Question

In a study of red/green color blindness, 1000 men and 2100 women are randomly selected and tested. Among the men, 88 have red/green color blindness. Among the women, 5 have red/green color blindness. Test the claim that men have a higher rate of red/green color blindness.
(Note: Type ‘‘p_m′′‘‘p_m″ for the symbol pmpm , for example p_mnot=p_wp_mnot=p_w for the proportions are not equal, p_m>p_wp_m>p_w for the proportion of men with color blindness is larger, p_m<p_wp_m<p_w , for the proportion of men is smaller. )

(a) State the null hypothesis:

(b) State the alternative hypothesis:

(c) The test statistic is

(d) Is there sufficient evidence to support the claim that men have a higher rate of red/green color blindness than women? Use a 10 % significance level.

A. Yes
B. No

(e) Construct the 90% confidence interval for the difference between the color blindness rates of men and women.

_ <(pm−pw)<_

Answer 1

For males : n1 = 1000, x1 = 88

p̂1 = x1/n1 = 0.088

For Females : n2 = 2100, x2 = 5

p̂2 = x2/n2 = 0.0024

a) Null hypothesis:

Ho : p1 = p2

b) Alternative hypothesis:

H1 : p1 > p2

c) Pooled proportion:

p̄ = (x1+x2)/(n1+n2) = (88+5)/(1000+2100) = 0.03

Test statistic:

z = (p̂1 - p̂2)/√ [p̄*(1-p̄)*(1/n1+1/n2)] = (0.088 - 0.0024)/√[0.03*0.97*(1/1000+1/2100)] = 13.0633

d) p-value = 1- NORM.S.DIST(13.0633, 1) = 0.0000

Yes, there is sufficient evidence to support the claim that men have a higher rate of red/green color blindness than women at 0.10 significance level.

e)

90% Confidence interval for the difference between the color blindness rates of men and women:

At α = 0.1, two tailed critical value, z_c = NORM.S.INV(0.1/2) = 1.645

Lower Bound = (p̂1 - p̂2) - z_c*√ [(p̂1*(1-p̂1)/n1)+(p̂2*(1-p̂2)/n2) ]

= (0.088 - 0.0024) - 1.645*√[(0.088*0.912/1000) + (0.0024*0.9976/2100)] = 0.0708

Upper Bound = (p̂1 - p̂2) + z_c*√ [(p̂1*(1-p̂1)/n1)+(p̂2*(1-p̂2)/n2) ]

= (0.088 - 0.0024) + 1.645*√[(0.088*0.912/1000) + (0.0024*0.9976/2100)] = 0.1005

0.0708 < p1 -p2 < 0.1005