Answer to Question #342502 in Statistics and Probability for justinebieber

Question

It is claimed that the average age of working students in a certain university is 35 years. A researcher selected a random sample of 49 working students. The computation of their ages resulted to an average of 32 years with a standard deviation of 10 years. Does this mean that the average age of the working students is different from 35 years? Use 0.05 level of significance and assure normality of population.

Answer 1

The following null and alternative hypotheses need to be tested:

$H_0:\mu=35$

$H_1:\mu\not=35$

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 $\alpha = 0.05,$ $df=n-1=48$ and the critical value for a two-tailed test is $t_c =2.010635.$

The rejection region for this two-tailed test is $R = \{t:|t|>2.010635\}.$

The t-statistic is computed as follows:

$$t=\dfrac{\bar{x}-\mu}{s/\sqrt{n}}=\dfrac{32-35}{10/\sqrt{49}}=-2.1$$

Since it is observed that $|t|=2.1>2.010635=t_c,$ it is then concluded that the null hypothesis is rejected.

Using the P-value approach:

The p-value for two-tailed, $df=48$ degrees of freedom, $t=-2.1$ is $p=0.041009,$ and since $p=0.041009<0.05=\alpha,$ it is concluded that the null hypothesis is rejected.

Therefore, there is enough evidence to claim that the population mean $\mu$

is different than 35, at the $\alpha = 0.05$ significance level.