77231/assignments/295419?module item id-1084333 Previous Problem List Next (2 po

Question

77231/assignments/295419?module item id-1084333 Previous Problem List Next (2 points) 1. In a study of red/green color blindness, 600 men and 2700 women are randomly selected and tested. Among the men, 51 have red/green color bilindness. Among the women, 6 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% significance level? OA. Yes B. No 2. Construct the 99% confidence interval for the difference between the color blindness rates of men and women (Pi P2)

Explanation / Answer

p1 = 51/600 = 0.085

p2 = 6/2700 = 0.002

H0: p1 = p2

H1: p1 > p2

Pooled proportion (P) = (p1 * n1 + p2 * n2)/(n1 + n2)

                                   = (0.085 * 600 + 0.002 * 2700)/(600 + 2700) = 0.017

SE = sqrt(P * (1 - P) * (1/n1 + 1/n2))

      = sqrt(0.017 * (1 - 0.017) * (1/600 + 1/2700))

      = 0.0058

The test statistic z = (p1 - p2)/SE

                              = (0.085 - 0.002)/0.0085 = 9.76

P-value = P(Z > 9.76)

              = 1 - P(Z < 9.76) = 1 - 1 = 0

At 0.01% significance level as the P-value is less than the significance level (0 < 0.0001), so the null hypothesis is rected.

Yes there is sufficient evidence to support the claim.

2) The 99% confidence interval is

(p1 - p2) +/- z* * SE

= (0.085 - 0.002) +/- 2.58 * 0.0058

= 0.0065 +/- 0.015

= -0.0085, 0.0215

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