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. a. 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. b. 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)null hypothesis: mean =24000

alternate hypothesis:mean>24000

rejection region at 0.05 level z>1.64485

here std error =std deviaiton/(n)1/2 =400

here test stat z=(X-mean)/std deviaiton =(24600-24000)/400=1.5

as test stat does not fall in rejection region we can not reject null hypothesis

b)for critical region is =sample mean +z*std deviation =24000+1.64485*400=24657.94

hence type 2 error probability =P(X<25257.94) =P(Z<(24657.94-25000)/400)=P(Z<-0.8552)=0.1962

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