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?

- State the null hypothesis and the alternate hypothesis.

- State the decision rule for 0.010 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 0.010 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 H_{0} follows a t distribution with n-1 df.

We reject H_{0} 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 H_{0} 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.

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