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 38 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 5% level of significance.

(a) What is the level of significance?____

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 Student's t, 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 standard normal, 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.)_____

(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 = 0.05 level, we reject the null hypothesis and conclude the data are statistically significant.

__At the α = 0.05 level, we reject the null hypothesis and conclude the data are not statistically significant.    

__At the α = 0.05 level, we fail to reject the null hypothesis and conclude the data are statistically significant.

__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.

__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.

__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

Solution:

Claim : p is less than 67%

n = 38

x = 21

Let denotes the sample proportion.

     = x/n  = 21/ 38 = 0.5526

a)

H₀: p = 0.67; H₁: p < 0.67    

b)

np = 38 * 0.67 = 25.46

nq = 38 * (1 - 0.67) = 12.54

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

The test statistic z is

z =   

=  (0.5526 - 0.67)/[0.67*(1 - 0.67)/38]

= -1.54

The value of the test statistic z = -1.54

c)

Left tailed test.

So ,

P value = P( Z< -1.54) = 0.0618

p value = 0.0618

d)

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

(Because p value is greater than 0.05)

e)

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.