Answer to Question #250980 in Statistics and Probability for Biostat101

Question

The SRTC gets its budget for a training program on the assumption that the average of the supplies is to be P12, 000 or less. To see if this claim is realistic, 10 supply figures are randomly obtained from similar training programs. The sample has yielded a mean =P12, 900 and s = P1, 100. Is the mean greater than P12, 000? Use  to test the hypothesis.

Answer 1

The following null and alternative hypotheses need to be tested:

$H_0:\mu\leq 12000$

$H_1:\mu>12000$

This corresponds to a right-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 $\alpha = 0.05, df=n-1$

$=10-1=9$ ​degrees of freedom, and the critical value for a right-tailed test is$t_c = 1.833113.$

The rejection region for this right-tailed test is $R = \{t: t > 1.833113\}.$

The t-statistic is computed as follows:

$$t=\dfrac{\bar{x}-\mu}{s/\sqrt{n}}=\dfrac{12900-12000}{1100/\sqrt{10}}=2.587318$$

Since it is observed that $t = 2.587318 > 1.833113=t_c ,$ it is then concluded that the null hypothesis is rejected.

Using the P-value approach: The p-value for right-tailed, $\alpha=0.05, df=9,$ $t=2.587318$ is $p=0.014671,$ and since $p=0.014671<0.05=\alpha,$ it is concluded that the null hypothesis is rejected.

Therefore, there is enough evidence to claim that the population mean $\mu$ is greater than 12000, at the $\alpha = 0.05$ significance level.