Chapter 10, Problem 86CE

a.

To determine

State the hypotheses.

Explanation of Solution

The given information is that the summary statistic for Sonando High School students is ${\overline{x}}_1=14.51$, $s_1=2.69$ and $n_1=11$. The summary statistic for Gedacht High School is ${\overline{x}}_2=11.88$, $s_2=2.66$ and $n_2=16$.

Here, the direction of the test is right tailed.

Assume ${\mu }_S$ is the average number of times Sonando High School students ordered fast-food and ${\mu }_G$ is the average number of times Gedacht High School students ordered fast-food.

State the hypotheses:

Null hypothesis:

$H_0:{\mu }_S-{\mu }_G\le 0$

That is, Sonando High School students did not choose fast-food restaurants more often than Gedacht High School.

Alternative hypothesis:

$H_1:{\mu }_S-{\mu }_G>0$

That is, Sonando High School students choose fast-food restaurants more often than Gedacht High School.

b.

To determine

Find the test statistic, assuming equal population variances.

Answer to Problem 86CE

The test statistic is 2.513.

Explanation of Solution

Calculation:

The formula for test statistic is,

$t_{\text{calc}}=\frac{\left({\overline{x}}_1-{\overline{x}}_2\right)}{\sqrt{\frac{s_p^2}{n_1}+\frac{s_p^2}{n_2}}}$

Where, $s_p^2=\frac{\left(n_1-1\right)s_1^2+\left(n_2-1\right)s_2^2}{n_1+n_2-2}$

The pooled variance is,

$\begin{array}{c}{s}_p^2=\frac{\left(11-1\right){\left(2.69\right)}^2+\left(16-1\right){\left(2.66\right)}^2}{11+16-2}\\ =\frac{\left(10\right)\left(7.2361\right)+\left(15\right)\left(7.0756\right)}{25}\\ =\frac{178.495}{25}\\ =7.1398\end{array}$

The test statistic is,

$\begin{array}{c}{t}_{\text{calc}}=\frac{\left(14.51-11.88\right)}{\sqrt{\frac{7.1398}{11}+\frac{7.1398}{16}}}\\ =\frac{2.63}{\sqrt{0.6491+0.4462}}\\ =\frac{2.63}{\sqrt{1.0953}}\\ =\frac{2.63}{1.0466}\\ =2.513\end{array}$

Thus, the test statistic is 2.513.

c.

To determine

State the decision rule and the quick rule for degrees of freedom.

Answer to Problem 86CE

The decision rule is “reject the null hypothesis, if $t_{\text{calc}}>+2.896$”.

The degrees of freedom is 8.

Explanation of Solution

Calculation:

The degrees of freedom is,

$\begin{array}{c}df=n_1+n_2-2\\ =11+16-2\\ =25\end{array}$

Critical value:

Software procedure:

Step-by-step software procedure to obtain critical value using EXCEL is as follows:

• Open an EXCEL file.
• In cell A1, enter the formula “=T.INV(0.99, 25)”
• Output using Excel software is given below:

From the output, the critical value is 2.485.

Decision rule:

If $t_{\text{calc}}>+2.485$, then reject the null hypothesis.

d.

To determine

State the conclusion.

Answer to Problem 86CE

There is evidence to infer that Sonando High School students choose fast-food restaurants more often than Gedacht High School.

Explanation of Solution

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

That is, $t_{\text{calc}}\left(=2.513\right)>+2.485$.

Therefore, the null hypothesis is rejected.

Thus, there is evidence to infer that Sonando High School students choose fast-food restaurants more often than Gedacht High School.

e.

To determine

Check whether it is reasonable to assume equal variances.

Answer to Problem 86CE

Yes, it is reasonable to assume equal variances.

Explanation of Solution

Calculation:

State the hypotheses:

Null hypothesis:

$H_0:\frac{{\sigma }_1^2}{{\sigma }_2^2}=1$

That is, the variances are equal.

Alternative hypothesis:

$H_1:\frac{{\sigma }_1^2}{{\sigma }_2^2}\ne 1$

That is, the variances are not equal.

Degrees of freedom for numerator:

$\begin{array}{c}d{f}_1=n_1-1\\ =11-1\\ =10\end{array}$

Degrees of freedom for denominator:

$\begin{array}{c}d{f}_2=n_2-1\\ =16-1\\ =15\end{array}$

Level of significance:

$\begin{array}{c}\frac{\alpha }{2}=\frac{0.01}{2}\\ =0.005\end{array}$

Lower critical value:

Software procedure:

Step-by-step software procedure to obtain lower critical value using EXCEL is as follows:

• Open an EXCEL file.
• In cell A1, enter the formula “=F.INV(0.005,10,15)”
• Output using EXCEL is given below:

From the output, the critical value is $F_L=0.1828$.

Upper critical value:

Software procedure:

Step-by-step software procedure to obtain upper critical value using EXCEL is as follows:

• Open an EXCEL file.
• In cell A1, enter the formula “=F.INV.RT(0.005,10,15)”
• Output using EXCEL is given below:

From the output, the critical value is $F_R=4.4235$.

Decision rule:

• If $F_{\text{calc}}<0.1828$, then reject the null hypothesis
• If $F_{\text{calc}}>4.4235$, then reject the null hypothesis.

Test statistic:

$\begin{array}{c}F=\frac{s_1^2}{s_2^2}\\ =\frac{{\left(2.69\right)}^2}{{\left(2.66\right)}^2}\\ =\frac{7.2361}{7.0756}\\ =1.023\end{array}$

Thus, the test statistic is 1.023.

Conclusion:

Here, the test statistic is lies between lower and upper critical values.

That is, $0.1828<F_{\text{calc}}\left(=1.023\right)<4.4235$.

Therefore, the null hypothesis is not rejected.

Thus, there is no evidence to infer that the variance has changed. That is, the population variances are equal.

Hence, it is reasonable to assume equal variances.