Homework SourseThe mean income per person in the United States is 41500 and
The mean income per person in the United States is $41,500, and the distribution of incomes follows a normal distribution. A random sample of 10 residents of Wilmington, Delaware, had a mean of $47,500 with a standard deviation of $10,600. At the .01 level of significance, is that enough evidence to conclude that residents of Wilmington, Delaware, have more income than the national average?
Is there enough evidence to substantiate that residents of Wilmington, Delaware have more income than the national average at the .01 significance level?
(Click to select)RejectFail to reject H0 and conclude that the mean income in Wilmington is (Click to select)greater thanless than
$41,500.
| The mean income per person in the United States is $41,500, and the distribution of incomes follows a normal distribution. A random sample of 10 residents of Wilmington, Delaware, had a mean of $47,500 with a standard deviation of $10,600. At the .01 level of significance, is that enough evidence to conclude that residents of Wilmington, Delaware, have more income than the national average? |
Solution
(a) Ho: mu=41,500
H1: mu>41,500
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(b) The degree of freedom =n-1=10-1=9
Given a=0.01, the critical value is t(0.01, df=9)=2.821 (from student t table)
Reject H0 if t > 2.821
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(c) test statistic is
t=(xbar-mu)/(s/vn)
=(47500-41500)/(10600/sqrt(10))
=1.79
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(d)Fail to reject H0 and conclude that the mean income in Wilmington is less than
$41,500.
