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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 :

Answer:

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

Step-by-step explanation:

We are given the following in the question:  

Population mean, μ = $50,000

Sample mean, [tex]\bar{x}[/tex] = $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

[tex]H_{0}: \mu = 50,000\text{ dollars}\\H_A: \mu \> 50,000\text{ dollars}[/tex]

We use One-tailed t test to perform this hypothesis.

c)  Formula:

[tex]t_{stat} = \displaystyle\frac{\bar{x} - \mu}{\frac{s}{\sqrt{n-1}} }[/tex]

Putting all the values, we have

[tex]t_{stat} = \displaystyle\frac{60000 - 50000}{\frac{10000}{\sqrt{9}} } = 3[/tex]

Now, [tex]t_{critical} \text{ at 0.05 level of significance, 9 degree of freedom } = 1.833[/tex]

Since,              

b) [tex]t_{stat} > t_{critical}[/tex]

We reject the null hypothesis and fail to accept it.

d) We accept the alternate hypothesis and conclude that residents of Wilmington, Delaware, have more income than the national average.

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