The mean income per person in the United States is $44,500, and the distribution of incomes follows a normal distribution. A random sample of 16 residents of Wilmington, Delaware, had a mean of $52,500 with a standard deviation of $9,500.
At the 0.050 level of significance, is that enough evidence to conclude that residents of Wilmington, Delaware, have more income than the national average?

At 5% level of significance there is enough evidence to concluded that the residents of Wilmington, Delaware, have more income than the national average.

Step-by-step explanation:

The national average income of United states is, $44,500.

The hypothesis to test whether the residents of Wilmington, Delaware, have more income than the national average, can be defined as:

H₀: The residents of Wilmington, Delaware, does not have more income than the national average, i.e. μ < $44,500.

Hₐ: The residents of Wilmington, Delaware, have more income than the national average, i.e. μ > $44,500.

It is provided that the distribution of incomes follows a normal distribution.

Given:

The significance level of the test is, α = 0.05.

As the sample size is small and the population standard deviation is not provided, use the t-test for single mean.

The test statistic is:

Decision Rule:

If the p value of the test is less than the significance level then the null hypothesis is rejected.

Compute the p value as follows:

Use the t-table for probability.

The p-value = 0.002 < α = 0.05.

The null hypothesis will be rejected at 5% significance level.

Conclusion:

As the null hypothesis is rejected it implies that the residents of Wilmington, Delaware, have more income than the national average.

okay so 15.15 *()5: +5/6

step-by-step explanation: