Answer to Question #110913 in Statistics and Probability for Khatrie Loh

Question

A researcher claims that a patient takes on the average 12.8 months to fully recover. Random sample of 60 patients were studied, the mean 11.2months and standard deviation of 3.5 month. Test the researcher’s claim at 0.01 level of significant

Answer 1

The following null and alternative hypotheses need to be tested:

$H_0:\mu=12.8$

$H_1:\mu\not=12.8$

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 $\alpha=0.01, df=60-1=59,$ and the critical value for a two-tailed test is $t_c=2.662.$

The rejection region for this two-tailed test is $R=\{t:|t|>2.662\}$

The t-statistic is computed as follows:

$$t={\bar{X}-\mu_0\over s/\sqrt{n}}={11.2-12.8\over 3.5/\sqrt{60}}\approx-3.541$$

Since it is observed that $|t|=3.541>t_c=2.662,$ it is then concluded that the null hypothesis is rejected. Therefore, there is enough evidence to claim that the population $\mu$  is different than 12.8, at the 0.01 significance level.

Using the P-value approach: The p-value is $p=0.000786,$ and since $p=0.000786<0.01,$ it is concluded that the null hypothesis is rejected. Therefore, there is enough evidence to claim that the population $\mu$  is different than 12.8, at the 0.01 significance level.