Question

(21 point) A company is criticized because only 13 of 43 people in executive-level positions are women. The company complains that although this proportion is lower than it might wish, it’s not a surprising value given that only 40% of all its employees are women.

(a) Test an appropriate hypothesis and state your conclusion. Use α = 0.05. Be sure the appropriate assumptions and conditions are satisfied before you proceed. (Hint: conduct a oneproportion z-test)

Step 1: State null and alternative hypothesis.

Step 2: Assumptions and conditions check, and decide to conduct a one-proportion z-test.

Step 3: Compute the sample statistics and find p-value.

Step 4: Interpret you p-value, compare it with α = 0.05 and make your decision.

(b) Construct a 95% confidence interval for the proportion that women are in executive-level positions and interpret it. (Hint: construct a one-proportion z-interval and be sure the appropriate assumptions and conditions are satisfied before you proceed. )

Answer 1

Given :

x = 13

n = 43

p = 40% = 0.40

(a)

step 1 : Null and alternative hypothesis :

    (claim)

step 2 :Assumptions and conditions

Assuming that the sample is a random sample and 43 people are less than 10% of all people.

np = 43*0.40 = 17.2 >10

n(1-p) = 43*(1-0.40) = 25.8>10

so both np ≥ 10 and n(1-p) ≥ 10

Therefore all the comditions are satisfied to conduct one proportion z test .

step 3 :sample statistic and p-value

Therefore sample statistic is -1.31

To get p-value using excel function ,

=NORMSDIST(z)

=NORMSDIST(-1.31)

= 0.0951

Therefore p-value = 0.0951

step 4 :

Decision rule : If p-value is less than α , then we reject Ho.

As p-value (0.0951) is greater than α (0.05) , we fail to reject Ho .

Conclusion :

Therefore there is not sufficient evidence to support the claim that the proportion is lower .

(b)

A sample is simple randaom sample.

Therefoe both ≥ 10 and ≥ 10

As the conditions are satisfied , we can use one proportion z-interval.

95% confidence interval for the proportion

c=95% = 0.95

Using excel function ,

Now plug all the values in the formula ,

The 95% confidence interval for the proportion that women are in executive-level positions is (0.165053 , 0.439599)

Therefore we are 95% confident that the true proportion of all women are in executive-level positions is between 0.165053 and 0.439599