Chapter 8, Problem 8.2.1RE

To determine

To test: Whether there is any sufficient evidence to infer that the average time spent in internet by teenagers differs from 18.3hours per week.

Answer to Problem 8.2.1RE

There is enough evidence to infer that the average time spent in internet by teenagers is significantly different from 18.3hours per week.

Explanation of Solution

Given info:

To examine the average number of hours spent per week by teenagers in internet, the results based on the random sample of 48 teenagers is $\overline{x}=20.9\text{hours}$. The additional information is about the population parameters $\mu =18.3\text{hours}$ ${\sigma }^2=32.49\text{hours}$ and level of significance $\alpha =0.02$.

Calculation:

The testing hypotheses are given below:

Null hypothesis $H_0$:

$H_0:\mu =18.3$

Alternative hypothesis $H_1$:

$H_1:\mu \ne 18.3$

Requirements for z-test about mean when population standard deviation $\left(\sigma \right)$known:

• The sample must be drawn randomly from the desired population.
• The sample must be greater than or equal to 30, if not the population must be approximately normal.

Here, the sample size is greater than 30 and the sample is drawn randomly.

Thus, z-test about mean is valid in this case.

Decision rule based on classical approach:

If $z>z_{\frac{\alpha }{2}}$, then reject the null hypothesis $H_0$.

Critical value:

For the level of significance $\alpha =0.02$,

$\begin{array}{c}\text{Two}\text{tail}\text{area=}\frac{\alpha }{2}\\ =\frac{0.02}{2}\\ =0.01\end{array}$

Hence, the cumulative area to the left is,

$\begin{array}{c}\text{Area}\text{to}\text{the}\text{left}=1-\text{Area}\text{to}\text{the}\text{right}\\ =1-0.01\\ =0.99\end{array}$

From Table E of the standard normal distribution from Appendix A, the critical value is 2.33.

Thus, the critical value of z is $\left(z_{\frac{\alpha }{2}}\right)=2.33$.

Test statistic:

The test statistic for the large sample single mean test is,

$z=\frac{\left(\overline{x}-\mu \right)}{\frac{\sigma }{\sqrt{n}}}$

The test statistic is obtained as follows:

Here, $\overline{x}=20.9\text{hours}$ $n=48$ and $\mu =18.3\text{hours}$ ${\sigma }^2=32.49\text{hours}$

$\begin{array}{c}z=\frac{\left(\overline{x}-\mu \right)}{\sqrt{\frac{{\sigma }^2}{n}}}\\ =\frac{20.9-18.3}{\sqrt{\frac{32.49}{48}}}\\ =\frac{2.6}{0.823}\\ =3.159\end{array}$

Thus, the test statistic is $z=3.159$.

Conclusion based on classical approach:

The test statistic value is $z=3.159$ and the critical value is $\left(z_{\frac{\alpha }{2}}\right)=2.33$.

Here, test statistic value is less than the critical value.

That is, $3.159\left(=z\right)>2.33\left(=z_{\frac{\alpha }{2}}\right)$

Hence, reject the null hypothesis $H_0$.

Thus, it can be concluded that there is enough evidence to infer that the average time spent in internet by teenagers is significantly different from 18.3hours per week.