Business, 26.06.2020 16:01 hungtistic

A student believes that the average grade on the statistics final examination is 87. A sample of 36 final examinations is taken. The average grade in the sample is 83.96. The population variance is 144.
a. State the null and alternative hypotheses.
b. Using a critical value, test the hypothesis at the 5% level of significance.
c. Using a p-value, test the hypothesis at the 5% level of significance.
d. Using a confidence interval, test the hypothesis at the 5% level of significance.

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Answer from: korirosekc

(a) H₀: μ = 87 vs. Hₐ: μ ≠ 87.

(b) $z=-1.52-z_{\alpha/2}=-1.96$ . The average grade on the statistics final examination is 87.

(d) The 95% confidence interval for the average grade on the statistics final examination is (80.04, 87.88).

Explanation:

A statistical hypothesis test is to be performed to determine whether the average grade on the statistics final examination is 87.

(a)

The hypothesis can be defined as follows:

H₀: The average grade on the statistics final examination is 87, i.e. μ = 87.

Hₐ: The average grade on the statistics final examination is not 87, i.e. μ ≠ 87.

(b)

The information provided is:

$n=36\\\bar x=83.96\\\sigma^{2}=144$

As the population variance is provided, we will use a z-test for single mean.

Compute the test statistic value as follows:

$z=\frac{\bar x-\mu}{\sqrt{\sigma^{2}/n}}=\frac{83.96-87}{\sqrt{144/36}}=-1.52$

The test statistic value is -1.52.

Decision rule:

Reject H₀ if:

The calculated value of $z=-1.52-z_{\alpha/2}=-1.96$ .

The null hypothesis will not be rejected.

Conclusion:

The average grade on the statistics final examination is 87.

(c)

Decision Rule:

If the p-value of the test is less than the significance level then the null hypothesis will be rejected.

Compute the p-value for the two-tailed test as follows:

$p-value=2\cdot P(Z$

*Use a z-table for the probability.

The p-value of the test is 0.1286.

p-value = 0.1286 > α = 0.05

The null hypothesis will not be rejected.

Thus, it can be concluded that average grade on the statistics final examination is 87.

(d)

Compute the 95% confidence interval for the average grade on the statistics final examination as follows:

$CI=\bar x\pm z_{\alpha/2}\cdot \sqrt{\frac{\sigma^{2}}{n}}$

$=83.96\pm 1.96\cdot \sqrt{\frac{144}{36}}\\\\=83.96\pm 3.92\\\\=(80.04, 87.88)$

The 95% confidence interval for the average grade on the statistics final examination is (80.04, 87.88).

As the 95% confidence interval consists of the null value, i.e. 87, the null hypothesis will not be rejected.

Hence, concluding that the average grade on the statistics final examination is 87.

Answer from: Quest

answer is i don't know

Answer from: Quest

prokaryotic cells do not contain a nucleus or any other membrane-bound organelle. eukaryotic cells contain membrane-bound organelles, including a nucleus. so i belive the answer is a : p hope it

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