# The mean income per person in the United States is \$44,000, and the distribution of incomes follows a normal distribution. A random sample of 15 residents of Wilmington, Delaware, had a mean of \$48,000 with a standard deviation of \$10,500. 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?a) State the null hypothesis and the alternate hypothesis.b) State the decision rule for 0.100 significance level. (Round your answer to 3 decimal places.) Reject Ho if t >_____________c) Compute the value of the test statistic. (Round your answer to 2 decimal places.) Value of the test statistic _______________

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The mean income per person in the United States is \$44,000, and the distribution of incomes follows a normal distribution. A random sample of 15 residents of Wilmington, Delaware, had a mean of \$48,000 with a standard deviation of \$10,500. 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?

a) State the null hypothesis and the alternate hypothesis.

b) State the decision rule for 0.100 significance level. (Round your answer to 3 decimal places.) Reject Ho if t >_____________

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

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Step 1

Part (a):Null and alternative hypothesis:

The hypotheses to be tested are:

H0: μ ≤ 44,000, that is, the mean income of residence of Wilmington, Delaware is less than or equal to 44,000.

Versus:

H0: μ > 44,000, that is, the residence of Wilmington, Delaware have more income than the national average.

Step 2

Part (b): Decision rule:

The decision rule is given by:

Reject the null hypothesis if t >critical value. Otherwise fail to reject the null hypothesis.

The level of significance is 0.10.

It is given that test is right tailed. α = 0.10, n=15.

Note that the Excel formula =T....

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