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:

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

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.

H0: p = 0.50

This means that:

$\mu = 0.5, \sigma = \sqrt{0.5*0.5} = 0.5$

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 $n = 305, X = \frac{128}{305} = 0.4197$

a) Value of the test statistic:

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

$z = \frac{0.4197 - 0.5}{\frac{0.5}{\sqrt{305}}}$

$z = -2.8$

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).

6 0

3 years ago

Maya drew a scale drawing of her room where 1.5 cm represented 4 ft. If the room was actually 16 ft by 12 ft what were the dimen

Butoxors [25]

If 1.5cm = 4ft

1.5÷4 =0.375

0.375cm= 1ft

16×0.375=6
12×0.375=4.5

hopefully this is correct and helped! ♡

7 0

3 years ago