Answer to Question #215968 in Statistics and Probability for Mohammad Hasan

Question

A researcher claims that 10-year old children watch 6.6 hours of TV daily on average. You try to verify this for a sample of size 100 with mean 6.1 hours and standard deviation 2.5 hours. 

Answer 1

The following null and alternative hypotheses need to be tested:

$H_0:\mu=6.6$

$H_1:\mu\not=6.6$

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 $\alpha=0.05,$ $df=n-1=100-1=99$ degrees of freedom and the critical value for a two-tailed test is $t_c=1.9842.$

The rejection region for this two-tailed test is $R=\{t: |t|>1.9842\}$

The t-statistic is computed as follows:

$$t=\dfrac{\bar{x}-\mu}{s/\sqrt{n}}=\dfrac{6.1-6.6}{2.5/\sqrt{100}}=-2$$

Since it is observed that $|t|=2>1.9842=t_c,$ it is then concluded that the null hypothesis is rejected.

Using the P-value approach: The p-value for $\alpha=0.05, df=99$ degrees of freedom, two-tailed, $t=-2$ is $p=0.04824,$ and since $p=0.04824<0.05=\alpha,$ it is concluded that the null hypothesis is rejected.

Therefore, there is enough evidence to claim that the population mean $\mu$  is different than $6.6,$ at the $\alpha=0.05$ significance level.