In: Statistics and Probability
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) 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 standard normal, since np > 5 and nq > 5.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.
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.
(a) What is the level of significance?
Level of significance = α = 1% = 0.01
State the null and alternate hypotheses.
H0: p = 0.67; H1: p < 0.67
(b) What sampling distribution will you use?
We have
n = 40
p = 0.67
q = 1 – 0.67 = 0.33
np = 40*0.67 = 26.8
nq = 40*0.33 = 13.2
np > 5, nq > 5
The standard normal, since np > 5 and nq > 5.
What is the value of the sample test statistic?
Test statistic formula for this test is given as below:
Z = (p̂ - p)/sqrt(pq/n)
Where, p̂ = Sample proportion, p is population proportion, q = 1 - p, and n is sample size
x = number of items of interest = 21
n = sample size = 40
p̂ = x/n = 21/40 = 0.525
p = 0.67
q = 1 - p = 0.33
Z = (p̂ - p)/sqrt(pq/n)
Z = (0.525 – 0.67)/sqrt(0.67*0.33/40)
Z = -1.9503
Test statistic = -1.95
Part c
P-value = 0.0256
(by using z-table)
P-value > α = 0.01
So, we do not reject the null hypothesis
There is not sufficient evidence to conclude that the population proportion of women athletes who graduate from the university is now less than 67%.