The mean income per person in the United States is $41,500, and the distribution

Question

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 0.01 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. H0: H1: > b. State the decision rule for 0.010 significance level. (Round your answer to 3 decimal places.) Reject H0 if t > c. Compute the value of the test statistic. (Round your answer to 2 decimal places.) Value of the test statistic d. Is there enough evidence to substantiate that residents of Wilmington, Delaware have more income than the national average at the 0.010 significance level? H0. There is evidence to conclude that the mean income in Wilmington is greater than $41,500.

Explanation / Answer

t-test For Single Mean
Set Up Hypothesis
Null, H0: U<=41500
Alternate, H1: U>41500
Test Statistic
Population Mean(U)=41500
Sample X(Mean)=47500
Standard Deviation(S.D)=10600
Number (n)=10
we use Test Statistic (t) = x-U/(s.d/Sqrt(n))
to =47500-41500/(10600/Sqrt(10))
to =1.79
| to | =1.79
Critical Value
The Value of |t | with n-1 = 9 d.f is 2.821
We got |to| =1.79 & | t | =2.821
Make Decision
Hence Value of |to | < | t | and Here we Do not Reject Ho
P-Value :Right Tail - Ha : ( P > 1.79 ) = 0.05354
Hence Value of P0.01 < 0.05354,Here We Do not Reject Ho

test satstic value is 1.79
there is no enough evidence to substantiate that residents of Wilmington, Delaware have more income than the national average

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