Answered

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A recent study at UGA (March 5, 2008) suggests that students with cell phones may take more risks than students that do not have cell phones. 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. Use the above survey results to test the claim 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.
H0: p = 0.50
HA: p < 0.50
(a) What is the value of the test statistic for this hypothesis test? (Use 2 decimal places in your answer).
(b) What is the correct conclusion for this hypothesis test at the 0.05 level of significance? (Multiple Choice: Is the answer 1, 2, 3, or 4?)
(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.
(2) Do not reject H0: p = 0.50; there is insufficient 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.
(3) Do not reject H0: 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.
(4) Reject H0: p = 0.50 in favor of HA: p < 0.50; there is insufficient 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.

Sagot :

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:

[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.

H0: p = 0.50

This means that:

[tex]\mu = 0.5, \sigma = \sqrt{0.5*0.5} = 0.5[/tex]

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 [tex]n = 305, X = \frac{128}{305} = 0.4197[/tex]

a) Value of the test statistic:

[tex]z = \frac{X - \mu}{\frac{\sigma}{\sqrt{n}}}[/tex]

[tex]z = \frac{0.4197 - 0.5}{\frac{0.5}{\sqrt{305}}}[/tex]

[tex]z = -2.8[/tex]

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).

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