where

E; Rixos N

df = ( R - 1) X (C - 1)

R = Number of Rows = 2

C = Number of Columns = 2

Type of Aberration	One Parent	Neither Parent	Total
Presumably Innocuous	27	20	R1 =47
Substantially Unbalanced	36	49	R2 =85
Total	C1 = 63	C2 =69	N =132

thus

R1 x C E11 = 47 x 63 -= 22.432 132

R1 x C2 E12= 47 x 69 -= 24.568 132

R2 x C E21 = 1 85 x 63 -= 40.568 132 N

E227 R2 x C2 N 85 x 69 -= 44.432 132

Thus

Oij: Observed frequency	Eij: Expected Frequency	Oij2/Eij
27	22.432	32.498
20	24.568	16.281
36	40.568	31.946
49	44.432	54.038
N = 132		== 134.764 乙日

Thus

C] มี

x2 = 134.764 – 132

Test statistic:

x2 = 2.764

the P-value is bounded by:

df = ( R - 1) X (C - 1)

df = ( 2 - 1) X (2 - 1)

df = 1 X 1

df = 1

X2975 X2100 X2050 X2025 A www X2995 0.000 0.010 0.072 0.207 0.412 X2990 0.000 0.020 0.115 0.297 0.554 0.001 0.051 0.216 0.484 0.831 X950 0.004 0.103 0.352 0.711 1.145 X2000 0.010 0.211 0.584 1.064 1.610 2.706 4.605 6.251 7.779 9.236 3.841 5.991 7.815 9.488 11.070 5.024 7.378 9.348 11.143 12.833

fall between 2.706 and 3.841 , corresponding right tail area is between 0.050 and 0.100

x2 = 2.764

thus

0.050 < p-value < 0.100

Is there sufficient evidence to reject independence?

Since p-value > 0.05 significance level , we fail to reject H0,

That there is not sufficient evidence to reject independence.

Thus answer is: No.