Chapter 11.4, Problem 7P

(a)

To determine

To test: The verification of ${\displaystyle \sum x=439},{\displaystyle \sum y=280,{\displaystyle \sum x^2=32.393,{\displaystyle \sum y^2=13.142}}},$

${\displaystyle \sum xy=20.599,r\approx \mathrm{0.784.}}$

Answer to Problem 7P

Solution: It is verified that, ${\displaystyle \sum x=439},{\displaystyle \sum y=280,{\displaystyle \sum x^2=32.393,{\displaystyle \sum y^2=13.142}}},$

${\displaystyle \sum xy=20.599,r\approx \mathrm{0.784.}}$

Explanation of Solution

Calculation: Let x be a random variable that represents the percentage of successful free throws a professional basketball player makes in a season. Let y be a random variable that represents the percentage of successful field goals a professional basketball player makes in a season.

The following table shows the $x,y,x^2,y^2,xy$:

$x$	$y$	$x^2$	$y^2$	$xy$
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
439	280	32393	13142	20599

In above table, the last column shows the total of corresponding column.

So, ${\displaystyle \sum x=439},{\displaystyle \sum y=280,{\displaystyle \sum x^2=32,393,{\displaystyle \sum y^2=13,142}}},$ ${\displaystyle \sum xy=20,599.}$

The correlation coefficient, r is calculated as follows:

$\begin{array}{c}r=\frac{n({\displaystyle \sum xy})-({\displaystyle \sum x})({\displaystyle \sum y})}{\sqrt{[n{\displaystyle \sum x^2-{({\displaystyle \sum x})}^2]}[n{\displaystyle \sum y^2-{({\displaystyle \sum y})}^2}]}}\\ =\frac{6(20599)-(439)(280)}{\sqrt{[6(32393)-{(439)}^2][6(13142)-{(280)}^2]}}\\ =\frac{674}{860.1884}\\ \approx 0.784\end{array}$

Conclusion: Hence, it is verified that,

${\displaystyle \sum x=439},{\displaystyle \sum y=280,{\displaystyle \sum x^2=32.393,{\displaystyle \sum y^2=13.142}}},$ ${\displaystyle \sum xy=20.599,r\approx \mathrm{0.784.}}$

(b)

To determine

To test: The claim that $\rho >0$ at 0.05 level of significance.

Answer to Problem 7P

Solution: We have sufficient evidence to conclude that the population correlation coefficient between x and y is positive at 5% level of significance.

Explanation of Solution

Calculation: Using the level of significance, $\alpha =0.05$.

The null hypothesis for testing is defined as,

$H_0:\rho =0$

The alternative hypothesis is defined as,

$H_1:\rho >0$

The sample test statistic is,

$\begin{array}{c}t=\frac{r\sqrt{n-2}}{\sqrt{1-r^2}}\\ =\frac{0.784\sqrt{6-2}}{\sqrt{1-{(0.784)}^2}}\\ \approx 2.526\end{array}$

The degrees of freedom are $d.f.=n-2=6-2=4$

The abovetest is right tailed test, so we can use the one-tail area in the student’s distributiontable (Table 4 of the Appendix). From table, the range of p-value for the sample test statistic $t=2.526$ is $0.025<P-value<0.05$. Since the interval containing the P-value is less than 0.05, hencewe have to reject the null hypothesis at $\alpha =0.05$. Therefore, we can conclude that the data is statistically significant at 0.05 level of significance.

Conclusion: We have sufficient evidence to conclude that the population correlation coefficient between x and y is positive at 5% level of significance.

(c)

To determine

To test: The verification of $S_e\approx 2.6964,a\approx 16.542,b\approx 0.4117\text{ and }\overline{x}=73.167$

Answer to Problem 7P

Solution: It is verified that, $S_e\approx 2.6964,a\approx 16.542,b\approx 0.4117\text{ and }\overline{x}=73.167$.

Explanation of Solution

Calculation: The results obtained in part (a) are,

$n=6,{\displaystyle \sum x=439},{\displaystyle \sum y=280,{\displaystyle \sum x^2=32,393,{\displaystyle \sum y^2=13,142}}},$ ${\displaystyle \sum xy=20,599.}$

Now, $a,b,S_e$ are calculated as follows:

$\begin{array}{c}b=\frac{n({\displaystyle \sum xy})-({\displaystyle \sum x})({\displaystyle \sum y})}{n{\displaystyle \sum x^2-{({\displaystyle \sum x})}^2}}\\ =\frac{6(20599)-(439)(280)}{6(32393)-{(439)}^2}\\ =\frac{674}{1637}\\ =0.41173\\ \approx 0.4117\end{array}$

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

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

$\begin{array}{c}a=\overline{y}-b\overline{x}\\ =46.667-0.4117(73.167)\\ \approx 16.54\end{array}$

$\begin{array}{c}{S}_e=\sqrt{\frac{{{\displaystyle \sum y}}^2-a{\displaystyle \sum y}-b{\displaystyle \sum xy}}{n-2}}\\ =\sqrt{\frac{13142-16.54(280)-0.4117(20599)}{6-2}}\\ =0.26959\\ \approx 2.696\end{array}$

Conclusion: Hence, it is verified that, $S_e\approx 2.6964,a\approx 16.542,b\approx 0.4117\text{ and }\overline{x}=73.167$

(d)

To determine

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

Answer to Problem 7P

Solution: 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 results obtained in above part are $a\approx 16.54,b\approx 0.4117$

The regression line equation is $\widehat{y}=a+bx$,

$\widehat{y}=16.54+0.4117x$

Now to find the predicted $\widehat{y}$ when $x=70\%$

$\begin{array}{c}\widehat{y}=16.54+0.4117(70)\\ \approx 45.36\end{array}$

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

(e)

To determine

To find: The 90% confidence interval for $y$ when $x=70$.

Answer to Problem 7P

Solution: The 90% confidence interval for $y$ when $x=70$ is $(39.06,51.67)$.

Explanation of Solution

Calculation: The find 90% confidence interval for $y$ when $x=70$ by using MINITAB software is as:

Step 1: Go to Stat >Regression>Regression > Predict.

Step 2: Select ‘y’ in Response and write 70 in ‘x’ box.

Step 3: Click on Options write 90 in ‘Confidence level’ and select ‘Two-sided’ in Type of interval. Then click on OK.

The 90% confidence interval is obtained as: $(39.06,51.67)$.

Interpretation: The 90% confidence interval for $y$ when $x=70$ is $(39.06,51.67)$.

(f)

To determine

To test: The claim that $\beta >0$ at 0.05 level of significance.

Answer to Problem 7P

Solution: We have sufficient evidence to conclude that the slopeis positive at 5% level of significance.

Explanation of Solution

Calculation: Using the level of significance, $\alpha =0.05$.

The null hypothesis for testing is defined as,

$H_0:\beta =0$

The alternative hypothesis is defined as,

$H_1:\beta >0$

The find t statistic and P-value using MINITAB software is as:

Step 1: Enter x and y in Minitab worksheet.

Step 2: Go to Stat > Regression > Regression >Fit Regression Model.

Step 2: Select ‘y’ in Response and ‘x’ in ‘Continuous predictors’ box. Then click on OK.

The sample test statistic is $t=2.52$ and P-value is 0.065.

The software gives the P-value for two-tailed test, for finding the p-value for one tailed testwe can divide the obtained P-value by 2.

$P-\text{value for one tailed test}=\frac{0.065}{2}=0.0325$.

Since P-value is less than 0.05, hence we can reject the null hypothesis at $\alpha =0.05$. Therefore, we can conclude that the data is statistically significant at 0.05 level of significance.

Conclusion: We have sufficient evidence to conclude that the slopeis positive at 5% level of significance.