Question

A random sample of n₁ = 22 summer days in Los Angeles gave a sample mean pollution index of x-bar₁ = 49. Previous studies show that σ₁ = 12. For San Francisco, a random sample of n₂ = 17 summer days gave a sample mean pollution index of x-bar₂ = 45. Previous studies show that σ₂ = 15. Assume the pollution index is normally distributed. Do these data indicate the mean population pollution index of Los Angeles is greater than San Francisco? Use a 1% level of significance.

Show all work and calculations.

Answer 1

Solution:

Here, we have to use two sample z test for the difference between two population means.

The null and alternative hypotheses for this test are given as below:

Null hypothesis: H₀: the mean population pollution index of Los Angeles is same as that of San Francisco.

Alternative hypothesis: H_a: the mean population pollution index of Los Angeles is greater than San Francisco.

H₀: µ₁ = µ₂ versus H_a: µ₁ > µ₂

This is an upper tailed test.

We are given

Level of significance = α = 0.01

The test statistic formula for this test is given as below:

Z = (X1bar – X2bar) / sqrt[(σ₁² / n₁)+(σ₂² / n₂)]

From given data, we have

X1bar = 49

X2bar = 45

σ₁ = 12

σ₂ = 15

n₁= 22

n₂= 17

X1bar – X2bar = 49 - 45 = 4

Z = (X1bar – X2bar) / sqrt[(σ₁² / n₁)+(σ₂² / n₂)]

Z = 4 / sqrt[(12^2 / 22)+( 15^2 / 17)]

Z = 4/4.4476

Z = 0.8994

Test statistic = Z = 0.8994

Critical value = 2.3263

(by using z-table)

P-value = 0.1842

(by using z-table)

P-value > α = 0.01

So, we do not reject the null hypothesis

There is not sufficient evidence to conclude that the mean population pollution index of Los Angeles is greater than San Francisco.