Respuesta :
Answer:
a)
The null hypothesis is [tex]H_0: p = 0.91[/tex].
The alternate hypothesis is [tex]H_1: p < 0.91[/tex].
The decision rule is: accept the null hypothesis for [tex]z > -1.645[/tex], reject the null hypothesis for [tex]z < -1.645[/tex].
Since [tex]z = -1.79 < -1.645[/tex], we reject the null hypothesis and accept the alternate hypothesis that that percentage of athletes who graduate is less than for the student population at large.
b)
The p-value for this test is 0.0367. Since this p-value is less than the significance level of [tex]\alpha = 0.05[/tex], we reject the null hypothesis and accept the alternate hypothesis that that percentage of athletes who graduate is less than for the student population at large.
Step-by-step explanation:
Question a:
Perform the appropriate hypothesis test to determine whether this is significant evidence that the percentage of athletes who graduate is less than for the student population at large:
At the null hypothesis, we test if the proportion is the same as the student population, of 91%. Thus:
[tex]H_0: p = 0.91[/tex]
At the alternate hypothesis, we test that the proportion for athletes is less than 91%, that is:
[tex]H_1: p < 0.91[/tex]
The test statistic is:
[tex]z = \frac{X - \mu}{\frac{\sigma}{\sqrt{n}}}[/tex]
In which X is the sample mean, [tex]\mu[/tex] is the value tested at the null hypothesis, [tex]\sigma[/tex] is the standard deviation and n is the size of the sample.
Test if the proportion is less at the 0.05 level:
The critical value is z with a p-value of 0.05, that is, z = -1.645. Thus, the decision rule is: accept the null hypothesis for [tex]z > -1.645[/tex], reject the null hypothesis for [tex]z < -1.645[/tex].
0.91 is tested at the null hypothesis:
This means that [tex]\mu = 0.91, \sigma = \sqrt{0.91*0.09}[/tex]
A sports reporter contacted 152 athletes randomly sampled from that same university and time period and found that 132 of them had graduated within 6 years.
This means that [tex]n = 152, X = \frac{132}{152} = 0.8684[/tex]
Value of the test statistic:
[tex]z = \frac{X - \mu}{\frac{\sigma}{\sqrt{n}}}[/tex]
[tex]z = \frac{0.8684 - 0.91}{\frac{\sqrt{0.91*0.09}}{\sqrt{152}}}[/tex]
[tex]z = -1.79[/tex]
Since [tex]z = -1.79 < -1.645[/tex], we reject the null hypothesis and accept the alternate hypothesis that that percentage of athletes who graduate is less than for the student population at large.
(b) (3 points) Calculate the P-value for this test. Explain how this P-value can be use to test the hypotheses in part (a).
The p-value of the test is the probability of finding a sample proportion of 0.8684 or below. This is the p-value of z = -1.79.
Looking a the z-table, z = -1.79 has a p-value of 0.0367.
The p-value for this test is 0.0367. Since this p-value is less than the significance level of [tex]\alpha = 0.05[/tex], we reject the null hypothesis and accept the alternate hypothesis that that percentage of athletes who graduate is less than for the student population at large.
