Business: 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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A student believes that the average grade on the statistics

Ответ:

sindy35111

29.01.2022

Business

(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.

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