Question

The mean income per person in the United States is $37,000, and the distribution of incomes follows a normal distribution. A random sample of 11 residents of Wilmington, Delaware, had a mean of $43,000 with a standard deviation of $8,800. At the 0.025 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 0.025 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 0.025 significance level?

Answer #1

a)

H0: <= 37000

Ha: > 37000

b)

df = n - 1 = 11 - 1 = 10

t critical value at 0.025 level with 10 df = 2.228

Decision rule= Reject H0 if test statistics t > 2.228

c)

Test statistics

t = ( - ) / (S / sqrt(n) )

= (43000 - 37000) / (8800 / sqrt(11) )

= 2.26

d)

Since test statistics value falls in rejection region, Reject H0.

We conclude at 0.025 level that we have sufficient evidence to support the claim.

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