Question

To meet the standards in the shipment of a chemical product, the percentage of pollution
the variance must not exceed 4.

H₀: ?² = 4
H₁: ?² > 4

In a random sample of 20 shipments, variance of pollution percentages have been found as 5.62. Accordingly, find the power of the test for ?² = 7

Answer 1

This is test for one variance. So we will use chi-square distribution.This is right tailed test because we are checking for the greater than side.

Sx^2 = sample variance =5.62 n = 20

H₀: ?² = 4
H₁: ?² > 4

Test Stat=

Where the null variance = 4

Test Stat = 26.695

p-value = P( > Test stat) ...................df = n-1

=P( > 26.70)

p-value = 0.112

level of significance = 0.05

Since p-value > 0.05

We do not reject the null hypothesis. There is not sufficient evidence to support that percentage exceeds 4.

To find the power of test we need to find the probability of type 2 error ()

Type 2 error is failing to reject a false null hypothesis.

= P( Non rejection of null | null hypothesis is false)

Not reject null hypothesis when

p-value > 0.05

P(( > ) > 0.05

We find the critical value at p = 0.05 with df = 19

> 30.1435

We sub all values except the sample variance

Sx² < 6.346 ...........sign changes because we are dividing by n -1

= P( Non rejection of null | null hypothesis is false)

= P( < | at ?² = 7)

=1 - P( > )

= 1 - 0.5746

= 0.4254

Power of test = 1 - ...................using chi-square tables

power of test = 0.57434