Question

The mean income per person in the United States is $44,500, and the distribution of incomes follows a normal distribution. A random sample of 16 residents of Wilmington, Delaware, had a mean of $52,500 with a standard deviation of $9,500. At the .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.



H0: µ =

H1: µ >


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(b) State the decision rule for .05 significance level. (Round your answer to 3 decimal places.)


Reject H0 if t >


(c) Compute the value of the test statistic. (Round your answer to 2 decimal places.)


Value of the test statistic

Answer 1

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

Explanation:

We are given the following in the question:

Population mean, μ = $44,500

Sample mean,
`\bar{x}` = $52,500

Sample size, n = 16

Alpha, α = 0.05

Sample standard deviation, s = $9,500

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

`H_(0): \mu = 44,500\text{ dollars}\\H_A: \mu > 44,500\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(52500 - 44500)/((9500)/(√(16)) ) = 3.37`

b) Rejection rule

If the calculated t-statistic is greater than the t-critical value, we fail to accept the null hypothesis and reject it.

Now,

`t_(critical) \text{ at 0.05 level of significance, 15 degree of freedom } = 1.753`

Since,

`t_(stat) > t_(critical)`

We fail to accept the null hypothesis and reject it. We accept the alternate hypothesis. We conclude that the residents of Wilmington, Delaware, have higher income than the national average