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

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

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_{(\alpha )/(2) } = t_(0.025,14) = 2.145`

Degrees of freedom ν = n-1 = 15-1 =14

Step(iii):-

Test statistic

`t = (x^(-) -mean)/((s)/(√(n) ) )`

`t = (50,000 -39,000)/((9600)/(√(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