Question

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?

Answer 1

Yes, at the 0.05 level of significance we have enough evidence to conclude that residents of Wilmington, Delaware, have more income than the national average.

Explanation:

We are given that the mean income per person in the United States is $44,500, and the distribution of incomes follows a normal distribution.

Null Hypothesis,
`H_0` :
`\mu` <= $44,500 {means that residents of Wilmington, Delaware does not have more income than the national average}

Alternate Hypothesis,
`H_1` :
`\mu` > $44,500 {means that residents of Wilmington, Delaware have more income than the national average}

Since, we know nothing about population standard deviation so the test statistics used here will be;

`(Xbar-\mu)/((s)/(√(n) ) )` ~
`t_n_-_1`

where,
`Xbar` = Sample mean of $52,500

s = Sample standard deviation of $9,500

n = sample size of 16 residents

Test Statistics =
`(52500-44500)/((9500)/(√(16) ) )` ~
`t_1_5`

= 3.368

At 0.05 level of significance, t table gives the value of 1.753 . Since our test statistics is higher than the critical value as 3.368 > 1.753 so we have sufficient to reject null hypothesis as our test statistics will fall in the rejection region.

Hence, we have enough evidence to conclude that residents of Wilmington, Delaware have more income than the national average.

Answer 2

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

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:

`\bar x = \$52,500\\s= \$9,500\\n = 16`

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:

`t=(\bar x-\mu)/(s/√(n)) =(52500-44500)/(9500/√(16) )=3.37`

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:

`p-value=P(t>3.37)=0.002`

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.