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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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