Suppose course evaluation ratings for four college instructors are shown in the following table. Instructor Black...

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Question

Suppose course evaluation ratings for four college instructors are shown in the following table. Instructor Black...

Suppose course evaluation ratings for four college instructors are shown in the following table.

Instructor
Black	Jennings	Swanson	Wilson
89	86	89	80
80	79	77	83
78	81	69	57
69	83	81	72
97	98	83	88
68	98	81	86
	83	84
	94	85
		82

Use α = 0.05 and test for a significant difference among the rating for these instructors.

State the null and alternative hypotheses.

H₀: Not all populations of teaching evaluations are identical.
H_a: All populations of teaching evaluations are identical.

H₀: Median_B ≠ Median_J ≠ Median_S ≠ Median_W
H_a: Median_B = Median_J = Median_S = Median_W

H₀: Median_B = Median_J = Median_S = Median_W
H_a: Median_B < Median_J > Median_S > Median_W

H₀: Median_B = Median_J = Median_S = Median_W
H_a: Median_B ≠ Median_J ≠ Median_S ≠ Median_W

H₀: All populations of teaching evaluations are identical.
H_a: Not all populations of teaching evaluations are identical.

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

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

p-value =

What is your conclusion?

Do not reject H₀. There is sufficient evidence to conclude that there is a significant difference among the rating for these instructors.

Do not reject H₀. There is not sufficient evidence to conclude that there is a significant difference among the rating for these instructors.

Reject H₀. There is sufficient evidence to conclude that there is a significant difference among the rating for these instructors.

Reject H₀. There is not sufficient evidence to conclude that there is a significant difference among the rating for these instructors.

math Statistics-And-Probability

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Answer 1

Answer #1

The statistical software output for this problem is:

Kruskal-Wallis results: Data stored in separate columns. Results adjusted for ties DF Chi-Square P-value 3 3.5723212 0.3115 S

Hence,

Hypotheses:

H₀: All populations of teaching evaluations are identical.
H_a: Not all populations of teaching evaluations are identical.

Test statistic = 3.57

P - value = 0.312

Conclusion: Option B is correct.

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Answer 2

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