Question

Women athletes at the a certain university have a long-term graduation rate of 67%. Over the past several years, a random sample of 40 women athletes at the school showed that 21 eventually graduated. Does this indicate that the population proportion of women athletes who graduate from the university is now less than 67%? Use a 1% level of significance.

(a) State the null and alternate hypotheses.

H0: p = 0.67; H1: p ≠ 0.67

H0: p < 0.67; H1: p = 0.67

H0: p = 0.67; H1: p > 0.67

H0: p = 0.67; H1: p < 0.67

(b) What sampling distribution will you use?

The standard normal, since np < 5 and nq < 5.

The Student's t, since np > 5 and nq > 5.

The standard normal, since np > 5 and nq > 5.

The Student's t, since np < 5 and nq < 5.

What is the value of the sample test statistic? (Round your answer to two decimal places.)

(c) Find the P-value of the test statistic. (Round your answer to four decimal places.)

Sketch the sampling distribution and show the area corresponding to the P-value.

(d) Based on your answers in parts (a) to (c), will you reject or fail to reject the null hypothesis? Are the data statistically significant at level α?

At the α = 0.01 level, we reject the null hypothesis and conclude the data are statistically significant.

At the α = 0.01 level, we reject the null hypothesis and conclude the data are not statistically significant.

At the α = 0.01 level, we fail to reject the null hypothesis and conclude the data are statistically significant.

At the α = 0.01 level, we fail to reject the null hypothesis and conclude the data are not statistically significant.

(e) Interpret your conclusion in the context of the application.

There is sufficient evidence at the 0.01 level to conclude that the true proportion of women athletes who graduate is less than 0.67.

There is insufficient evidence at the 0.01 level to conclude that the true proportion of women athletes who graduate is less than 0.67.

Answer 1

Solution

[NOTE: Answers are given below. Detailed Working and Back-up Theory follow at the end.]

Part (a)

The null and alternate hypotheses.

H0: p = 0.67; H1: p < 0.67

Last Option Answer 1

Part (b)

Sampling distribution

The standard normal, since np > 5 and npq > 5.

Third Option Answer 1

Value of the sample test statistic = - 2.02 Answer 3

Part (c)

P-value of the test statistic = 0.0217 Answer 4

Part (d)

Decision

At the α = 0.01 level, we fail to reject the null hypothesis and conclude the data are not statistically significant.

Fourth Option Answer 5

Part (e)

Interpretation

There is insufficient evidence at the 0.01 level to conclude that the true proportion of women athletes who graduate is less than 0.67.

Second Option Answer 6

Detailed Working and Back-up Theory

Let X = Number of women athletes who graduate from the school.

Then, X ~ B(n, p), where n = sample size and p = probability or population proportion of women athletes who graduate.

Given n = 40, p = 0.67 and X = 21

Claim : Proportion of women athletes who graduate is less than 0.67.

Hypotheses:

Null H₀ : p = p₀ = 0.67 Vs H_A : p < 0.67

Test Statistic:

Z = (pcap - p₀)/√{p₀(1 - p₀)/n} where pcap = X/n = sample proportion and n = sample size.

Calculations:

Z = {(21/40) – 0.67}/√{0.67 x 0.33/40}

= - 0.15/0.0743

= - 2.0188

Distribution, Critical Value and p-value:

Under H₀, distribution of Z can be approximated by Standard Normal Distribution, provided

np₀ and np₀(1 - p₀) are both greater than 10.

So, given a level of significance of α = 1% (i.e., 0.01), Critical Value = lower 1% of N(0, 1), and

p-value = P(Z < Zcal)

Using Excel Functions of N(0, 1), Critical Value = - 2.32 and p-value = 0.0217

Decision:

Since Zcal > Zcrit, p-value > α, H₀ is accepted.

Conclusion :

There is NOT enough evidence to suggest that the claim is valid.

DONE