question archive in a study of red/green color blindness, 900 men and 3000 women are randomly selected and tested
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in a study of red/green color blindness, 900 men and 3000 women are randomly selected and tested. Among the men, 79 have red/green color blindness. Among the women, 9 have red/green color blindness. Test the claim that men have a higher rate of red/green color blindness. (Note: Type "p_m" for the symbol pm , for example p_m not = p_w for the proportions are not equal, p_m > p_w for the proportion of men with color blindness is larger, p_m < p_w , for the proportion of men is smaller. ) (a) State the null hypothesis: P_m=p_w (b) State the alternative hypothesis: P_m>p_w (c) The test statistic is [j (d) Is there sufficient evidence to support the claim that men have a higher rate of red/green color blindness than women? Use a 5 % significance level. 0 A. Yes OB.No (e) Construct the 95 % confidence interval for the difference between the color blindness rates of men and women.
Null hypothesis; Ho : P_m = P_w
Alternative Hypothesis; Ha : P_m > P_w
Sample proportion for males; p1 = 79 / 900 = 0.0878
Sample proportion for females; p2 = 9 / 3000 = 0.003
Combined proportion; p = (79 + 9) / (900 + 3000) = 88 / 3900 = 0.0226
We will calculate Z-test statistic for sample proportions
= (p1 - p2) / Sqrt( p * (1 - p) * (1 / n1 + 1 / n2)
= (0.0878 - 0.003) / Sqrt(0.0226 * (1 - 0.0226) * (1 / 3000 + 1/900) )
Test statistic = 15.0126
p-value < 0.0001
Since p-value < significance level which means it is significant and we can reject the null hypothesis
Yes, there is sufficient evidence to support the claim
95% confidence interval for the difference between the proportions; CI = (p1 - p2) +- Z * Sqrt( p * (1 - p) * (1 / n1 + 1 / n2)
Lower bound = (p1 - p2) - Z * Sqrt( p * (1 - p) * (1 / n1 + 1 / n2)
= (0.0878 - 0.003) - 1.96 * Sqrt(0.0226 * (1 - 0.0226) * (1 / 3000 + 1/900)) = 0.07373
Upper bound = (p1 - p2) + Z * Sqrt( p * (1 - p) * (1 / n1 + 1 / n2)
= (0.0878 - 0.003) + 1.96 * Sqrt(0.0226 * (1 - 0.0226) * (1 / 3000 + 1/900)) = 0.09587
95% confidence interval is (0.07373, 0.09587)
Please let me know in the comments section in case of any confusion