Answer to Question #336489 in Statistics and Probability for Snow

Question

ACTIVITY 8. APPLYING HYPOTHESIS TESTING

Solve the problem:

A recent survey says that Filipino children spend an average of 4 hr a day playing online games with a standard deviation of 30 minutes. A random sample of 9 children is taken from a normally distributed population of children who spend an average of 4 hr and 30 min playing online games. Using the 1% level of significance, would you conclude that the statement given in the survey is correct?

Answer 1

The following null and alternative hypotheses need to be tested:

$H_0:\mu=4$

$H_a:\mu\not=4$

This corresponds to a two-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 $\alpha = 0.01,$ and the critical value for a two-tailed test is$z_c = 2.5758.$

The rejection region for this two-tailed test is $R = \{z: |z| > 2.5758\}.$

The z-statistic is computed as follows:

$$z=\dfrac{\bar{x}-\mu}{\sigma/\sqrt{n}}=\dfrac{4.5-4}{0.5/\sqrt{9}}=3$$

Since it is observed that $|z| = 3 >2.5758= z_c,$ it is then concluded that the null hypothesis is rejected.

Using the P-value approach:

The p-value is $p =2P(Z>3)=2(0.00135)=0.0027,$ and since $p = 0.0027 < 0.01=\alpha,$ it is concluded that the null hypothesis is rejected.

Therefore, there is enough evidence to claim that the population mean $\mu$  is different than 4, at the $\alpha = 0.01$ significance level.