Question

a random sample of 81 executives (these 81 include both male and female) is drawn for the purpose of estimating the population proportion of females and the mean age of all female executives. The sample contains 33 female executives for those ladies, the sample mean and standard deviation are 46.5 years and 6.8 years, respectively. We first want to build a confidence interval for the proportion of female executives in the population of all executives.

a.:check that the conditions to build a confidence interval are satisfied.

b.: find the margin of error for a 95% confidence interval

c.: find a 95% confidence interval for proportion of female executives

d.:interpret the confidence interval

e. based on your confidence interval, is it reasonable to state that half of all executives are female?

Answer 1

Answer:

a)

Given,

sample n = 81

n > 30

The assumptions are the arbitrary variable X must be binomial .

And we can normalize X by taking away mean and by partitioning with standard deviation and structure confidence interval utilizing standard ordinary distribution. since n>30.

Critical esteem is from Z distribution.

b)

p = x/n = 33/81 = 0.4074

Here at 95% CI, z value is 1.96

Margin of error E = z*sqrt(pqn)

= 1.96*sqrt(0.4074(1-0.4074)/81)

= 0.1070

c)

95% CI = xbar +/- z*s/sqrt(n)

= 0.4074 +/- 0.1070

= (0.3004 , 0.5144)

d)

Here we can say that, the proportion lies in the 95% confidence interval.

e)

Here we can't carefully says that half of the executes are females utilizing the determined interim. May be we can say that 30 % to 51% of executes are females.