Question

In Texas, state law requires that for regular classroom settings in elementary grades K-4, the student-to-teacher ratio can be no more than 22:1. Schools may request a waiver to exceed this ratio. A hypothesis test is performed to determine if the proportion of schools needing such a waiver is above 25% using a random sample of 85 schools and a significance level of 5%. What conclusion should be drawn if 32 of the schools surveyed need a waiver to exceed the required student-to-teacher ratio

Answer 1

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:

$H_0: p = 0.25$

At the alternate hypothesis, we test that the proportion is above 25%, that is:

$H_a: p > 0.25$

The test statistic is:

$z = \frac{X - \mu}{\frac{\sigma}{\sqrt{n}}}$

In which X is the sample mean, $\mu$ is the value tested at the null hypothesis, $\sigma$ is the standard deviation and n is the size of the sample.

0.25 is tested at the null hypothesis:

This means that $\mu = 0.25, \sigma = \sqrt{0.25*0.75}$

Sample of 85 schools, 32 need a waiver:

This means that $n = 85, p = \frac{32}{85} = 0.3765$

Value of the test statistic:

$z = \frac{X - \mu}{\frac{\sigma}{\sqrt{n}}}$

$z = \frac{0.3765 - 0.25}{\frac{\sqrt{0.25*0.75}}{\sqrt{85}}}$

$z = 2.69$

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%.