Answer to Question #196240 in Statistics and Probability for klent

Question

An officer of a certain agency claims that the mean monthly income of a family that lives in a depressed area in a certain town is Php7.500.00. A group of researchers conducted a survey in that area and found out that the mean monthly income of 25 selected families is Php6,000.00 with a standard deviation of Php150.00. Test the claim that = Php7,500.00 at 0.01 level of significance

Answer 1

Hypothesized Population Mean $\mu=7500$ 

Sample Standard Deviation $s=150$

Sample Size $n=25$

Sample Mean $\bar{X}=6000$

Significance Level $\alpha=0.01$ 

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

$H_0:\mu=7500$

$H_1:\mu\not=7500$

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 $\alpha=0.01,$ and $df=n-1=24$ degrees of freedom. The critical value for a two-tailed test is $t_c=2.79694.$ 

The rejection region for this left-tailed test is $R=\{t:|t|>2.79694\}.$ 

The t-statistic is computed as follows:

$$t=\dfrac{\bar{X}-\mu_0}{s/\sqrt{n}}=\dfrac{6000-7500}{150/\sqrt{25}}=-50$$

Since it is observed that $|t|=50>2.79694=t_c,$ it is then concluded that the null hypothesis is rejected.

Therefore, there is enough evidence to claim that the population mean $\mu$ is different than $7500,$ at the $\alpha=0.01$ significance level.

Using the P-value approach: The p-value 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 $7500,$ at the $\alpha=0.01$ significance level.