Question

The mean income per person in the United States is $35,500, and the distribution of incomes follows a normal distribution. A random sample of 8 residents of Wilmington, Delaware, had a mean of $39,500 with a standard deviation of $8,200. At the 0.010 level of significance, is that enough evidence to conclude that residents of Wilmington, Delaware, have more income than the national average?

1. State the null hypothesis and the alternate hypothesis.

2. State the decision rule for 0.010 significance level. (Round your answer to 3 decimal places.)

3. Compute the value of the test statistic. (Round your answer to 2 decimal places.)

4. Is there enough evidence to substantiate that residents of Wilmington, Delaware, have more income than the national average at the 0.010 significance level?

Answer 1

Solution:

Given: The mean income per person in the United States is $35,500, and the distribution of incomes follows a normal distribution.

$\mu = 35500$

Sample size = n = 8

Sample mean = $\bar{x} =39500$

Sample standard deviation = s = 8200

level of significance = 0.01

We have to test if there is evidence to conclude that residents of Wilmington, Delaware, have more income than the national average.

Part a) State the null hypothesis and the alternate hypothesis.

$H_{0}: \mu = 35500$

$H_{1}: \mu > 35500$

Part b) State the decision rule for 0.010 significance level.

Since sample size n is small and population standard deviation is unknown and population is Normally distributed, we use t test.

thus find t critical value

df = n - 1 = 8 -1 = 7

One tail area = 0.01

t Table cum. probl one-tail two-tails t 50 to 0.01 0.50 1.00 0.15 0.30 0.10 0.20 0.05 0.10 0.025 0.05 0.50 0.40 0.000 0.000 0.000 0.000 0.000 0.000 U.UUU 1.000 1.376 1.963 0.816 1.061 1.386 0.765 0.978 1 .250 0.741 0.941 1.190 0.727 0.920 1.156 0.718 0.906 1 .134 UITD.9961119 3.078 1.886 1.638 1.533 1.476 1.440 1.415 6.314 2.920 2.353 2.132 2.015 1.943 1.895 12.71 4.303 3.182 2.776 2.571 2.447 2365 31.82 .965 .541 1.747 1.365 3.143 2.998

t critical value = 2.998

Thus reject null hypothesis H0 , if t test statistic value $\geq 2.998$

Part c) Compute the value of the test statistic.

$t=\frac{\bar{x}-\mu }{s /\sqrt{n}}$

$t=\frac{39500-35500 }{8200 /\sqrt{8}}$

$t=\frac{4000 }{8200 /2.828427 }$

$t=\frac{4000 }{2899.1378 }$

Part d) there enough evidence to substantiate that residents of Wilmington, Delaware, have more income than the national average at the 0.010 significance level?

Since < t critical value = 2.998, we fail to reject H0,

thus there is not enough evidence to substantiate that residents of Wilmington, Delaware, have more income than the national average at the 0.010 significance level