Question

Offspring of fruit flies may have red or brown bodies and normal wings or short wings. Genetic theory predicts that these traits will appear in the ratio​ 9:3:3:1 (9 red​, ​normal: 3 red​, ​short: 3 brown​, ​normal: 1 brown​, ​short). A researcher checks 100 such flies and finds the distribution of the traits to be 55​, 20​, 14​, and 11​, respectively.

Compute the​ chi-square statistic. x²=

critical value?

P-value?

Suppose you double the amount of flies to 200 and the distribution of the traits respectively

what would the new chi-square statistic be?

critical value?

P-Value?

Answer 1

Genetic theory predicts that these traits will appear in the ratio​ 9:3:3:1 (9 red​, ​normal: 3 red​, ​short: 3 brown​, ​normal: 1 brown​, ​short) so proportion of each color will be

P(red,normal) = 9 / (9+3+3+1) = 0.5625

P(red,short) = 3 / (9+3+3+1) = 0.1875

P(brown, normal) = 3 / (9+3+3+1) = 0.1875

P(brown, short) = 1 / (9+3+3+1) = 0.0625

Following table shows the calculations for chi-square statistic. x²:

	O	p	E=p*100	(O-E)^2/E
Red, normal	55	0.5625	56.25	0.02777778
Red, short	20	0.1875	18.75	0.08333333
Brown, normal	14	0.1875	18.75	1.20333333
Brown, short	11	0.0625	6.25	3.61
Total	100	1	100	4.92444444

The test statitics is

Degree of freedom: df = 4-1=3

Let level of significance is alpha=0.05

The critical value of chi square test statistics using exce function "=CHIINV(0.05,3)" is 7.81.

The p-value using excel function "=CHIDIST(4.92,3)" is 0.1777.

Since p-value is greater than level of significance so we fail to reject the null hypothesis.

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For sample size 200:

Following table shows the calculations for chi-square statistic. x²:

	O	p	E=p*200	(O-E)^2/E
Red, normal	110	0.5625	112.5	0.05555556
Red, short	40	0.1875	37.5	0.16666667
Brown, normal	28	0.1875	37.5	2.40666667
Brown, short	22	0.0625	12.5	7.22
Total	200	1	200	9.84888889

The test statitics is

Degree of freedom: df = 4-1=3

Let level of significance is alpha=0.05

The critical value of chi square test statistics using exce function "=CHIINV(0.05,3)" is 7.81.

The p-value using excel function "=CHIDIST(9.84,3)" is 0.0200

Since p-value is less than level of significance so we reject the null hypothesis.