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 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.
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.)
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 = 0.05 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.
A long-term graduation rate is 67% at the a certain university.
So this is our population proportion P.
Sample data will have : random sample = n = 40 , number of women athletes graduated = x = 23
Sample proportion :
Claim : The population proportion of women athletes who graduate from the university is now less than 67%.
a) Hence the hypothesis are :
H0: P = 0.67 v/s H1 : P < 0.67
b)
Here we will use the standard normal distribution, since np > 5 and nq > 5.
The test statistic is,
= -1.21
c)
P value = p ( z < -1.21 )
= 0.1131 ------------( using excel formula " =norm.s.dist(-1.21,1)" )
d) Here p value > ( 0.05 ) .
.At the α = 0.05 level, we fail to reject the null hypothesis and conclude the data are not statistically significant.
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.