In a study of red/green color blindness. 1000 men and 2800 women are randomly se
ID: 3155019 • Letter: I
Question
In a study of red/green color blindness. 1000 men and 2800 women are randomly selected and tested Among the men. 85 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 Is there sufficient evidence to support the claim that men have a higher rate of red/green color blindness than women using the 0.05% significance level? Construct the 95% confidence interval for the difference between the color blindness rates of men and women Which of the following is the correct interpretation for your answer in part 2?Explanation / Answer
1.
Formulating the hypotheses
Ho: p1 - p2 <= 0
Ha: p1 - p2 > 0
Here, we see that pdo = 0 , the hypothesized population proportion difference.
Getting p1^ and p2^,
p1^ = x1/n1 = 0.085
p2 = x2/n2 = 0.002142857
Also, the standard error of the difference is
sd = sqrt[ p1 (1 - p1) / n1 + p2 (1 - p2) / n2] = 0.008862204
Thus,
z = [p1 - p2 - pdo]/sd = 9.349495843 [ANSWER, TEST STATISTIC]
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As significance level = 0.0005 , then the critical z is
zcrit = 3.290526731
Also, the p value is
P = 4.40332*10^-21 [ANSWER, P VALUE]
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As z > 3.29, and P < 0.0005, we REJECT THE NULL HYPOTHESIS.
Hence, YES, THERE IS SUFFICIENT EVIDENCE. [ANSWER]
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2.
For the 95% confidence level, then
alpha/2 = (1 - confidence level)/2 = 0.025
z(alpha/2) = 1.959963985
Margin of error = z(alpha/2)*sd = 0.017369601
lower bound = p1^ - p2^ - z(alpha/2) * sd = 0.065487541
upper bound = p1^ - p2^ + z(alpha/2) * sd = 0.100226744
Thus, the confidence interval is
( 0.065487541 , 0.100226744 ) [ANSWER]
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OPTION A: We can be 95% confident that the difference between the rates of ref/green color blindness for men and women lies in the interval. [CONCLUSION]
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