Women athletes at the a certain university have a long-term graduation rate of 6

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 37 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 10% 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 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.

(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 ?

At the = 0.10 level, we reject the null hypothesis and conclude the data are statistically significant.At the = 0.10 level, we reject the null hypothesis and conclude the data are not statistically significant.    At the = 0.10 level, we fail to reject the null hypothesis and conclude the data are statistically significant.At the = 0.10 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 0.10 level to conclude that the true proportion of women athletes who graduate is less than 0.67.There is insufficient evidence at the 0.10 level to conclude that the true proportion of women athletes who graduate is less than 0.67.  

Explanation / Answer

Given that,
possibile chances (x)=23
sample size(n)=37
success rate ( p )= x/n = 0.6216
success probability,( po )=0.67
failure probability,( qo) = 0.33
null, Ho:p=0.67
alternate, women athletes who graduate from the university is now less than 67, H1: p<0.67
level of significance, = 0.1
from standard normal table,left tailed z /2 =1.28
since our test is left-tailed
reject Ho, if zo < -1.28
we use test statistic z proportion = p-po/sqrt(poqo/n)
zo=0.62162-0.67/(sqrt(0.2211)/37)
zo =-0.6258
| zo | =0.6258
critical value
the value of |z | at los 0.1% is 1.28
we got |zo| =0.626 & | z | =1.28
make decision
hence value of |zo | < | z | and here we do not reject Ho
p-value: left tail - Ha : ( p < -0.62583 ) = 0.26571
hence value of p0.1 < 0.26571,here we do not reject Ho
ANSWERS
---------------
null, Ho:p=0.67
alternate, H1: p<0.67
test statistic: -0.6258
critical value: -1.28
decision: do not reject Ho
p-value: 0.26571

we don't have evidence that proportion of women athletes who graduate from the university
is now less than 67

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