Question

The mean income per person in the United States is $39,000, and the distribution of incomes follows a normal distribution. A random sample of 15 residents of Wilmington, Delaware, had a mean of $50,000 with a standard deviation of $9,600. At the 0.025 level of significance, is that enough evidence to conclude that residents of Wilmington, Delaware, have more income than the national average

Answer 1

Answer:

The calculated value t = 4.43 > 2.145 at 0.025 level of significance

Alternative hypothesis is accepted

The residents of Wilmington, Delaware, have not more income than the national average

Step-by-step explanation:

Step(i):-

Given data The mean income per person in the United States is $39,000

Mean of the Population = $39,000

Sample size 'n' = 15

mean of the sample 'x⁻' = $50,000

Standard deviation of the sample 'S' = $9,600

Step(ii):-

Null hypothesis:H₀: The residents of Wilmington, Delaware, have more income than the national average

Alternative Hypothesis H₁: The residents of Wilmington, Delaware, have not more income than the national average

level of significance α = 0.025

$t_{\frac{\alpha }{2} } = t_{0.025,14} = 2.145$

Degrees of freedom ν = n-1 = 15-1 =14

Step(iii):-

Test statistic

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

               $t = \frac{50,000 -39,000}{\frac{9600}{\sqrt{15} } }$

              t = 4.43

Conclusion:-

The calculated value t = 4.43 > 2.145 at 0.025 level of significance

null hypothesis is rejected

Alternative hypothesis is accepted

The residents of Wilmington, Delaware, have not more income than the national average