Question

According to the Federal Reserve Board, the mean net worth of U.S. households headed by persons 75 years or older is $640,000. Suppose a simple random sample of 50 households in this age group is obtained from a certain region of the United States and is found to have a mean net worth of $615,000, with a standard deviation of $120,000. From these sample results, and using the 0.05 level of significance in a two-tail test, comment on whether the mean net worth for all the region’s households in this age category might not be the same as the mean value reported for their counterparts across the nation.

Answer 1

Hypothesized Population Mean $\mu=\$640,000$ 

Sample Standard Deviation $s=\$120,000$

Sample Size $n=50$

Sample Mean $\bar{X}=\$ 615,000$

Significance Level $\alpha=0.05$ 

The following null and alternative hypotheses for the population proportion needs to be tested:

$H_0:\mu=640,000$

$H_1:\mu\not=640,000$

This corresponds to a two-tailed test, for which a t-test for one mean, with unknown population standard deviation will be used.

Based on the information provided, the significance level is $\alpha=0.05,$ and $df=n-1=49$ degrees of freedom. The critical value for a two-tailed test is $t_c=2.009575.$ 

The rejection region for this left-tailed test is $R=\{t:|t|>2.009575\}.$ 

The t-statistic is computed as follows:

$$t=\dfrac{\bar{X}-\mu_0}{s/\sqrt{n}}=\dfrac{615,000-640,000}{120,000/\sqrt{50}}\approx-1.473139$$

Since it is observed that $|t|=1.473139<2.009575=t_c,$ it is then concluded that the null hypothesis is not rejected.

Therefore, there is not enough evidence to claim that the population mean $\mu$ is different than $640,000$ at the $\alpha=0.05$ significance level.

Using the P-value approach:

The p-value for $\alpha=0.05, df=49, t=1.473139,$ two-tailed is $p=0.147119,$ and since $p=0.147119>0.05=\alpha,$ it is concluded that the null hypothesis is not rejected.

Therefore, there is not enough evidence to claim that the population mean $\mu$ is different than $640,000,$ at the $\alpha=0.05$ significance level.