In a random sample of 810 women employees it is found that 8

In a random sample of 810 women employees, it is found that 81 would prefer working for a female boss. The width of the 95 percent confidence interval for the proportion of women who prefer a female boss is:

± .0288

± .0105

± .0207

± .0196

How do you find he answer? And why is it that?

Solution

In a random sample of 810 women employees, it is found that 81 would prefer working for a female boss. The width of the 95 percent confidence interval for the proportion of women who prefer a female boss is:

± .0288

± .0105

± .0207

± .0196

How do you find the answer? And why is it that?

p=81/810=0.10

n=810

standard error = sqrt(p*(1-p)/n) =sqrt(0.1*(1-0.1)/810) =0.01054

z value for 95% =1.96

margin of error = z*se =1.96*0.01054 =0.0207

Answer is ± .0207