Suppose that the mean income of 35-year-olds in the United States is $24,000. A

Question

Suppose that the mean income of 35-year-olds in the United States is $24,000. A random sample of 100 35-year-olds in California results in a sample mean income of $24, 600 and a sample standard deviation of $4000. Although we don't know the population standard deviation sigma, the sample is large, so it is reasonable to use the sample standard deviation as an estimate of it. At the 5% level of significance, should we conclude that 35-year-olds in California have a higher average income than the national average? State the null and alternative hypotheses, find the rejection region, and state your conclusion in common English. Suppose that the true mean income of 35-year-olds in California is actually $24, 500. For the decision rule found in part a, find the probability beta of committing a Type II error.

Explanation / Answer

(A)

Below are the null and alternate hypothesis,
H0: u <= 24000
H1: u > 24000

Test statistics,
z = (24600 - 24000)/(4000/sqrt(100)) = 1.5

p-value = P(z>1.5) = 0.0668

Here the significance level is 0.05.
As p-value is greater than 0.05, we fail to reject the null hypothesis.
This means there is not significant evidence that 35-year old in California have a higher average income than the nationanl average.

(B)

For the above computation, critical value will be
24000 + 1.64*(4000/sqrt(100)) = 24656

If true mean is 24500,
P(X>24656) = P(z<(24656-24500)/(4000/sqrt(100))) = P(z<0.39) = 0.6517
Hence beta i.e. probability of type II error = 0.6517

Related Questions

Navigate

• Browse (All)
• Browse S
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.