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Sagot :
Answer:
The p-value for the one-sided Hypothesis test described in this example is 0.3121.
Step-by-step explanation:
Test the hypothesis that more than 50% of people plan on voting for the levy.
At the null hypothesis, we test that the proportion is 50%, that is:
[tex]H_0: p = 0.5[/tex]
At the alternate hypothesis, we test if this proportion is above 50%, that is:
[tex]H_a: p > 0.5[/tex]
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.
0.5 is tested at the null hypothesis:
This means that [tex]\mu = 0.5, \sigma = \sqrt{0.5*0.5} = 0.5[/tex]
Of these 150 respondents, 78 people say they plan on voting for the levy.
This means that [tex]n = 150, X = \frac{78}{150} = 0.52[/tex]
Value of the test statistic:
[tex]z = \frac{X - \mu}{\frac{\sigma}{\sqrt{n}}}[/tex]
[tex]z = \frac{0.52 - 0.5}{\frac{0.5}{\sqrt{150}}}[/tex]
[tex]z = 0.49[/tex]
Pvalue of the test:
The pvalue of the test is the probability of finding a proportion above 0.52, which is 1 subtracted by the pvalue of z = 0.49.
Looking at the z-table, z = 0.49 has a pvalue of 0.6879.
1 - 0.6879 = 0.3121
The p-value for the one-sided Hypothesis test described in this example is 0.3121.
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