In a study of red/green color blindness, 650 men and 2700 women are randomly sel

Question

In a study of red/green color blindness, 650 men and 2700 women are randomly selected and tested. Among the men, 55 have red/green color blindness. Among the women, 6 have red/green color blindness. Test the claim that men have a higher rate of red/green color blindness.

1) The test statistic is:
2) 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 % significance level?

A. Yes
B. No

2. Construct the 99 % confidence interval for the difference between the color blindness rates of men and women.
_______<(p1p2)<_________

Which of the following is the correct interpretation for your answer in part 2?
A. We can be 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 % 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 % 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

Explanation / Answer

(a)

Data:    

n1 = 650   

n2 = 2700   

p1 = 55/650 = 0.084615385   

p2 = 6/2700 = 0.002222222   

Hypotheses:    

Ho: p1 p2    

Ha: p1 > p2    

Decision Rule:    

= 0.01   

Critical z- score = 2.326347874   

Reject Ho if z > 2.326347874   

Test Statistic:    

Average proportion, p = (n1p1 + n2p2)/(n1 + n2) = (650 * 0.0846153846153846 + 2700 * 0.00222222222222222)/(650 + 2700) = 0.018208955

q = 1 - p = 1 - 0.0182089552238806 = 0.981791045

SE = [pq * {(1/n1) + (1/n2)}] = (0.0182089552238806 * 0.981791044776119 * ((1/650) + (1/2700))) = 0.005841655

z = (p1 - p2)/SE = (0.0846153846153846 - 0.00222222222222222)/0.00584165480844686 = 14.10442162

p- value = 0   

Decision (in terms of the hypotheses):    

Since 14.10442162 > 2.326347874 we reject Ho and accept Ha

Conclusion (in terms of the problem):

There is sufficient evidence that p1 > p2   

(b)

n1 = 650

n2 = 2700

p1 = 0.084615385

p2 = 0.002222222

% = 99

Pooled Proportion, p = (n1 p1 + n2 p2)/(n1 + n2) = (650 * 0.0846153846153846 + 2700 * 0.00222222222222222)/(650 + 2700) = 0.018208955

q = 1 - p = 1 - 0.0182089552238806 = 0.981791045

SE = (pq * ((1/n1) + (1/n2))) = (0.0182089552238806 * 0.981791044776119 * ((1/650) + (1/2700))) = 0.005841655

z- score = 2.575829304

Width of the confidence interval = z * SE = 2.57582930354892 * 0.00584165480844686 = 0.015047106

Lower Limit of the confidence interval = (p1 - p2) - width = 0.0823931623931624 - 0.0150471056368148 = 0.067346057

Upper Limit of the confidence interval = (p1 - p2) + width = 0.0823931623931624 + 0.0150471056368148 = 0.097440268

The 99% confidence interval is [0.067, 0.097]

A. We can be 99 % confident that the difference between the rates of red/green color blindness for men and women lies in the interval.

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