Question

The mean income per person in the United States is $43,500, and the distribution of incomes follows a normal distribution. A random sample of 14 residents of Wilmington, Delaware, had a mean of $50,500 with a standard deviation of $11,400. 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?
(a) State the null hypothesis and the alternate hypothesis.
H0: µ = =
H1: µ > =
(b) State the decision rule for .01 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

Answer:

a)

Null hypothesis : H₀:μ = $43,500

Alternative Hypothesis : H₁ : μ ≠ $43,500

b)

The calculated value t = 2.2975 < 1.7709 at 0.01 level of significance

Null hypothesis is accepted

c)

The calculated value t = 2.2975 < 1.7709

The residents of Wilmington, Delaware, have more income than the national average

Step-by-step explanation:

Step(i):-

Given mean of the Population = $43,500,

Given mean of the sample  = $50,500

Given standard deviation of the sample  = $11,400.

level of significance = 0.01

Step(ii):-

a)

Null hypothesis : H₀:μ = $43,500

Alternative Hypothesis : H₁ : μ ≠ $43,500

b)

Test statistic

$t = \frac{x^{-} -mean}{\frac{S}{\sqrt{n} } }$

$t = \frac{50,500 -43,500}{\frac{11400}{\sqrt{14} } }= 2.2975$

Degrees of freedom

                           ν =n-1 = 14-1 =13

The critical value

$Z_{\frac{0.01}{2} } = Z_{0.05} = 1.7709$

c)

The calculated value t = 2.2975 < 1.7709 at 0.01 level of significance

Null hypothesis is accepted

Conclusion:-

The residents of Wilmington, Delaware, have more income than the national average