Let x represent the average annual salary of college and university professors (in thousands of dollars) in the United States. For all colleges and universities in the United States, the population variance of x is approximately σ2 = 47.1. However, a random sample of 15 colleges and universities in Kansas showed that x has a sample variance σ2 = 83.2. Use a 5% level of significance to test the claim that the variance for colleges and universities in Kansas is greater than 47.1. Use the traditional method. Assume that a simple random sample is selected from a normally distributed population.
a. Check requirements.
b. Establish H0 and H1 and note the level of significance.
c. Find the sample test statistic.
d. Find Critical Value.
e. Conclude the test and interpret results.

Kindly check explanation

Step-by-step explanation:

Given that :

The hypothesis :

H0 : σ²= 47.1

H1 : σ² > 47.1

α = 5% = 0.05

Population variance, σ² = 47.1

Sample variance, s² = 83.2

Sample size, n = 15

The test statistic = (n-1)*s²/σ²

Test statistic, T = [(15 - 1) * 83.2] ÷ 47.1

Test statistic = T = [(14 * 83.2)] * 47.1

Test statistic = 1164.8 / 47.1

Test statistic = 24.73

The degree of freedom, df = n - 1 ; 10 = 9

Critical value (0.05, 9) = 16.92 (Chisquare distribution table)

Reject H0 ; If Test statistic > Critical value

Since ; 24.73 > 16.92 ; Reject H0 and conclude that variance is greater.

n=9

step-by-step explanation:

7 times a number = 7n

decreased by 13 = 7n-13

same as is =

5 times the number = 5n

increased by 5 = 5n+5

7n-13 = 5n+5

2n=18

n=9

answer: do we write it

step-by-step explanation: