Respuesta :
Answer:
Step-by-step explanation:
A hypothesis test is performed to determine if the proportion of schools needing such a waiver is above 25%
At the null hypothesis, we test that the proportion is of 25%, that is:
[tex]H_0: p = 0.25[/tex]
At the alternate hypothesis, we test that the proportion is above 25%, that is:
[tex]H_a: p > 0.25[/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.
0.25 is tested at the null hypothesis:
This means that [tex]\mu = 0.25, \sigma = \sqrt{0.25*0.75}[/tex]
Sample of 85 schools, 32 need a waiver:
This means that [tex]n = 85, p = \frac{32}{85} = 0.3765[/tex]
Value of the test statistic:
[tex]z = \frac{X - \mu}{\frac{\sigma}{\sqrt{n}}}[/tex]
[tex]z = \frac{0.3765 - 0.25}{\frac{\sqrt{0.25*0.75}}{\sqrt{85}}}[/tex]
[tex]z = 2.69[/tex]
P-value of the test and decision:
The p-value of the test is the probability of finding a sample proportion above 0.3765, which is 1 subtracted by the p-value of z = 2.69.
Looking at the z-table, z = 2.69 has a p-value of 0.9964
1 - 0.9964 = 0.0036
The p-value of the test is 0.0036 < 0.05, thus we reject the null hypothesis that the proportion is 25%, and accept the alternate hypothesis that the proportion of schools needing such a waiver is above 25%.
