Answer to Question #343180 in Statistics and Probability for judea

Question

The SSG president claims that the average number of hours the students study their  lesson is more than 25 hours per week with standard deviation of 4 hours. The average of 30  students surveyed is 36 hours per week, is there sufficient evidence to reject the SSG  president’s claim at �� = 0.10 level of significance?  

(a) What is the parameter to be tested?  

(b) Is the population standard deviation given?  

(c) How many samples do we have in this problem?  

(d) What is the given level of significance?  

(e) What is the appropriate statistical test to be used? 

Answer 1

(a) the parameter to be tested is the average number of hours.

(b) the population standard deviation $\sigma$ is given

(c) there are $n=30$ samples

(d) the significance level is $\alpha = 0.10$ 

(e)

The following null and alternative hypotheses need to be tested:

$H_0:\mu\le25$

$H_1:\mu>25$

This corresponds to a right-tailed test, for which a z-test for one mean, with known population standard deviation will be used.

Based on the information provided, the significance level is $\alpha = 0.10,$ and the critical value for a right-tailed test is $z_c =1.2816.$

The rejection region for this right-tailed test is $R = \{z:z>1.2816\}.$

The z-statistic is computed as follows:

$$z=\dfrac{\bar{x}-\mu}{\sigma/\sqrt{n}}=\dfrac{36-25}{4/\sqrt{30}}\approx15.0624$$

Since it is observed that $z=15.0624>1.2816=z_c,$ it is then concluded that the null hypothesis is rejected.

Using the P-value approach:

The p-value is $p=P(z>15.0624)=0,$ and since $p= 0<0.10=\alpha,$ it is concluded that the null hypothesis is rejected.

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

is greater than 25, at the $\alpha = 0.10$ significance level.