Answer to Question #339396 in Statistics and Probability for Vhan

Question

A researcher claims that 10 year olds watch 6.6 hours of TV daily with SD = 2.5 hours. You try to verify this with the following sample data of 100 and a sample mean of 6.1 hours. Test the claim of the researcher. Test at α = .01

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 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:

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

Since it is observed that $|z| = 2<2.5758=z_c ,$ it is then concluded that the null hypothesis is not rejected.

Using the P-value approach: The p-value is $p=2P(Z<-2)=0.0455,$ and since $p = 0.0455 > 0.01=\alpha,$ it is concluded that the null hypothesis is not rejected.

Therefore, there is not enough evidence to claim that the population mean $\mu$

is different than 6.6, at the $\alpha = 0.01$ significance level.