Answer to Question #338026 in Statistics and Probability for KLTino

Question

Determine the given and compute the test statistic of the problem below using Central Limit Theorem, and construct the rejection region for each.

A certain group of welfare recipients receives relief goods with a mean amount of Php 500 per week. A random sample of 75 recipients is surveyed and found that the mean amount of relief goods they received in a week is Php 600 and a standard deviation of Php50.00 . Test the claim at 1% level of significance is not Php 500 per week and assume that the population is approximately normally distributed.

Answer 1

The following null and alternative hypotheses need to be tested:

$H_0:\mu=500$

$H_a:\mu\not=500$

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 $\alpha = 0.01,$ $df=n-1=74$ degrees of freedom, and the critical value for a two-tailed test is $t_c = 2.643913.$

The rejection region for this two-tailed test is $R = \{t: |t| > 2.643913\}.$

The t-statistic is computed as follows:

$$t=\dfrac{\bar{x}-\mu}{s/\sqrt{n}}=\dfrac{600-500}{50/\sqrt{75}}\approx17.3205$$

5. Since it is observed that $|t| =17.3205> 2.643913=t_c,$ it is then concluded that the null hypothesis is rejected.

Using the P-value approach:

The p-value for two-tailed, $df=74$ degrees of freedom, $t=17.3205$ 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 500, at the $\alpha = 0.01$ significance level.