Answer:
(a) -2.8
b) (1) Reject H0: p = 0.50 in favor of HA: p < 0.50; there is sufficient evidence to conclude that less than half of all UGA female students who, if they had a cell phone, would be willing to walk somewhere after dark that they would normally not go.
Step-by-step explanation:
The test statistic is:
![z = \frac{X - \mu}{\frac{\sigma}{\sqrt{n}}}](https://tex.z-dn.net/?f=z%20%3D%20%5Cfrac%7BX%20-%20%5Cmu%7D%7B%5Cfrac%7B%5Csigma%7D%7B%5Csqrt%7Bn%7D%7D%7D)
In which X is the sample mean,
is the value tested at the null hypothesis,
is the standard deviation and n is the size of the sample.
H0: p = 0.50
This means that:
![\mu = 0.5, \sigma = \sqrt{0.5*0.5} = 0.5](https://tex.z-dn.net/?f=%5Cmu%20%3D%200.5%2C%20%5Csigma%20%3D%20%5Csqrt%7B0.5%2A0.5%7D%20%3D%200.5)
In a random sample of 305 UGA female students, 128 responded that, if they had a cell phone, they would be willing to walk somewhere after dark that they would normally not go.
This means that ![n = 305, X = \frac{128}{305} = 0.4197](https://tex.z-dn.net/?f=n%20%3D%20305%2C%20X%20%3D%20%5Cfrac%7B128%7D%7B305%7D%20%3D%200.4197)
a) Value of the test statistic:
![z = \frac{X - \mu}{\frac{\sigma}{\sqrt{n}}}](https://tex.z-dn.net/?f=z%20%3D%20%5Cfrac%7BX%20-%20%5Cmu%7D%7B%5Cfrac%7B%5Csigma%7D%7B%5Csqrt%7Bn%7D%7D%7D)
![z = \frac{0.4197 - 0.5}{\frac{0.5}{\sqrt{305}}}](https://tex.z-dn.net/?f=z%20%3D%20%5Cfrac%7B0.4197%20-%200.5%7D%7B%5Cfrac%7B0.5%7D%7B%5Csqrt%7B305%7D%7D%7D)
![z = -2.8](https://tex.z-dn.net/?f=z%20%3D%20-2.8)
Pvalue:
We are testing the hypothesis that the proportion is less than 0.5, which means that the pvalue of the test is the pvalue of z = -2.81.
Looking at the z-table, z = -2.8 has a pvalue of 0.0026
(b) What is the correct conclusion for this hypothesis test at the 0.05 level of significance?
0.0026 < 0.05, which means that we reject the null hypothesis, that the proportion is 0.5, and accept the alternate hypothesis, that the proportion is less than 0.5, option (1).