Question

A sample of 9 days over the past six months showed that a dentist treated the following numbers of patients at his dental clinic: 22, 26, 20, 18, 15, 22, 25, 19, and 26. If the number of patients seen per day is normally distributed, would an analysis of these sample data reject the hypothesis that the variance in the number of patients seen per day is equal to 10? Use a 0.10 level of significance.

State the null and alternative hypotheses.

H0: σ2 ≥ 10 Ha: σ2 < 10

H0: σ2 = 10 Ha: σ2 ≠ 10

H0: σ2 ≤ 10 Ha: σ2 > 10

H0: σ2 < 10 Ha: σ2 ≥ 10

H0: σ2 > 10 Ha: σ2 ≤ 10

Find the value of the test statistic. (Round your answer to three decimal places.)

Find the p-value. (Round your answer to four decimal places.) p-value =

State your conclusion.

Reject H0. We can conclude that the variance in the number of patients seen per day is not equal to 10.

Do not reject H0. We cannot conclude that the variance in the number of patients seen per day is not equal to 10.

Do not reject H0. We can conclude that the variance in the number of patients seen per day is not equal to 10.

Reject H0. We cannot conclude that the variance in the number of patients seen per day is not equal to 10.

Answer 1

Below are the null and alternative Hypothesis,
Null Hypothesis, H0: σ^2 = 10
Alternative Hypothesis, Ha: σ^2 ≠ 10

Test statistic,
Χ^2 = (n-1)*s^2/σ^2
Χ^2 = (8 - 1)*16.5536/10
Χ^2 = 11.588

P-value Approach
P-value = 0.2299

As P-value >= 0.1, fail to reject null hypothesis.

Do not reject H0. We cannot conclude that the variance in the number of patients seen per day is not equal to 10.