Question

The mean income per person in the USA is $50,000, and the distribution of incomes follows a normal distribution. A random sample of 10 residents of Wilmington, DE, had a mean of $60,000 with a standard deviation of $10,000. At the .05 level of significance, is that enough evidence to conclude that residents of Wilmington, DE, have more income than the national average? Please include your critical value, test statistic, and decision in your answer.

a) Yes
b) No

Answer 1

Using a one-sample t-test comparing the sample mean income of Wilmington residents to the US mean, we find a t-value greater than the critical value, which indicates enough evidence to conclude that Wilmington residents have a higher income at the .05 significance level.

Step-by-step explanation:

To determine if the mean income of residents in Wilmington, DE, is statistically significantly higher than the national average, we can use a one-sample t-test. The national mean income, referred to as the null hypothesis value (μ0), is $50,000. The sample from Wilmington has a mean (μ) of $60,000 and a standard deviation (s) of $10,000 with a sample size (n) of 10.

The test statistic can be calculated using the formula:

t = ( μ - μ0 ) / ( s / √n )

t = (60000 - 50000) / (10000 / √10) = 10000 / 3162.28 ≈ 3.16

At the .05 level of significance, for a one-tailed test with 9 degrees of freedom (n-1), the critical value can be found in the t-distribution table or calculated using a statistical software or calculator. It is approximately 1.833. Our calculated t-value of 3.16 is greater than the critical value of 1.833.

Since our t-value exceeds the critical value, we reject the null hypothesis. This suggests there is enough evidence to conclude that the average income in Wilmington, DE is significantly more than the national average at the .05 level of significance.

Therefore, the answer to the question is (a) Yes, there is enough evidence to conclude that residents of Wilmington, DE, have more income than the national average.