Question

2. The CEO of a software company is committed to expanding the proportion of highly qualified women in the organization’s staff of salespersons. He believes that the proportion of women in similar sales positions across the country in 2004 is less than 45%. Hoping to find support for his belief, he directs his assistant to collect a random sample of salespersons employed by his company, which is thought to be representative of sales staffs of competing organizations in the industry. The collected random sample of size 50 showed that only 18 were women.

(A) State the appropriate null and alternative hypotheses in this situation.

(B) Test this CEO’s belief at the =.05 significance level and report the p-value. Do you find statistical support for his hypothesis that the proportion of women in similar sales positions across the country is less than 40%?

(C) Suppose the sample size above is 100, instead of 50, and the sample proportion is again 0.36. Would this change your results in (B)? Explain your answer.

Answer 1

2)We are asked to test the hypothesis that proportion of women in similar sales positions across the country in 2004 is less than 45%.

A) Given that n=50, x=18

$p=\frac{x}{n}=\frac{18}{50}=0.36$(sample proportion of women in sales positions)

P₀=45%=0.45(population proportion of women in sales positions)

Q₀=1-P=1-0.45=0.55

Null hypothesis

H₀:P₀=0.45

i.e., proportion of women in similar sales positions across the country is 45%

Alternate hypothesis

H₁:P₀<0.45

i.e., proportion of women in similar sales positions across the country is less than 45%

B)Test statistic under H₀

$z=\frac{p-P_0}{\sqrt{\frac{P_0Q_0}{n}}}\sim N(0,1)=\frac{0.36-0.45}{\sqrt{\frac{0.45*0.55}{50}}}=-1.279$

Given significance level=.05, we have obtained a p-value 0.1004.

Conclusion

Since p-value is greater than significance level, we accept the null hypothesis.

Hence we conclude that proportion of women in similar sales positions across the country is 45%.

We are now asked to conduct the hypothesis for the alternate hypothesis that proportion of women in similar sales positions across the country is less than 40%.

Now, P=0.40 and Q=1-P=1-0.40=0.60

Null hypothesis

H₀:P=0.40

i.e., proportion of women in similar sales positions is equal to 40%

Alternate hypothesis

H₁:P<0.40

Test statistic under H₀

P - PO 0.36 – 0.40 N(0,1) = 0.40*0.60 50 = -0.577 PQ n

Conclusion

p-value obtained here is 0.2819 at significance level of 0.05

Since p-value is greater than significance level, we accept the null hypothesis.

Hence we conclude that the proportion of women in similar sales positions across the country is 40%.

C) Increasing the sample size increases the test statistic which may lead to change in results since test statistic value and sample size are directly proportional to each other. Hence increasing the sample size would change the test statistic leading to difference in results.