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?

1. State the null hypothesis and the alternate hypothesis.

2. State the decision rule for 0.025 significance level. (Round your answer to 3 decimal places.)
3. Compute the value of the test statistic. (Round your answer to 2 decimal places.)
4. 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.