Question

An opinion poll was conducted at a hospital in Texarkana. There were 451 males and 550 females. One of the question asked during the poll was: “Should nicotine test be made mandatory for all employees? – Yes/No” 410 males and 505 females responded “Yes” to this question. Construct a 95% confidence interval for the difference in proportion of males vs females who favor mandatory nicotine testing? Interpret the confidence interval in the context of this question.

Answer 1

Here, , n1 = 451 , n2 = 550
p1cap = 0.9091 , p2cap = 0.9182

Standard Error, sigma(p1cap - p2cap),
SE = sqrt(p1cap * (1-p1cap)/n1 + p2cap * (1-p2cap)/n2)
SE = sqrt(0.9091 * (1-0.9091)/451 + 0.9182*(1-0.9182)/550)
SE = 0.0179

For 0.95 CI, z-value = 1.96
Confidence Interval,
CI = (p1cap - p2cap - z*SE, p1cap - p2cap + z*SE)
CI = (0.9091 - 0.9182 - 1.96*0.0179, 0.9091 - 0.9182 + 1.96*0.0179)
CI = (-0.0442 , 0.0260)

we are 95% confident that the difference in proportion of males vs females who favor mandatory nicotine testing is between -0.0442 and 0.0260