Chapter 9.3, Problem 7P

(a)

To determine

Verify the values of ${\displaystyle \sum x}$, ${\displaystyle \sum x^2}$, ${\displaystyle \sum y}$, ${\displaystyle \sum y^2}$, ${\displaystyle \sum xy}$ and r.

Explanation of Solution

Calculation:

The variable x denotes a random variable that represents the percentage of successful free throws a professional basketball player makes in a season and y denotes a random variable that represents the percentage of successful field goals a professional basketball player makes in a season.

The formula for correlation coefficient is,

$r=\frac{n{\displaystyle \sum xy}-\left({\displaystyle \sum x}\right)\left({\displaystyle \sum y}\right)}{\sqrt{n{\displaystyle \sum x^2}-{\left({\displaystyle \sum x}\right)}^2}\sqrt{n{\displaystyle \sum y^2}-{\left({\displaystyle \sum y}\right)}^2}}$

In the formula, n is the sample size.

The values are verified in the table below,

x	y	$x^2$	$y^2$	xy
67	44	4489	1936	2948
65	42	4225	1764	2730
75	48	5625	2304	3600
86	51	7396	2601	4386
73	44	5329	1936	3212
73	51	5329	2601	3723
${\displaystyle \sum x}=439$	${\displaystyle \sum y}=280$	${\displaystyle \sum x^2}=32,393$	${\displaystyle \sum y^2}=13,142$	${\displaystyle \sum xy}=20,599$

Hence, the values are verified.

The number of data pairs are $n=6$. The value of correlation r is,

$\begin{array}{c}r=\frac{6\left(20599\right)-\left(439\times 280\right)}{\sqrt{6\left(32393\right)-{\left(439\right)}^2}\sqrt{6\left(13142\right)-{\left(280\right)}^2}}\\ =\frac{123594-122920}{\sqrt{1637}\sqrt{452}}\\ =\frac{674}{860.18835}\\ =0.784\end{array}$

Hence, the value of r is verified as approximately 0.784.

(b)

To determine

Check whether the claim $\rho >0$ is significant or not.

Answer to Problem 7P

The claim $\rho >0$ is significant.

Explanation of Solution

Calculation:

Null hypothesis:

$H_0:\rho =0$

Alternative hypothesis:

$H_1:\rho >0$

Test statistic:

The test statistic formula for test correlation r is,

$t=\frac{r\sqrt{n-2}}{\sqrt{1-r^2}}$

Where r is the sample correlation coefficient, n is the sample size with degrees of freedom $d.f.=n-2$.

Substitute r as 0.784, and n as 6 in the test statistic formula.

$\begin{array}{c}t=\frac{0.784\sqrt{6-2}}{\sqrt{1-{\left(0.784\right)}^2}}\\ =\frac{1.568}{0.6208}\\ =2.526\end{array}$

The test statistic value is 2.526.

The degrees of freedom is,

$\begin{array}{c}d.f.=6-2\\ =4\end{array}$

Step by step procedure to obtain P-value using MINITAB software is given below:

• Choose Graph > Probability Distribution Plot choose View Probability > OK.
• From Distribution, choose ‘t’ distribution.
• Enter the Degrees of freedom as 4.
• Click the Shaded Area tab.
• Choose X Value and Right Tail, for the region of the curve to shade.
• Enter the X value as 2.526.
• Click OK.

Output using MINITAB software is given below:

From Minitab output, the P-value is 0.0325.

Rejection rule:

• If the P-value is less than or equal to $\alpha$, then reject the null hypothesis and the test is statistically significant. That is, $P\text{-value}\le \alpha$.

Conclusion:

The P-value is 0.0325 and the level of significance is 0.05.

The P-value is less than the level of significance.

That is, $0.0325\left(=P\text{-value}\right)<0.05\left(=\alpha \right)$.

By the rejection rule, the null hypothesis is rejected.

Hence, the claim $\rho >0$ is significant at level 0.05. That is, there is a positive correlation between x and y.

(c)

To determine

Verify the values of $S_e$, a, b, and $\overline{x}$.

Explanation of Solution

Calculation:

The value of $\overline{x}$ is,

$\begin{array}{c}\overline{x}=\frac{{\displaystyle \sum x}}{n}\\ =\frac{439}{6}\\ =73.167\end{array}$

The value of $\overline{x}$ is 73.167.

The value of $\overline{y}$ is,

$\begin{array}{c}\overline{y}=\frac{{\displaystyle \sum y}}{n}\\ =\frac{280}{6}\\ =46.667\end{array}$

The value of $\overline{y}$ is 46.667.

The value of b is,

$\begin{array}{c}b=\frac{n{\displaystyle \sum xy}-\left({\displaystyle \sum x}\right)\left({\displaystyle \sum y}\right)}{n{\displaystyle \sum x^2}-{\left({\displaystyle \sum x}\right)}^2}\\ =\frac{6\left(20599\right)-\left(439\times 280\right)}{6\left(32393\right)-{\left(439\right)}^2}\\ =\frac{674}{1637}\\ =0.4117\end{array}$

The value of b is 0.4117.

The value of a is,

$\begin{array}{c}a=\overline{y}-b\overline{x}\\ =46.667-\left(0.41173\right)73.167\\ =46.667-30.125\\ =16.542\end{array}$

The value of a is 16.542.

The value of $S_e$ is,

$\begin{array}{c}{S}_e=\sqrt{\frac{{\displaystyle \sum y^2-a{\displaystyle \sum y}-b{\displaystyle \sum xy}}}{n-2}}\\ =\sqrt{\frac{13142-\left(16.542\times 280\right)-\left(0.41173\times 20599\right)}{6-2}}\\ =\sqrt{\frac{29.01373}{4}}\end{array}$

     $\begin{array}{c}=\sqrt{7.2706}\\ =2.6964\end{array}$

The value of $S_e$ is 2.6964.

The equation of the least-squares line is,

$\begin{array}{c}\widehat{y}=a+bx\\ =16.542+0.4117x\end{array}$

The regression equation is $\widehat{y}=16.542+0.4117x$.

(d)

To determine

Find the predicted percentage $\widehat{y}$ of successful field goals for a player with $x=70\%$ successful free throws.

Answer to Problem 7P

The predicted percentage $\widehat{y}$ of successful field goals for a player with $x=70\%$ successful free throws is 45.36%.

Explanation of Solution

Calculation:

The regression equation is $\widehat{y}=16.542+0.4117x$.

Substitute x as 70 in the regression equation

$\begin{array}{c}\widehat{y}=16.542+0.4117\left(70\right)\\ =16.542+28.819\\ =45.361\\ \approx 45.36\%\end{array}$

Hence, the predicted percentage $\widehat{y}$ of successful field goals for a player with $x=70\%$ successful free throws is 45.36%.

(e)

To determine

Find the 90% confidence interval for y when $x=70\%$.

Answer to Problem 7P

The 90% confidence interval for y when $x=70\%$ is $39.06<y<51.67$.

Explanation of Solution

Calculation:

Step by step procedure to obtain confidence interval using MINITAB software is given below:

• Choose Stat > Regression > Regression.
• In Response, enter the column containing the response as y.
• In Predictors, enter the columns containing the predictor as x.
• Choose Options.
• In Prediction intervals for new observations, enter the value as 70.
• In Confidence level, enter value as 90.
• Click OK.

Output using MINITAB software is given below:

From Minitab output, the confidence interval is $39.06<y<51.67$.

Hence, the 90% confidence interval for y when $x=70\%$ is $39.06<y<51.67$.

(f)

To determine

Check whether the claim $\beta >0$ is significant or not.

Answer to Problem 7P

The claim $\beta >0$ is significant.

Explanation of Solution

Calculation:

Null hypothesis:

$H_0:\beta =0$

Alternative hypothesis:

$H_1:\beta >0$

Test statistic:

The test statistic formula for slope $\beta$ is,

$t=\frac{b-\beta }{\frac{S_e}{\sqrt{{\displaystyle \sum x^2}-\frac{1}{n}{\left({\displaystyle \sum x}\right)}^2}}}$

Where $S_e$ is the standard error of estimate computed from the sample, n is the sample size with degrees of freedom $d.f.=n-2$.

Substitute $S_e$ as 2.6964, ${\displaystyle \sum x^2}$ as 32393, ${\displaystyle \sum x}$ as 439 and n as 6 in the test statistic formula

$\begin{array}{c}t=\frac{0.4117}{\frac{2.6964}{\sqrt{32393-\frac{1}{6}{\left(439\right)}^2}}}\\ =\frac{6.8003}{0.6208}\\ =2.522\end{array}$

The test statistic value is 2.522.

The degrees of freedom is,

$\begin{array}{c}d.f.=6-2\\ =4\end{array}$

Step by step procedure to obtain P-value using MINITAB software is given below:

• Choose Graph > Probability Distribution Plot choose View Probability > OK.
• From Distribution, choose ‘t’ distribution.
• Enter the Degrees of freedom as 4.
• Click the Shaded Area tab.
• Choose X Value and Right Tail, for the region of the curve to shade.
• Enter the X value as 2.522.
• Click OK.

Output using MINITAB software is given below:

From Minitab output, the P-value is 0.0326.

Rejection rule:

• If the P-value is less than or equal to $\alpha$, then reject the null hypothesis and the test is statistically significant. That is, $P\text{-value}\le \alpha$.

Conclusion:

The P-value is 0.0326 and the level of significance is 0.05.

The P-value is less than the level of significance.

That is, $0.0326\left(=P\text{-value}\right)<0.05\left(=\alpha \right)$.

By the rejection rule, the null hypothesis is rejected.

Hence, the claim $\beta >0$ is significant at level 0.05. That is, there is a positive correlation between x and y.

(g)

To determine

Find a 90% confidence interval for $\beta$.

Interpret the confidence interval.

Answer to Problem 7P

The 90% confidence interval for $\beta$ is $0.064<\beta <0.760$.

Explanation of Solution

Calculation:

Confidence interval for slope:

The confidence interval formula for slope $\beta$ is,

$b-E<\beta <b+E$

Where $E=\frac{t_c{S}_e}{\sqrt{{\displaystyle \sum x^2-\frac{1}{n}{\left({\displaystyle \sum x}\right)}^2}}}$, $t_c$ is the critical value, $S_e$ is the standard error of estimate computed from the sample, n is the sample size with degrees of freedom $d.f.=n-2$.

Critical value:

Use the Appendix II: Tables, Table 6: Critical Values for Student’s t Distribution:

• In d.f. column locate the value 4.
• In the row of two-tail area locate the level of significance $\alpha =0.10$.
• The intersecting value of row and columns is 2.132.

The critical value is $\pm 2.132$.

The margin of error is,

$\begin{array}{c}E=\frac{2.132\times 2.6964}{\sqrt{32393-\frac{1}{6}{\left(439\right)}^2}}\\ =\frac{5.7487}{16.5177}\\ =0.3480\end{array}$

The 90% confidence interval for $\beta$ is,

$\begin{array}{c}0.4117-0.3480<\beta <0.4117+0.3480\\ 0.064<\beta <0.760\end{array}$

Hence, the 90% confidence interval for $\beta$ is $0.064<\beta <0.760$.

The percentage of successful field goals for a basketball player increases by an amount that ranges between 0.064 and 0.760, if percentage of successful free throws increases by one unit.