Respuesta :
Testing the hypothesis, we can conclude that the increase in the West is greater than in the Midwest, as the p-value of the test is 0.0053 < 0.05.
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- The comparison is with the Midwest, in which the increase was of 6.6%. Thus, at the null hypothesis, we test if the increase in the West was also of 6.6%, that is:
[tex]H_0: \mu = 6.6[/tex]
- We want to test if the increase is greater, thus, at the alternative hypothesis, we test if the mean is greater than 6.6%, that is:
[tex]H_1: \mu > 6.6[/tex]
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- We have the standard deviation for the sample, which means that the t-distribution is used.
The test statistic is:
[tex]t = \frac{X - \mu}{\frac{s}{\sqrt{n}}}[/tex]
In which
- X is the sample mean.
- [tex]\mu[/tex] is the value tested at the null hypothesis.
- s is the standard deviation of the sample.
- n is the size of the sample.
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- 6.6 is tested at the null hypothesis, thus [tex]\mu = 6.6[/tex]
- Sample of 36, thus [tex]n = 36[/tex]
- Sample standard deviation of 2%, thus [tex]s = 2[/tex]
- Sample mean of 7.5%, thus [tex]X = 7.5[/tex]
The value of the test statistic is:
[tex]t = \frac{X - \mu}{\frac{s}{\sqrt{n}}}[/tex]
[tex]t = \frac{7.5 - 6.6}{\frac{2}{\sqrt{36}}}[/tex]
[tex]t = 2.7[/tex]
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- We are testing if the mean is more than a value, thus, the p-value of the test is a right-tailed test, with 36 - 1 = 35 degrees of freedom and t = 2.7.
- Using a t-distribution calculator, this p-value is of 0.0053.
We can conclude that the increase in the West is greater than in the Midwest, as the p-value of the test is 0.0053 < 0.05.
A similar problem is given at https://brainly.com/question/24146681
