Question

1. In a study of red/green color blindness, 850 850 men and 2950 2950 women are randomly selected and tested. Among the men, 79 79 have red/green color blindness. Among the women, 8 8 have red/green color blindness. Test the claim that men have a higher rate of red/green color blindness. The test statistic is The p-value is Is there sufficient evidence to support the claim that men have a higher rate of red/green color blindness than women using the 0.01 0.01 % significance level? A. Yes B. No 2. Construct the 99 99 % confidence interval for the difference between the color blindness rates of men and women. <( p 1 − p 2 )< <(p1−p2)< Which of the following is the correct interpretation for your answer in part 2? A. We can be 99 99 % confident that the difference between the rates of red/green color blindness for men and women lies in the interval B. We can be 99 99 % confident that that the difference between the rates of red/green color blindness for men and women in the sample lies in the interval C. There is a 99 99 % chance that that the difference between the rates of red/green color blindness for men and women lies in the interval D. None of the above

Answer 1

Solution:

Here, we have to use two sample z test for population proportions.

H₀: p₁ = p₂ versus H_a: p₁ > p₂

Where, p₁ is proportion of men having red/green color blindness and p₂ is proportion of women having red/green color blindness.

This is upper tailed or right tailed test.

WE are given

Level of significance = α = 0.01

X1=79, X2=8, n1=850, n2=2950

Sample proportions are given as below:

P1=X1/n1 = 79/850= 0.092941176

P2=X2/n2=8/2950= 0.002711864

(P1 – P2) = 0.092941176 - 0.002711864 = 0.090229312

Test statistic is given as below:

Z = (P1 – P2) / sqrt[(P1*(1 – P1)/N1) + (P2*(1 – P2)/N2)]

Z = (0.092941176 - 0.002711864) / sqrt(0.092941176*(1 - 0.092941176)/850)+( 0.002711864*( 1 - 0.002711864)/2950)

Z = 15.4966

P-value = 0.00

P-value < α = 0.01

So, we reject H₀

Yes, there is sufficient evidence to support the claim that men have a higher rate of red/green color blindness than women using the 0.01level of significance.

Confidence interval for difference between two population proportions:

Confidence interval = (P1 – P2) ± Z*sqrt[(P1*(1 – P1)/N1) + (P2*(1 – P2)/N2)]

Z = 2.5758 (by using z-table)

Confidence interval = (0.092941176 - 0.002711864) ± 2.5758*sqrt(0.092941176*(1 - 0.092941176)/850)+( 0.002711864*( 1 - 0.002711864)/2950)

Confidence interval = (0.092941176 - 0.002711864) ± 2.5758* 0.0100

Confidence interval = (0.092941176 - 0.002711864) ± 0.0258

Confidence interval = 0.090229312 ± 0.0258

Lower limit = 0.090229312 - 0.0258 =0.0645

Upper limit = 0.090229312 + 0.0258 = 0.1160

Interpretation:

C. There is a 99 % chance that that the difference between the rates of red/green color blindness for men and women lies in the interval.