Question

Assume we know that the population standard deviation of income in the United States (σ) is $6,000. A labor economist wishes to test the hypothesis that average income for the United States equals $53,000. A random sample of 2500 individuals is taken and the sample mean is found to be $53,300. Test the hypothesis at the 0.05 and 0.01 levels of significance. Provide the p-value.

Answer 1

Given that,

Population standard deviation =

n= randomly selected sample = 2500

Sample mean =

Claim : To test the hypothesis that average income for the united states equals $53000

We write it as

The null hypothesis =H0 :

Alternative hypotheis = H1 :

To find the test statistic use below formula.

So test statistic becomes,

So the test statistic is 2.5

We have two tailed test.

So two tailed p value = 2* right tailed p value

Where right tailed p value = 1- left tailed p value

Use below excel function to find the left tailed p value.

=NORM.S.DIST(Z,TRUE)

=NORM.S.DIST(2.5,TRUE)

So left tailed p value = 0.99379

Right tailed p value = 1- 0.99379 = 0.00621

Then the two tailed p value = 2*0.00621 = 0.012419

Decision Rule : If p value is less than level of significance then we reject H0 and if p value is greater than level of significance then we fail to reject H0

Here given that level of significance = 0.01

So P value 0.012419 > 0.01

So we fail to reject H0

Conclusion : By using decision we conclude that average income for the united states equals $53000.