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 36 women athletes at the school showed that 18 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) What is the level of significance?


State the null and alternate hypotheses.

H₀: p < 0.67; H₁: p = 0.67H₀: p = 0.67; H₁: p < 0.67    H₀: p = 0.67; H₁: p > 0.67H₀: 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 standard normal, 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.)

Answer 1

Solution:

Given:Women athletes at the a certain university have a long-term graduation rate of 67%

that is p = proportion of women athletes who graduate from the university = 0.67

Sample size = n= 36

x = Number of of women athletes who graduated from the university = 18

level of significance = 0.01

Part a) What is the level of significance?

The  level of significance = 0.01

State the null and alternative hypotheses.

Since we have to test if  the population proportion of women athletes who graduate from the university is now less than 67%, this is left tailed test.

Thus hypothesis are:

H₀: p = 0.67; H₁: p < 0.67

Part b) What sampling distribution will you use?

n*p = 36*0.67 = 24.12 > 5

n*q = n*(1-p) = 36 * ( 1-0.67) =11.88 > 5

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

What is the value of the sample test statistic?

where

thus

Part c) Find the P-value of the test statistic.

For left tailed test , p-value is:

p-value = P(Z < z test statistic)

p-value = P(Z < -2.17 )

Look in z table for z = -2.1 and 0.07 and find corresponding area.

P( Z< -2.17) = 0.0150

thus

p-value = P(Z < -2.17 )

p-value =0.0150

Conclusion:

Since p-value= 0.0150 > 0.01  level of significance., we fail to reject H0 and

thus at 0.01  level of significance, this data  does not indicate that the population proportion of women athletes who graduate from the university is now less than 67%.