The mean income per person in the United States is $40,000. And the distribution

Question

The mean income per person in the United States is $40,000. And the distribution of incomes follows the normal distribution. A random sample of 10 residents of Wilmington Delaware, had a mean of $50,000 with a standard deviation of $10,000. At the 0.05 level of significance, is that enough evidence to conclude that residents of Wilmington Delaware have income greater than the national average? a. Set up the null and alternative hypothesis for this problem. b. Which distribution (normal or t) should be used in this problem? WHY? c. State the level of significance. d. Create a decision rule for this problem. (INCLUDE the critical value.) e. Calculate the test statistic. f. What is your conclusion? (IN DETAIL.) g. Estimate the p-value. Restate the conclusion.

Explanation / Answer

Answers to the question:

Sample mean = 50000
Standard deviation = 10000
Pop. mean = 40000

a.

Ho: Mu <= 40000
Ha: Mu > 40000

b. t because n<30 . The value of sample size is , n=30

c. Level of significance is .05

d. If t>1.8333 then we reject null hypo

e. t = (50000-40000)/(10000/sqrt(10)) = 3.16

f. We reject null hypo and conclude that Mu>40000. Or the claim is right

g. The P-Value is .005775.The result is significant at p < .05.

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