Chapter 9.1, Problem 21E

a.

To determine

State the null and alternate hypotheses.

Answer to Problem 21E

The hypotheses are given below:

Null hypothesis:

$H_0:{\mu }_1={\mu }_2$

That is, there is no significant difference between the mean number of hours spent on internet for the first survey and the second survey.

Alternate hypothesis:

$H_1:{\mu }_1<{\mu }_2$

That is, the mean number of hours spent on internet is significantly increased from the first survey to second survey.

Explanation of Solution

Calculation:

It is given that, the statistics regarding the number of hours spent on internet per week in the first survey is $n_1=1,048$, ${\overline{x}}_1=9.79$, $s_1=13.41$ and the statistics regarding the number of hours spent on internet per week in the second survey is $n_2=1,018$, ${\overline{x}}_2=10.47$, $s_2=14.52$. Furthermore, it is given that both the samples in the surveys are simple random samples of adults and $\alpha =0.05$.

Hypothesis:

Hypothesis is an assumption about the parameter of the population, and the assumption may or may not be true.

Let ${\mu }_1$ be the population mean number of hours spent on internet in the first survey and ${\mu }_2$ be the population mean number of hours spent on internet in the second survey.

Claim:

Here, the claim is whether the mean number of hours spent on internet is been increased in the second survey or not.

The hypotheses are given below:

Null hypothesis:

Null hypothesis is a statement which is tested for statistical significance in the test. The decision criterion indicates whether the null hypothesis will be rejected or not in the favor of alternate hypothesis.

$H_0:{\mu }_1={\mu }_2$

That is, there is no significant difference between the mean number of hours spent on internet for the first survey and the second survey.

Alternate hypothesis:

Alternate hypothesis is contradictory statement of the null hypothesis

$H_0:{\mu }_1<{\mu }_2$

That is, the mean number of hours spent on internet is significantly increased from the first survey to second survey.

b.

To determine

Find the value of t-test statistic.

Answer to Problem 21E

The value of t-test statistic is –1.105006.

Explanation of Solution

Calculation:

Requirements for a small sample t-test:

• The samples must be independent and drawn randomly from the populations.
• Either the sample size must be greater than 30 or the population must be approximately normal.

Here, it is given that both the samples are simple random samples.

Thus, the conditions are satisfied.

The test statistic for the small sample t is obtained as follows:

$t=\frac{\left({\overline{x}}_1-{\overline{x}}_2\right)-\left({\mu }_1-{\mu }_2\right)}{\sqrt{\frac{s_1{}^2}{n_1}+\frac{s_2{}^2}{n_2}}}$

Under the null hypothesis, $\left({\mu }_1-{\mu }_2\right)=0$.

Therefore the test statistic is,

$\begin{array}{c}t=\frac{\left({\overline{x}}_1-{\overline{x}}_2\right)-0}{\sqrt{\frac{s_1{}^2}{n_1}+\frac{s_2{}^2}{n_2}}}\\ =\frac{9.79-10.47}{\sqrt{\frac{{13.41}^2}{1,048}+\frac{{14.52}^2}{1,018}}}\\ =\frac{-0.68}{0.615381}\\ =-1.105006\end{array}$

Thus, the test statistic is –1.105006.

c.

To determine

Find the number of degrees of freedom for the test statistic.

Answer to Problem 21E

The number of degrees of freedom for the test statistic is 1,017.

Explanation of Solution

Calculation:

Degrees of freedom:

The degrees of freedom for t using computer package is,

$\Delta =\frac{{\left[\left(\frac{s_1^2}{n_1}+\frac{s_2^2}{n_2}\right)\right]}^2}{\frac{{\left(\frac{s_1^2}{n_1}\right)}^2}{n_1-1}+\frac{{\left(\frac{s_2^2}{n_2}\right)}^2}{n_2-1}}$

The degrees of freedom, when computing by hand is smaller of $n_1-1$ and $n_2-1$.

The degrees of freedom is,

$\begin{array}{c}d.f=\min \left(n_1-1,n_2-1\right)\\ =\min \left(1,048-1,1,018-1\right)\\ =\min \left(1,047,1,017\right)\\ =1,017\end{array}$

Thus, the degree of freedom is 1,017.

d.

To determine

Check whether the null hypothesis is rejected at $\alpha =0.05$.

Answer to Problem 21E

There is not enough evidence to reject the null hypothesis at $\alpha =0.05$.

There is no significant difference between the mean number of hours spent on internet for the first survey and the second survey.

Explanation of Solution

P-value:

Step-by-step procedure to obtain the P-value using the MINITAB software:

• Choose Graph > Probability Distribution Plot
• Choose View Probability > OK.
• From Distribution, choose‘t’ distribution.
• In Degrees of freedom, enter 1,017.
• Click the Shaded Area tab.
• Choose X value and Left Tail for the region of the curve to shade.
• In X-value enter –1.105006.
• Click OK.

Output using the MINITAB software is given below:

From the MINITAB output, the P-value is 0.1347.

Thus, the P-value is 0.1347.

Decision rule based on P-value:

If $P-\text{value}\le \alpha$, then reject the null hypothesis $H_0$.

If $P-\text{value}>\alpha$, then fail to reject the null hypothesis $H_0$.

Here, the level of significance is $\alpha =0.05$.

Conclusion based on P-value approach:

The P-value is 0.1347 and $\alpha$ value is 0.05.

Here, P-value is greater than the $\alpha$ value.

That is, $0.1347\left(=P-\text{value}\right)>0.05\left(=\alpha \right)$.

By the rejection rule, fail to reject the null hypothesis.

Hence, it can be concluded that there is no significant difference between the mean number of hours spent on internet for the first survey and the second survey.

Thus, it can be concluded that there is sufficient evidence to infer the equality of two population means at $\underset{\_}{\alpha =0.05}$.