Question

(1 pt) In a study of red/green color blindness, 950 men and 2050 women are randomly selected and tested. Among the men, 89 have red/green color blindness. Among the women, 6 have red/green color blindness. Construct the 99% confidence interval for the difference between the color blindness rates of men and women.

Answer 1

Answer: (0.066,0.116)

Step-by-step explanation:

The confidence interval for proportion is given by :-

$p_1-p_2\pm z_{\alpha/2}\sqrt{\dfrac{p_1(1-p_1)}{n_1}+\dfrac{p_2(1-p_2)}{n_2}}$

Given : The proportion of men have red/green color blindness = $p_1=\dfrac{89}{950}\approx0.094$

The proportion of women have red/green color blindness = $p_2=\dfrac{6}{2050}\approx0.003$

Significance level : $\alpha=1-0.99=0.01$

Critical value : $z_{\alpha/2}=z_{0.005}=\pm2.576$

Now, the 99% confidence interval for the difference between the color blindness rates of men and women will be:-

$(0.094-0.003)\pm (2.576)\sqrt{\dfrac{0.094(1-0.094)}{950}+\dfrac{0.003(1-0.003)}{2050}}\approx0.091\pm 0.025\\\\=(0.09-0.025,0.09+0.025)=(0.066,\ 0.116)$

Hence, the 99% confidence interval for the difference between the color blindness rates of men and women= (0.066,0.116)