Answer to Question #338718 in Statistics and Probability for Seb

Question

The treasurer of a municipality claims that the average net worth of families living in this municipality is ₱590,000. A random sample of 50 families selected from this area produced a mean net worth of ₱720,000 with standard deviation of ₱65,000. Using 1% significance level, can we conclude that the claim is true? Also, find the 99% confidence interval of the true mean.

Answer 1

We need to construct the $99\%$ confidence interval for the population mean $\mu.$

The critical value for $\alpha = 0.05$ and $df = n-1 =49$ degrees of freedom is $t_c = z_{1-\alpha/2; n-1} = 2.679952.$

The corresponding confidence interval is computed as shown below:

$$CI=(\bar{x}-t_c\times\dfrac{s}{\sqrt{n}},\bar{x}+t_c\times\dfrac{s}{\sqrt{n}})$$

$$=(720000-2.679952\times\dfrac{65000}{\sqrt{50}},$$

$$720000+2.679952\times\dfrac{65000}{\sqrt{50}})$$

$$=(695364.841,744635.159)$$

Therefore, based on the data provided, the 99% confidence interval for the population mean is $695364.841 < \mu < 744635.159,$ which indicates that we are 99% confident that the true population mean $\mu$ is contained by the interval $(695364.841, 744635.159).$

The following null and alternative hypotheses need to be tested:

$H_0:\mu=590000$

$H_a:\mu\not=590000$

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

Based on the information provided, the significance level is $\alpha = 0.01,$ $df=n-1=49$ degrees of freedom, and the critical value for a two-tailed test is $t_c =2.679952.$

The rejection region for this two-tailed test is $R = \{t: |t|> 2.679952\}.$

The t-statistic is computed as follows:

$$t=\dfrac{\bar{x}-\mu}{s/\sqrt{n}}=\dfrac{720000-590000}{65000/\sqrt{50}}\approx14.142$$

Since it is observed that $|t |= 14.142 \ge2.679952=t_c,$ it is then concluded that the null hypothesis is rejected.

Using the P-value approach:

The p-value for two-tailed test, $df=49$ degrees of freedom, $t=14.142$ is $p = 0,$ and since $p = 0< 0.01=\alpha,$ it is concluded that the null hypothesis is rejected.

Therefore, there is enough evidence to claim that the population mean $\mu$

is different than 590000, at the $\alpha = 0.01$ significance level.