Question

The mean income per person in the United States is $50,000, and the distribution of incomes follows a normal distribution. A random sample of 10 residents of Wilmington, Delaware, had a mean of $60,000 with a standard deviation of $10,000. At the 0.05 level of significance, is that enough evidence to conclude that residents of Wilmington, Delaware, have more income than the national average

Answer 1

Answer:

Yes. There is sufficient eveidence to support  the claim that Wilmington resident's earn more than the national average

Step-by-step explanation:

Given the following;

$\alpha=Significance \ Level=0.05\\n=10\\\bar X=60000\\\ s=sample\ \sigma=10000$

#We set our hypothesis as:

$H_o:\mu\leq 50000\\H_a:\mu>50000$

The rejection region of either hypothesis:

$P(t>t_o)=0.05$, we determine the critical values for this probability in the T distribution table;

$t=1.833$

=>We reject for all values greater than 1.833, t>1.833.

#We now determine the value of the test statistic as:

$t=\frac{\bar X-\mu_o}{s/\sqrt{n}}\\\\=\frac{60000-50000}{10000/\sqrt{10}}\\\\=3.1623$

We find that:

$t=3.1623>1.833,\ \ \ Reject \ H_o$

Hence, there is sufficient eveidence to support  the claim that Wilmington resident's earn more than the national average.