Question

The mean income per person in the United States is $45,000, and the distribution of incomes follows a normal distribution. A random sample of 17 residents of Wilmington, Delaware, had a mean of $55,000 with a standard deviation of $8,500. At the 0.100 level of significance, is that enough evidence to conclude that residents of Wilmington, Delaware, have more income than the national average? State the null hypothesis and the alternate hypothesis. State the decision rule for 0.100 significance level. (Round your answer to 3 decimal places.) Compute the value of the test statistic. (Round your answer to 2 decimal places.) Is there enough evidence to substantiate that residents of Wilmington, Delaware, have more income than the national average at the 0.100 significance level?

Answer 1

The null and alternative hypothesis are:

; i.e., the mean income of Wilmington, Delaware is not different than the national average income.

; i.e., the mean income of Wilmington, Delaware is more than that of national average income.

Now, we have given a sample of residents of Wilmington, Delaware with mean income, and sample standard deviation, . And we need to test the hypothesis for the given significance level of .

Test-statistic: with degrees of freedom,

So, the test-statistic is calculated as to be

P-value:

Since, it is a right-tailed hypothesis.

Decision:

Since,

So, at the sample data provide enough evidence to support the alternative hypothesis, i.e., . Hence we conclude that, the mean income of Wilmington, Delaware is greater than the national average income of $45000.