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

Solution

From the given data we find that

n1 = 451

p1 = 410/451 = 0.9091

p2 = 505/550 = 0.9182

Standard error is computed as

= sqrt(0.9091 * (1-0.9091)/451 + 0.9182*(1-0.9182)/550)

= 0.01788

Confidence interval is computed using the formula as

Here z for 95% 1.96

Hence CI

=>  (0.9091 - 0.9182) - 1.96*0.01788, AND (0.9091 - 0.9182 )+ 1.96*0.01788

=>  = (-0.0442 , 0.0260)

Interpretation:

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