Answer to Question #350646 in Statistics and Probability for Matt

Question

A certain group of welfare recipients receives relief goods with a mean amount of Php 500.00 per week. A random sample of 75 recipients is survived and found that the mean amount of relief goods they received in a week is Php 600 and a standard decision of Php. 50.00. Test the claim at 1% level of significance is not Php 500.00 per week and assume that the population is approximately normally distributed. What type of test does the problem represent?

Answer 1

The following null and alternative hypotheses need to be tested:

$H_0:\mu=500$

$H_1:\mu\not=500$

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 $\alpha = 0.01,$ $df=n-1=74$ 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:

$$t=\dfrac{\bar{x}-\mu}{s/\sqrt{n}}=\dfrac{600-500}{50/\sqrt{75}}=17.3205$$

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 500.00, at the $\alpha = 0.01$ significance level.