Question

The mean income per person in the United States is $50,000, and the distribution of incomes follows a normal distribution. A random sample of 10 residents of Wilmington, Delaware, had a mean of $60,000 with a standard deviation of $10,000. At the 0.05 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.b. State the decision rule for 0.05 significance level.Reject H0 if t > ____c. Compute the 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 0.05 significance level?

Answer 1

We conclude that the residents of Wilmington, Delaware, have more income than the national average.

Explanation:

We are given the following in the question:

Population mean, μ = $50,000

Sample mean,
`\bar{x}` = $60,000

Sample size, n = 10

Alpha, α = 0.05

Sample standard deviation, s = $10,000

a) First, we design the null and the alternate hypothesis

`H_(0): \mu = 50000\text{ dollars}\\H_A: \mu > 50000\text{ dollars}`

We use one-tailed(right) t test to perform this hypothesis.

c) Formula:

`t_(stat) = \displaystyle\frac{\bar{x} - \mu}{(s)/(√(n)) }`

Putting all the values, we have

`t_(stat) = \displaystyle(60000 - 50000)/((10000)/(√(10)) ) = 3.162`

Now,
`t_(critical) \text{ at 0.05 level of significance, 9 degree of freedom } = 1.833`

b) Rejection Rule:

If the calculated t-statistic is greater than the the critical value, we rect the null hypothesis.

Since,

`t_(stat) > t_(critical)`

We fail to accept the null hypothesis and reject it.We accept the alternate hypothesis.

d) There is enough evidence to conclude that the residents of Wilmington, Delaware, have more income than the national average.