Question

The mean income per person in the United States is $37,500, and the distribution of incomes follows a normal distribution. A random sample of 12 residents of Wilmington, Delaware, had a mean of $46,500 with a standard deviation of $9000. At the .010 level of significance, is that enough evidence to conclude that residents of Wilmington, Delaware, have more income than the national average?

(a) State the null hypothesis and the alternate hypothesis.

H0: u is less than or equal to	__________
H1: u>	__________

(b) State the decision rule for .010 significance level. (Round your answer to 3 decimal places.)

Reject H0 if t > ______

(c) Compute the value of the test statistic. (Round your answer to 2 decimal places.)

Value of the test statistic:

(d) Is there enough evidence to substantiate that residents of Wilmington, Delaware have more income than the national average at the .010 significance level?

(Reject/Do not reject) H0. There is (sufficient/insufficeint) evidence to conflcude WIlmington Deleware is (greater than/less than) 37,500

Answer 1

(a) State the null hypothesis and the alternate hypothesis.

H0: u is less than or equal to	37500
H1: u>	37500

(b) State the decision rule for .010 significance level. (Round your answer to 3 decimal places.)

Reject H0 if t > t(0.01,11)=2.718

(c) Compute the value of the test statistic. (Round your answer to 2 decimal places.)

Value of the test statistic:t=(x--mu)/(sd/sqrt(n))=(46500-37500)/(9000/sqrt(12))=3.46

(d) Is there enough evidence to substantiate that residents of Wilmington, Delaware have more income than the national average at the .010 significance level?

Reject H0. There is sufficient evidence to conflcude WIlmington Deleware is greater than 37,500

(Reject/Do not reject) H0. There is (sufficient/insufficeint) evidence to conflcude WIlmington Deleware is (greater than/less than) 37,500