Chapter 10, Problem 85CE

Male and female students in a finance class were asked how much their last tank of gas cost. Can you conclude that on average males spent more on gas than females? (a) State the hypotheses for this test. (b) Show the calculation of the test statistic, assuming unequal population variances. (c) State the decision rule, using α = .01 and the quick rule for degrees of freedom. (d) Draw the conclusion. (e) Is it reasonable to assume unequal variances? Explain.

To determine

State the hypotheses.

Explanation of Solution

The given information is that the summary statistic for male is ${\overline{x}}_1=\$43.20$, $s_1=\$8.30$ and $n_1=13$. The summary statistic for female is ${\overline{x}}_2=\$36.6$, $s_2=\$3.10$ and $n_2=9$.

Here, the direction of the test is right-tailed.

Assume ${\mu }_M$ is the average last tank cost spent by males and ${\mu }_F$ is the average last tank cost spent by females.

State the hypotheses:

Null hypothesis:

$H_0:{\mu }_M-{\mu }_F\le 0$

That is, on average males did not spent more on gas than females.

Alternative hypothesis:

$H_1:{\mu }_M-{\mu }_F>0$

That is, on average males spent more on gas than female.

To determine

Find the test statistic, assuming unequal population variances.

Answer to Problem 85CE

The test statistic is 2.616.

Explanation of Solution

Calculation:

Test statistic:

$\begin{array}{c}{t}_{\text{calc}}=\frac{{\overline{x}}_1-{\overline{x}}_2}{\sqrt{\frac{s_1^2}{n_1}+\frac{s_2^2}{n_2}}}\\ =\frac{43.20-36.60}{\sqrt{\frac{{\left(8.30\right)}^2}{13}+\frac{{\left(3.10\right)}^2}{9}}}\\ =\frac{6.6}{\sqrt{5.2992+1.0678}}\\ =\frac{6.6}{\sqrt{6.367}}\\ =\frac{6.6}{2.5233}\\ =2.616\end{array}$

Thus, the test statistic is 2.616.

To determine

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

Answer to Problem 85CE

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

The degrees of freedom is 8.

Explanation of Solution

Calculation:

By using the quick rule, the degrees of freedom is,

$\begin{array}{c}df=\text{min}\left(n_1-1,n_2-1\right)\\ =\text{min}\left(13-1,9-1\right)\\ =\text{min}\left(12,8\right)\\ =8\end{array}$

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, 8)”
• Output using Excel software is given below:

From the output, the critical value is 2.896.

Decision rule:

If $t_{\text{calc}}>+2.896$, then reject the null hypothesis.

To determine

State the conclusion.

Answer to Problem 85CE

There is no evidence to infer that on average males spent more on gas than female.

Explanation of Solution

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

That is, $t_{\text{calc}}\left(=2.616\right)<+2.896$.

Therefore, the null hypothesis is not rejected.

Thus, there is no evidence to infer that on average males spent more on gas than female.

To determine

Check whether it is reasonable to assume unequal variances.

Answer to Problem 85CE

Yes, it is reasonable to assume unequal 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:

$\begin{array}{c}d{f}_1=n_1-1\\ =13-1\\ =12\end{array}$

Degrees of freedom for denominator:

$\begin{array}{c}d{f}_2=n_2-1\\ =9-1\\ =8\end{array}$

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(0.005,12,8)”
• Output using EXCEL is given below:

From the output, the critical value is $F_L=0.1871$.

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(0.005,12,8)”
• Output using EXCEL is given below:

From the output, the critical value is $F_R=7.015$.

Decision rule:

• If $F_{\text{calc}}<0.1871$, then reject the null hypothesis
• If $F_{\text{calc}}>7.015$, then reject the null hypothesis.

Test statistic:

$\begin{array}{c}F=\frac{s_1^2}{s_2^2}\\ =\frac{{\left(8.30\right)}^2}{{\left(3.10\right)}^2}\\ =\frac{68.89}{9.61}\\ =7.169\end{array}$

Thus, the test statistic is 7.169.

Conclusion:

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

That is, $F_{\text{calc}}\left(=7.169\right)>7.015$.

Therefore, the null hypothesis is rejected.

Thus, there is evidence to infer that the variance has changed. That is, the population variances are unequal.

Hence, it is reasonable to assume unequal variances.