Chapter 10, Problem 10.1.1RE

a.

To determine

To construct: The scatterplot for the give data.

Answer to Problem 10.1.1RE

The scatterplot for the data is as follows,

Explanation of Solution

Given info:

The data represents the weekly average number of passengers per flight and the average one-way fare in dollars for common commercial routes.

Calculation:

Software procedure:

Step by step procedure to obtain scatterplot using the MINITAB software:

• Choose Graph > Scatterplot.
• Choose Simple and then click OK.
• Under Y variables, enter a column of Avg. one-way fare.
• Under X variables, enter a column of Avg. no.of passengers.
• Click OK.

b.

To determine

To compute: The value of the correlation coefficient.

Answer to Problem 10.1.1RE

The value of the correlation coefficient is –0.686.

Explanation of Solution

Calculation:

Correlation coefficient r:

Software Procedure:

Step-by-step procedure to obtain the ‘correlation coefficient’ using the MINITAB software:

• Select Stat >Basic Statistics > Correlation.
• In Variables, select Avg. number of passengers and Avg. one-way fare from the box on the left.
• Click OK.

Output using the MINITAB software is given below:

Thus, the value of the correlation is –0.686.

c.

To determine

To test: The significance of the correlation coefficient at $\alpha =0.01$ using Table I.

Answer to Problem 10.1.1RE

The conclusion is that, there is no linear relation between the Avg. number of passengers and Avg. one-way fare.

Explanation of Solution

Given info:

The level of significance is $\alpha =0.01$ and the sample size is 6.

Calculation:

The hypotheses are given below:

Null hypothesis:

$H_0:\rho =0$

That is, there is no linear relation between the Avg. number of passengers and Avg. one-way fare.

Alternative hypothesis:

$H_1:\rho \ne 0$

That is, there is a linear relation between the Avg. number of passengers and Avg. one-way fare.

The formula to obtain the degrees of the freedom is $n-2$.

That is,

$\begin{array}{c}n-2=6-2\\ =4\end{array}$

From the “TABLE –I: Critical Values for the PPMC”, the critical value for 4 degrees of freedom and $\alpha =0.01$ level of significance is 0.917.

Rejection Rule:

If the absolute value of r is greater than the critical value then reject the null hypothesis.

Conclusion:

From part (b), the value of the correlation of Avg. number of passengers and Avg. one-way fare is –0.686. That is the absolute value of r is 0.686.

Here, the absolute value of r is less than the critical value.

That is, $\left|r\right|\left(=0.686\right)<\text{critical value}\left(=0.917\right)$.

By the rejection rule, the null hypothesis is not rejected.

Therefore, there is no sufficient evidence to support the claim that “there is a linear relation between the Avg. number of passengers and Avg. one-way fare”.

d.

To determine

To find: The regression equation for the given data

Answer to Problem 10.1.1RE

The regression equation for the given data is not valid.

Explanation of Solution

It is observed that the r is not significant.

Thus, the regression is not valid.

e.

To determine

To plot: The regression line on the scatter plot if appropriate.

Answer to Problem 10.1.1RE

The regression line cannot be plotted in the scatterplot.

Explanation of Solution

From part (d), it is observed that the regression line is not valid.

Thus, it is inappropriate to plot the regression line on the scatter plot.

f.

To determine

To predict: The value of y for a specific value of x if appropriate.

Answer to Problem 10.1.1RE

The value of y for a specific value of x is not possible.

Explanation of Solution

From part (d) the regression line is not valid.

Therefore it is not possible to predict the value of y for a specific value of x.