Answer to Question #241093 in Statistics and Probability for Darv

Question

A sample survey on the average total yearly expenditure included 150 students of a certain university. The mean total expenditure per student per year for the sample was 3,000 with a standard deviation of 500. How likely is it that the students spend an average of 3500 per year as claimed by a parent at .01 significant level. 

Answer 1

The following null and alternative hypotheses need to be tested:

$H_0:\mu=3500$

$H_1:\mu\not=3500$

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 $\alpha = 0.01,$ $df=n-1=150-1=149$ degrees of freedom, and the critical value for a two-tailed test is $t_c = 2.609.$

The rejection region for this two-tailed test is $R = \{t: |t| > 2.609\}.$

The t-statistic is computed as follows:

$$t=\dfrac{\bar{x}-\mu}{s/\sqrt{n}}=\dfrac{3000-3500}{500/\sqrt{150}}=-12.247$$

Since it is observed that $|t| = 12.247 >2.609= t_c ,$ it is then concluded that the null hypothesis is rejected.

Using the P-value approach: The p-value for two-tailes, $\alpha=0.01, df=149,$ $t=-12.247$ is $p\approx0,$ 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 3500, at the $\alpha = 0.01$ significance level.