Chapter 11, Problem 11CR

(a)

To determine

To test: The verification of $S_e=0.518$.

Answer to Problem 11CR

Solution: It is verified that, $S_e\approx 0.647$.

Explanation of Solution

Calculation: Let x represent the batting average of a professional baseball player and y represent the home run percentage.

We have, ${\displaystyle \sum x=1957},{\displaystyle \sum y=30.1,{\displaystyle \sum x^2=0.553,{\displaystyle \sum y^2=150.15}}},$

${\displaystyle \sum xy=8.753}$.

We have to calculate first $a,b$ as follows:

$\begin{array}{c}b=\frac{n({\displaystyle \sum xy})-({\displaystyle \sum x})({\displaystyle \sum y})}{n{\displaystyle \sum x^2-{({\displaystyle \sum x})}^2}}\\ =\frac{7(8.753)-(1.957)(30.1)}{7(0.553)-{(1.957)}^2}\\ \approx 55.166\end{array}$

$\begin{array}{c}\overline{x}=\frac{{\displaystyle \sum x}}{n}\\ =\frac{1.957}{7}\\ \approx 0.2796\end{array}$

$\begin{array}{c}\overline{y}=\frac{{\displaystyle \sum y}}{n}\\ =\frac{30.1}{7}\\ \approx 4.3\end{array}$

$\begin{array}{c}a=\overline{y}-b\overline{x}\\ =4.3-55.166(0.2796)\\ \approx -11.1244\end{array}$

$\begin{array}{c}{S}_e=\sqrt{\frac{{{\displaystyle \sum y}}^2-a{\displaystyle \sum y}-b{\displaystyle \sum xy}}{n-2}}\\ =\sqrt{\frac{150.15+11.1244(30.1)-55.166(8.753)}{7-2}}\\ =0.64745\\ \approx 0.647\end{array}$

Conclusion: Hence, it is verified that, $S_e\approx 0.647$

(b)

To determine

To test: The verification of $r\approx 0.948$ and the claim that $\rho >0$ at 0.01 level of significance.

Answer to Problem 11CR

Solution: It is verified that $r\approx 0.948$. We have sufficient evidence to conclude that the population correlation coefficient between x and y is positive at 1% level of significance.

Explanation of Solution

Calculation:

The correlation coefficient, r is calculated as follows:

$\begin{array}{c}r=\frac{n({\displaystyle \sum xy})-({\displaystyle \sum x})({\displaystyle \sum y})}{\sqrt{[n{\displaystyle \sum x^2-{({\displaystyle \sum x})}^2]}[n{\displaystyle \sum y^2-{({\displaystyle \sum y})}^2}]}}\\ =\frac{7(8.753)-(1.957)(30.1)}{\sqrt{[7(0.553)-{(1.957)}^2][7(150.15)-{(30.1)}^2]}}\\ \approx 0.948\end{array}$

Using the level of significance, $\alpha =0.01$.

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,

$\begin{array}{c}t=\frac{r\sqrt{n-2}}{\sqrt{1-r^2}}\\ =\frac{0.948\sqrt{7-2}}{\sqrt{1-{(0.948)}^2}}\\ \approx 6.66\end{array}$

The degrees of freedom are $d.f.=n-2=7-2=5$

The above test is right tailed test, so we can use the one-tail area in the student’s distribution table (Table 4 of the Appendix). From table, the p-value for the sample test statistic $t=6.66$ is $0.0005<P-value<0.005$. Since the interval containing the P-value is less than 0.01, hence we have to reject the null hypothesis at $\alpha =0.01$. Therefore, we can conclude that the data is statistically significant at 0.01 level of significance.

Conclusion: It is verified that $r\approx 0.948$. We have sufficient evidence to conclude that the population correlation coefficient between x and y is positive at 1% level of significance.

(c)

To determine

To test: The verification of $b\approx 55.166$ and the claim that $\beta >0$ at 0.01 level of significance.

Answer to Problem 11CR

Solution: It is verified that, $b\approx 55.166$. We have sufficient evidence to conclude that the slope is positive at 1% level of significance.

Explanation of Solution

Calculation:

The value b is calculated as follows:

$\begin{array}{c}b=\frac{n({\displaystyle \sum xy})-({\displaystyle \sum x})({\displaystyle \sum y})}{n{\displaystyle \sum x^2-{({\displaystyle \sum x})}^2}}\\ =\frac{7(8.753)-(1.957)(30.1)}{7(0.553)-{(1.957)}^2}\\ \approx 55.166\end{array}$

Using the level of significance, $\alpha =0.01$.

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 $t=6.67$ and P-value is 0.001.

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

$P-\text{value for one tailed test}=\frac{0.001}{2}=0.0005$.

Since P-value is less than 0.01, hence we can reject the null hypothesis at $\alpha =0.01$. Therefore, we can conclude that the data is statistically significant at 0.01 level of significance.

Conclusion: Hence, it is verified that, $b\approx 55.166$. We have sufficient evidence to conclude that the slope is positive at 1% level of significance.

(d)

To determine

To test: The verification of $\widehat{y}\approx -11.123+55.166x$ and the 90% confidence interval for the predicted home run percentage for a player with a batting average of 0.310.

Answer to Problem 11CR

Solution: It is verified that, $\widehat{y}\approx -11.123+55.166x$. The 90% confidence interval for the predicted home run percentage for a player with a batting average of 0.310 is from 4.49 to 7.46.

Explanation of Solution

Calculation: The results obtained in above part are $a\approx -11.12,b\approx 55.166$

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

$\widehat{y}\approx -11.12+55.166x$

The find 90% confidence interval for $y$ when $x=0.310$ by using MINITAB software is as:

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

Step 2: Select ‘y’ in Response and write 0.310 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: $(4.49,7.46)$.

Conclusion: Hence, it is verified that, $\widehat{y}\approx 0.856+0.402x$. The 90% confidence interval for $y$ when $x=10$ is $(4.49,7.46)$.