Chapter 12, Problem 15CQ

To determine

To check: Whether there is sufficient evidence to conclude a difference in means.

To perform: The appropriate test to find out where the difference in means if there is sufficient evidence to conclude a difference in means

Answer to Problem 15CQ

Yes, there is sufficient evidence to conclude a difference in means.

There is significant difference between the means “Asia and Europe” and “Asia and Africa”.

Explanation of Solution

Given info:

The table shows the particulate matter for prominent cities of three continents. The level of significance is 0.05.

Calculation:

The hypotheses are given below:

Null hypothesis:

$H_0:{\mu }_1={\mu }_2={\mu }_3$

Alternative hypothesis:

$H_1:$ At least one mean is different from the others

Here, at least one mean is different from the others is tested. Hence, the claim is that, at least one mean is different from the others.

The level of significance is 0.05. The number of samples k is 3, the sample sizes $n_1$, $n_2$ and $n_3$ are 4, 4 and 3.

The degrees of freedom are $\text{d}\text{.f}\text{.N}=k-1\text{ and d}\text{.f}\text{.D}=N-k$.

Where

$\begin{array}{c}N=n_1+n_2+n_3\\ =4+4+3\\ =11\end{array}$

Substitute 3 for k in $\text{d}\text{.f}\text{.N}$

$\begin{array}{c}\text{d}\text{.f}\text{.N}=k-1\\ \text{=3-1}\\ \text{=2}\end{array}$

Substitute 11 for N and 3 for k in $\text{d}\text{.f}\text{.D}$

$\begin{array}{c}\text{d}\text{.f}\text{.D}=N-k\\ =11-3\\ =8\end{array}$

Critical value:

The critical F-value is obtained using the Table H: The F-Distribution with the level of significance $\alpha =0.05$.

Procedure:

• Locate 8 in the degrees of freedom, denominator row of the Table H.
• Obtain the value in the corresponding degrees of freedom, numerator column below 2.

That is, the critical value is 4.46.

Rejection region:

The null hypothesis would be rejected if $F>4.46$.

Software procedure:

Step-by-step procedure to obtain thetest statistic using the MINITAB software:

• Choose Stat > ANOVA > One-Way.
• In Response, enter the Gasoline prices.
• In Factor, enter the Factor.
• Click OK.

Output using the MINITAB software is given below:

From the MINITAB output, the test value F is 6.65.

Conclusion:

From the results, the test value is 6.65.

Here, the F-statistic value is greater than the critical value.

That is, $6.65>4.46$.

Thus, it can be concluding that, the null hypothesis is rejected.

Hence, the result concludes that, there is sufficient evidence to conclude a difference in means.

Consider, ${\overline{X}}_1$, ${\overline{X}}_2$ and ${\overline{X}}_3$ represents the means of Asia, Europe and Africa, and $s_1^2,s_2^2\text{ and }{s}_3^2$ represents the variances of samples of Asia, Europe and Africa.

Step-by-step procedure to obtain the test mean and standard deviation using the MINITAB software:

• Choose Stat > Basic Statistics > Display Descriptive Statistics.
• In Variables enter the columns Asia, Europe and Africa.
• Choose option statistics, and select Mean, Variance and N total.
• Click OK.

Output using the MINITAB software is given below:

The sample sizes $n_1,n_2{\text{ and n}}_3$ are 4, 4 and 3.

The means are ${\overline{X}}_1,{\overline{X}}_2\text{ and }{\overline{X}}_3$ are 74.0, 35.50 and 30.67.

The sample variances are $s_1^2,s_2^2\text{ and }{s}_3^2$ are 694.0, 29.67 and 186.33.

Here, the samples of sizes of three states are not equal. So, the test used here is Scheffe test.

Tukey test:

Critical value:

The formula for critical value F¹ for the Scheffe test is,

$F^1=\left(k-1\right)\left(\text{Critical value}\right)$

Here, the critical value of F test is 4.46.

Substitute 4.46 for critical value is of F and 2 for k-1 in $F^1$

$\begin{array}{c}{F}^1=2\left(\text{4}\text{.46}\right)\\ =8.92\end{array}$

Comparison of the means:

The formula for finding $s_W^2$ is,

$s_W^2=\frac{{\displaystyle \sum \left(n_i-1\right)s_i^2}}{{\displaystyle \sum \left(n_i-1\right)}}$

That is,

$\begin{array}{c}{s}_W^2=\frac{(4-1)694.0+(4-1)29.67+(3-1)186.33}{\left(4-1\right)+\left(4-1\right)+\left(3-1\right)}\\ =\frac{2082+89.01+372.66}{8}\\ =\frac{2543.67}{8}\\ =317.9588\end{array}$

Comparison between the means ${\overline{X}}_1$ and ${\overline{X}}_2$:

The hypotheses are given below:

Null hypothesis:

$H_0:$ There is no significant difference between ${\overline{X}}_1$ and ${\overline{X}}_2$.

Alternative hypothesis:

$H_1:$ There is significant difference between ${\overline{X}}_1$ and ${\overline{X}}_2$.

Rejection region:

The null hypothesis would be rejected if absolute value greater than the critical value.

The formula for comparing the means ${\overline{X}}_1$ and ${\overline{X}}_2$ is,

$F_s=\frac{{\left({\overline{X}}_1-{\overline{X}}_2\right)}^2}{s_W^2\left[\frac{1}{n_1}+\frac{1}{n_2}\right]}$

Substitute 74.0 and 35.50 for ${\overline{X}}_1$ and ${\overline{X}}_2$, 4 for n_1, 4 for n₂ and 317.9588 for $s_W^2$

$\begin{array}{c}{F}_s=\frac{{\left(74.0-35.50\right)}^2}{317.9588\left[\frac{1}{4}+\frac{1}{4}\right]}\\ =\frac{1482.25}{317.9588\left[0.25+0.25\right]}\\ =\frac{1482.25}{158.9794}\\ =9.32\end{array}$

Thus, the value of $F_s$ is 9.32.

Conclusion:

The value of $F_s$ is 9.32.

Here, the value of $F_s$ is greater than the critical value.

That is, $9.32>8.92$.

Thus, the null hypothesis is rejected.

Hence, there is significant difference between the means ${\overline{X}}_1$ and ${\overline{X}}_2$.

Comparison between the means ${\overline{X}}_1$ and ${\overline{X}}_3$:

The hypotheses are given below:

Null hypothesis:

$H_0:$ There is no significant difference between ${\overline{X}}_1$ and ${\overline{X}}_3$.

Alternative hypothesis:

$H_1:$ There is significant difference between ${\overline{X}}_1$ and ${\overline{X}}_3$.

Rejection region:

The null hypothesis would be rejected if absolute value greater than the critical value.

The formula for comparing the means ${\overline{X}}_1$ and ${\overline{X}}_3$.is,

$F_s=\frac{{\left({\overline{X}}_1-{\overline{X}}_3\right)}^2}{s_W^2\left[\frac{1}{n_1}+\frac{1}{n_3}\right]}$

Substitute 74.0 and 30.67 for ${\overline{X}}_1$ and ${\overline{X}}_3$, 4 for n_1, 3 for n₃ and 317.9588 for $s_W^2$

$\begin{array}{c}{F}_s=\frac{{\left(74.0-30.67\right)}^2}{317.9588\left[\frac{1}{4}+\frac{1}{3}\right]}\\ =\frac{1877.4889}{317.9588\left[0.25+0.33\right]}\\ =\frac{1877.4889}{184.42}\\ =10.18\end{array}$

Thus, the value of $F_s$ is 10.18.

Conclusion:

The value of $F_s$ is 10.18.

Here, the value of $F_s$ is greater than the critical value.

That is, $10.18>8.92$.

Thus, the null hypothesis is rejected.

Hence, there is significant difference between the means ${\overline{X}}_1$ and ${\overline{X}}_3$.

Comparison between the means ${\overline{X}}_2$ and ${\overline{X}}_3$:

The hypotheses are given below:

Null hypothesis:

$H_0:$ There is no significant difference between ${\overline{X}}_2$ and ${\overline{X}}_3$.

Alternative hypothesis:

$H_1:$ There is significant difference between ${\overline{X}}_2$ and ${\overline{X}}_3$.

Rejection region:

The null hypothesis would be rejected if absolute value greater than the critical value.

The formula for comparing the means ${\overline{X}}_2$ and ${\overline{X}}_3$.is,

$F_s=\frac{{\left({\overline{X}}_2-{\overline{X}}_3\right)}^2}{s_W^2\left[\frac{1}{n_2}+\frac{1}{n_3}\right]}$

Substitute 35.50 and 30.67 for ${\overline{X}}_2$ and ${\overline{X}}_3$, 4 for n_2, 3 for n₃ and 317.9588 for $s_W^2$

$\begin{array}{c}{F}_s=\frac{{\left(35.50-30.67\right)}^2}{317.9588\left[\frac{1}{4}+\frac{1}{3}\right]}\\ =\frac{23.3289}{317.9588\left[0.25+0.33\right]}\\ =\frac{23.3289}{184.42}\\ =0.13\end{array}$

Thus, the value of $F_s$ is 0.13.

Conclusion:

The value of $F_s$ is 0.13.

Here, the value of $F_s$ is lesser than the critical value.

That is, $0.13<8.92$.

Thus, the null hypothesis is not rejected.

Hence, there is no significant difference between the means ${\overline{X}}_2$ and ${\overline{X}}_3$.

Justification:

Here, there is significant difference between the means ${\overline{X}}_1$ and ${\overline{X}}_2$, ${\overline{X}}_1$ and ${\overline{X}}_3$. Thus, it can be concluding that there is significant difference between the means “Asia and Europe” and “Asia and Africa”.