Question

The mean income per person in the United States is $39,500, and the distribution of incomes follows a normal distribution. A random sample of 16 residents of Wilmington, Delaware, had a mean of $45,500 with a standard deviation of $9,800. At the 0.010 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 0.010 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 0.010 significance level?

Answer 1

To test null hypothesis against alternative hypothesis

Here

sample mean

sample standard deviation

and sample size

The test statistic can be written as

which under H₀ follows a t distribution with n-1 df.

We reject H₀ at 0.01 level of significance if P-value < 0.01

Now,

The value of the test statistic =

and P-value =

Since P-value < 0.01, so we reject H₀ at 0.01 level of significance and we can conclude that there is enough evidence to substantiate that residents of Wilmington, Delaware, have more income than the national average.