Answer to Question #346622 in Statistics and Probability for rei

Question

In a recent survey a researcher claimed that on average filipino teenagers watch k-dramas for 3.5 hours daily is the claim correct if a random sample of 50 filipino teenagers showed a mean of 2.8 hours with a standard deviation of 0.9 hours? use 99% confidence level

Answer 1

The following null and alternative hypotheses need to be tested:

$H_0:\mu=3.5$

$H_1:\mu\not=3.5$

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 $\alpha = 0.01,$ $df=n-1=49$ and the critical value for a two-tailed test is $t_c =2.679952.$

The rejection region for this two-tailed test is $R = \{t:|t|>2.679952\}.$

The t-statistic is computed as follows:

$$t=\dfrac{\bar{x}-\mu}{s/\sqrt{n}}=\dfrac{2.8-3.5}{0.9/\sqrt{50}}=-5.4997$$

Since it is observed that $|t|=5.4997>2.679952=t_c,$ it is then concluded that the null hypothesis is rejected.

Using the P-value approach:

The p-value for two-tailed, $df=39$ degrees of freedom, $t=-3.1623$ is $p=0.000001,$ and since $p= 0.000001<0.01=\alpha,$ it is concluded that the null hypothesis is rejected.

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

is different than 3.5, at the $\alpha = 0.01$ significance level.