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Sagot :
Answer:
The p-value for this test is 0.1375.
Step-by-step explanation:
A student believes that the average grade on the statistics final examination was 87. The student wants to test whether the average is different from 87 at 90% level of confidence.
At the null hypothesis, we test if the average is 87, that is:
[tex]H_0: \mu = 87[/tex]
At the alternate hypothesis, we test if the average is different from 87, that is:
[tex]H_a: \mu \neq 87[/tex]
The test statistic is:
We have the standard deviation for the sample, which means that the t-distribution is used to solve this question.
[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, [tex]s[/tex] is the standard deviation of the sample and n is the size of the sample.
87 is tested at the null hypothesis:
This means that [tex]\mu = 87[/tex]
The average grade in the sample was 83.96 with a standard deviation of 12. Sample of 36
This means that [tex]X = 83.96, s = 12, n = 36[/tex]
Value of the test-statistic:
[tex]t = \frac{X - \mu}{\frac{s}{\sqrt{n}}}[/tex]
[tex]t = \frac{83.96 - 87}{\frac{12}{\sqrt{36}}}[/tex]
[tex]t = -1.52[/tex]
Pvalue:
Probability of the sample mean differing of 87, which means that we have a two-tailed test, with t = -1.52 and 36 - 1 = 35 degrees of freedom.
With the help of a calculator, the pvalue is of 0.1375.
The p-value for this test is 0.1375.
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