Question

The mean income per person in the United States is $39,500, and the distribution of incomes follows a normal distribution. A random sample of 16 residents of Wilmington, Delaware, had a mean of $45,500 with a standard deviation of $9,800. At the 0.010 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.

B.)State the decision rule for 0.010 significance level. (Round your answer to 3 decimal places.)

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

D.)I s there enough evidence to substantiate that residents of Wilmington, Delaware, have more income than the national average at the 0.010 significance level?

___________HO. There is________ evidence to conclude that the mean income in Wilmington is________ $39,500.

Answer 1

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. - - - A) Null Hypothesis - Ho: MZ HO Alternative hypothesis → Ha: M.<MO B Since the sample size is small (< 30) we will - use the t- distribution. .. At 0.010 significance level, df = n-1= 15 T value = 2.602 Decision Rule . Reject is it u Rejer i Test statistic < 2,602 2,602 •Fail to Reject, if Test statistic > 2.602 fail to Reject - c) Value of Test Statistic = 39500 - 45 500 = -2.45 D) Since test statistic is <2,602, we will Reject to - » Reject to. There is enough evidence to I conclude that the mean inome in Wilmington I is greater than $39.500.