Question

Women athletes at a certain university have a long-term graduation rate of 67%. Over the past several years, a random sample of 36 women athletes at the school showed that 23 eventually graduated. Does this indicate that the population proportion of women athletes who graduate from the university is now less than 67%? Use a 5% level of significance.

(a) What is the level of significance?
State the null and alternate hypotheses.
A. H0: p = 0.67; H1: p < 0.67
B. H0: p = 0.67; H1: p ≠ 0.67
C. H0: p < 0.67; H1: p = 0.67
D. H0: p = 0.67; H1: p > 0.67
(b) What sampling distribution will you use?
A. The standard normal, since np > 5 and nq > 5.
B. The standard normal, since np < 5 and nq < 5.
C. The Student's t, since np > 5 and nq > 5.
D. The Student's t, since np < 5 and nq < 5.
What is the value of the sample test statistic?
(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?
A. At the α = 0.05 level, we reject the null hypothesis and conclude the data are statistically significant.
B. At the α = 0.05 level, we reject the null hypothesis and conclude the data are not statistically significant.
C. At the α = 0.05 level, we fail to reject the null hypothesis and conclude the data are statistically significant.
D. At the α = 0.05 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.
A. There is sufficient evidence at the 0.05 level to conclude that the true proportion of women athletes who graduate is less than 0.67.
B. There is insufficient evidence at the 0.05 level to conclude that the true proportion of women athletes who graduate is less than 0.67.

Answer 1

A. H0: p = 0.67; H1: p < 0.67

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

Test statistic = - 0.397 ;

Pvalue = 0.3457;

D. At the α = 0.05 level, we fail to reject the null hypothesis and conclude the data are not statistically significant.

B. There is insufficient evidence at the 0.05 level to conclude that the true proportion of women athletes who graduate is less than 0.67.

Explanation:

p = 0.67 ; q = 1 - p = 1 - 0.67

Sample size, n = 36

x = 23

Test for normality :

(36*0.67) = 24.12

(36 * (1-0.67)) = 11.88

For a normal distribution :

np ≥ 5 and n(1 - p) ≥ 5

The hypothesis :

H0 : p = 0.67

H1 : p < 0.67

The Test statistic :

Z = (phat - p) / √[(p(1 -p))/n]

Phat = x / n = 23 / 36

Z = ((24/36) - 0.67)) / √[(0.67(1 -0.67))/36]

Z = - 0.031 / 0.0783687

Z = - 0.396983

Z = - 0.397

Usong the Pvalue from Z calculator ;

Pvalue of Z = 0.3457

If Pvalue < α ; Reject H0 ; If otherwise, Fail to reject H0