Answer to Question #338015 in Statistics and Probability for KLTino

Question

In a study of television viewing, the mean number of television program they watched during daytime was 7. A survey was conducted on the random sample of 25 households and found that the mean number of television program they watched during daytime was 5 with a standard deviation of 1.5. Test the hypothesis at 10% level of significance.

Parameter: _______________________________________

Statistic: ________________________________

Null Hypothesis: ____________________________________________________

Alternative hypothesis: _____________________________________________

Answer 1

Parameter: $\mu,$ the mean number of television program they watched during daytime

Statistic: $t-$ statistic, two-tailed test

Null Hypothesis: $H_0:\mu=7$

Alternative hypothesis: $H_a:\mu\not=7$

1. The following null and alternative hypotheses need to be tested:

$H_0:\mu=7$

$H_a:\mu\not=7$

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.

2. Based on the information provided, the significance level is $\alpha = 0.10,$ $df=n-1=24$ degrees of freedom, and the critical value for a two-tailed test is $t_c = 1.710882.$

3. The rejection region for this two-tailed test is $R = \{t: |t| > 1.710882\}.$

4. The t-statistic is computed as follows:

$$t=\dfrac{\bar{x}-\mu}{s/\sqrt{n}}=\dfrac{5-7}{1.5/\sqrt{25}}\approx-6.6667$$

5. Since it is observed that $|t| =6.6667>1.710882=t_c,$ it is then concluded that the null hypothesis is rejected.

Using the P-value approach:

The p-value for two-tailed, $df=24$ degrees of freedom, $t=-6.6667$ is $p=0.000001,$ and since $p= 0.000001<0.10=\alpha,$ it is concluded that the null hypothesis is rejected.

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

is different than 7, at the $\alpha = 0.10$ significance level.