Chapter 12.3, Problem 35E

Refer back to the data in Exercise 4, in which y = ammonium concentration (mg/L) and x = transpiration (ml/h). Summary quantities include n = 13, Σx_i = 303.7, Σy_i = 52.8, S_xx = 1585.230769, S_v = −341.959231. and Syy = 77.270769.

1. a. Obtain the equation of the estimated regression line and use it to calculate a point prediction of ammonium concentration for a future observation made when ammonium concentration is 25 ml/h.
2. b. What happens if the estimated regression line is used to calculate a point estimate of true average concentration when transpiration is 45 ml/h? Why does it not make sense to calculate this point estimate?
3. c. Calculate and interpret s.
4. d. Do you think the simple linear regression model does a good job of explaining observed variation in concentration? Explain.

To determine

Find the interval estimate for the slope of the population regression.

Answer to Problem 35E

The 95% confidence interval for the slope of the population regression is $\underset{\_}{0.632\le {\beta }_1\le 2.440}$.

Explanation of Solution

Given info:

The summary statistics of the data correspond to the variables motion sickness dose $\left(x\right)$ and % reported nausea $\left(y\right)$. The results of the summary statistics are $n=17$, ${\displaystyle \sum _{i=1}^n{x}_i}=222.1$, ${\displaystyle \sum _{i=1}^n{y}_i}=193$, ${\displaystyle \sum _{i=1}^n{y}_i^2}=2,975$,${\displaystyle \sum _{i=1}^n{x}_i^2}=3,056.69$ and ${\displaystyle \sum _{i=1}^n{x}_i{y}_i}=2,759.6$. The range of the values of the variable motion sickness dose is 6.0 to 17.6.

Calculation:

Linear regression model:

In a linear equation $y=b_0+b_1{x}_i$ the constant $b_1$ be the slope and $b_0$ be the y-intercept and x is the independent variable and y is the independent variable.

A linear regression model is given as $\widehat{y}={\widehat{\beta }}_0+{\widehat{\beta }}_{1}x$ where $\widehat{y}$ be the predicted values of response variable and x be the predictor variable. The ${\widehat{\beta }}_1$ be the estimate of slope and ${\widehat{\beta }}_0$ be the estimate of intercept of the line.

Y-intercept:

In a linear equation $\widehat{y}={\widehat{\beta }}_0+{\widehat{\beta }}_{1}x$ the constant $b_1$ be the slope and $b_0$ be the y-intercept form and x is the independent variable and y is the independent variable.

The general formula to obtain y-intercept is,

$\begin{array}{c}{\widehat{\beta }}_0=\overline{y}-{\widehat{\beta }}_1\overline{x}\\ =\frac{{\displaystyle \sum _i{y}_i}-{\widehat{\beta }}_1{\displaystyle \sum _i{x}_i}}{n}\end{array}$

Slope:

In a linear equation $\widehat{y}={\widehat{\beta }}_0+{\widehat{\beta }}_{1}x$ the constant $b_1$ be the slope and $b_0$ be the y-intercept form and x is the independent variable and y is the independent variable.

The general formula to obtain slope is,

$\begin{array}{c}{\widehat{\beta }}_1=\frac{S_{xy}}{S_{xx}}\\ =\frac{\left[{\displaystyle \sum _i{x}_i{y}_i}-\frac{\left({\displaystyle \sum _i{x}_i}\right)\left({\displaystyle \sum _i{y}_i}\right)}{n}\right]}{{\displaystyle \sum _i{x}_i^2}-\frac{{\left({\displaystyle \sum _i{x}_i}\right)}^2}{n}}\end{array}$

The slope coefficient of the simple linear regression is,

$\begin{array}{c}{\widehat{\beta }}_1=\frac{\left[{\displaystyle \sum _i{x}_i{y}_i}-\frac{\left({\displaystyle \sum _i{x}_i}\right)\left({\displaystyle \sum _i{y}_i}\right)}{n}\right]}{{\displaystyle \sum _i{x}_i^2}-\frac{{\left({\displaystyle \sum _i{x}_i}\right)}^2}{n}}\\ =\frac{2759.6-\frac{222.1\times 193}{17}}{3,056.69-\frac{222.1\times 222.1}{17}}\\ =\frac{2759.6-2,521.488}{3,056.69-2,901.6712}\\ =\frac{238.112}{155.0188}\end{array}$

$=1.536$

Thus, the point estimate of the slope is ${\widehat{\beta }}_1=1.536$.

Total sum of square: (SST)

The total variation in the observed values of the response variable is defined as the total sum of squares. The formula for total sum of square is $SST={{\displaystyle \sum _i\left(y_i-\overline{y}\right)}}^2$ where $y_i$ be the i^th observation value and $\overline{y}$ be the sample mean.

The total sum of square is obtained as ,

$\begin{array}{c}SST=S_{yy}\\ ={\displaystyle \sum _i{y}_i^2}-\frac{{\left({\displaystyle \sum _i{y}_i}\right)}^2}{n}\\ =2,975-\frac{\left(193\times 193\right)}{17}\\ =783.8824\end{array}$

Therefore, the total sum of squares is $SST=S_{yy}=783.8824$

Regression sum of square: (SSR)

The variation in the observed values of the response variable explained by the regression is defined as the regression sum of squares. The formula for regression sum of square is $SSR={{\displaystyle \sum _i\left({\widehat{y}}_i-\overline{y}\right)}}^2$ where ${\widehat{y}}_i$ be the predicted value of the i^th observation and $\overline{y}$ be the sample mean.

The regression sum of squares is obtained as is,

$\begin{array}{c}SSR=\frac{S_{xy}^2}{S_{xx}}\\ =\frac{{\left[{\displaystyle \sum _i{x}_i{y}_i}-\frac{\left({\displaystyle \sum _i{x}_i}\right)\left({\displaystyle \sum _i{y}_i}\right)}{n}\right]}^2}{{\displaystyle \sum _i{x}_i^2}-\frac{{\left({\displaystyle \sum _i{x}_i}\right)}^2}{n}}\\ =\frac{{\left(2759.6-\frac{222.1\times 193}{17}\right)}^2}{3,056.69-\frac{222.1\times 222.1}{17}}\\ =\frac{{\left(2759.6-2,521.488\right)}^2}{3,056.69-2,901.6712}\\ =\frac{{238.112}^2}{155.0188}\end{array}$.

$=365.7448$

Error sum of square: (SSE)

The variation in the observed values of the response variable which is not explained by the regression is defined as the error sum of squares. The formula for error sum of square is $SSE={{\displaystyle \sum _i\left(y_i-\widehat{y}\right)}}^2$ where $y_i$ be the predicted value of the i^th observation and $\overline{y}$ be the predicted value for the i^th observation.

The general formula to obtain error sum of square is,

$SSE=SST-SSR$.

The error sum of squares is obtained as,

$\begin{array}{c}SSE=SST-SSR\\ =783.8824-365.7448\\ =418.1376\end{array}$

Therefore, the error sum of squares is $SSE=418.1376$

Estimate of error standard deviation:

The general formula for the estimate of error standard deviation is,

$\sigma =s=\sqrt{\frac{\text{SSE}}{n-2}}$.

The estimate of error standard deviation is obtained as,

$\begin{array}{c}s=\sqrt{\frac{\text{SSE}}{n-2}}\\ =\sqrt{\frac{418.1376}{17-2}}\\ =5.28\end{array}$

Thus, the estimate of error standard deviation is $\underset{\_}{s=5.28}$.

Error sum of square: (SSE)

The variation in the observed values of the response variable that is not explained by the regression is defined as the regression sum of squares. The formula for error sum of square is $SSE={{\displaystyle \sum _i\left(y_i-\widehat{y}\right)}}^2$ where $y_i$ be the predicted value of the i^th observation and $\overline{y}$ be the predicted value for the i^th observation.

Estimate of error standard deviation of slope coefficient:

The general formula for the estimate of error standard deviation of slope coefficient is,

${\sigma }_{{\widehat{\beta }}_1}=\frac{\sigma }{\sqrt{S_{xx}}}$,

The defining formula for $S_{\text{xx}}$ is,

$S_{xx}={\displaystyle \sum _i{x}_i^2}-\frac{{\left({\displaystyle \sum _i{x}_i}\right)}^2}{n}$

The estimate of error standard deviation of slope coefficient is,

$\begin{array}{c}{s}_{{\widehat{\beta }}_1}=\frac{s}{\sqrt{{\displaystyle \sum _i{x}_i^2}-\frac{{\left({\displaystyle \sum _i{x}_i}\right)}^2}{n}}}\\ =\frac{5.28}{\sqrt{3,056.69-\frac{222.1\times 222.1}{17}}}\\ =0.424\end{array}$

Thus, the estimate of error standard deviation of slope coefficient is $\underset{\_}{s_{{\widehat{\beta }}_1}=0.424}$.

Confidence interval:

The general formula for the confidence interval for the slope of the regression line is,

$CI={\widehat{\beta }}_1\pm t_{a/2,\left(n-2\right)}\times s_{{\widehat{\beta }}_1}$

Where, ${\widehat{\beta }}_1$ be the slope of the sample regression line, $s_{{\widehat{\beta }}_1}$ be the estimate of error standard deviation of slope coefficient.

Since, the level of confidence is not specified. The prior confidence level 95% can be used.

Critical value:

For 95% confidence level,

$\begin{array}{c}1-\alpha =1-0.95\\ \alpha =0.05\\ \frac{\alpha }{2}=\frac{0.05}{2}\\ =0.025\end{array}$

Degrees of freedom:

The sample size is $n=17$

The degrees of freedom is,

$\begin{array}{c}d.f=n-2\\ =17-2\\ =15\end{array}$

From Table A.5 of the t-distribution in Appendix A, the critical value corresponding to the right tail area 0.025 and 15 degrees of freedom is 2.131.

Thus, the critical value is $\left(t_{0.025,15}\right)=2.131$.

The 95% confidence interval is,

$\begin{array}{c}C.I={\widehat{\beta }}_1-\left(t_{a/2}\times s_{{\widehat{\beta }}_1}\right)\le {\beta }_1\le {\widehat{\beta }}_1+\left(t_{a/2}\times s_{{\widehat{\beta }}_1}\right)\\ =\left(1.536-\left(2.131\times 0.424\right)\le {\beta }_1\le 1.536+\left(2.131\times 0.424\right)\right)\\ =\left(1.536\pm 0.903544\right)\\ \simeq \left(0.632,2.440\right)\end{array}$

Thus, the 95% confidence interval for the slope of the population regression is $\underset{\_}{0.632\le {\beta }_1\le 2.440}$.

Interpretation:

There is 95% confident, that the expected change in % reported nausea associated with 1 unit increase in motion sickness dose lies between 0.632 and 2.440.

To determine

Test whether there is enough evidence to conclude that the predictor variable motion sickness dose is useful for predicting the value of the response variable % reported nausea.

Answer to Problem 35E

There is sufficient evidence to conclude that the predictor variable motion sickness dose is useful for predicting the value of the response variable % reported nausea.

Explanation of Solution

Calculation:

From part (a), the slope coefficient of the regression line is ${\widehat{\beta }}_1=1.536$.

The test hypotheses are given below:

Null hypothesis:

$H_0:{\beta }_1=0$

That is, there is no useful relationship between the variables motion sickness dose $\left(y\right)$ and % reported nausea $\left(x\right)$.

Alternative hypothesis:

$H_1:{\beta }_1\ne 0$

That is, there is useful relationship between the variables motion sickness dose $\left(y\right)$ and % reported nausea $\left(x\right)$.

T-test statistic:

The test statistic is,

$t=\frac{{\widehat{\beta }}_1-{\beta }_1}{s_{{\widehat{\beta }}_1}}\sim t\left(n-2\right)$

Degrees of freedom:

The sample size is $n=17$

The degrees of freedom is,

$\begin{array}{c}d.f=n-2\\ =17-2\\ =15\end{array}$

Thus, the degree of freedom is 15.

Level of significance:

Here, level of significance is not given.

So, the prior level of significance $\alpha =0.05$ can be used.

For the level of significance $\alpha =0.05$,

$\begin{array}{c}\frac{\alpha }{2}=\frac{0.05}{2}\\ =0.025\end{array}$

From Table A.5 of the t-distribution in Appendix A, the critical value corresponding to the right tail area 0.025 and 15 degrees of freedom is 2.131.

Thus, the critical value is $\left(t_{0.025,15}\right)=2.131$.

From part (a), the estimate of error standard deviation of slope coefficient is $s_{{\widehat{\beta }}_1}=0.424$.

Test statistic under null hypothesis:

Under the null hypothesis, the test statistic is obtained as follows:

$\begin{array}{c}t=\frac{{\widehat{\beta }}_1-{\beta }_1}{s_{{\widehat{\beta }}_1}}\\ =\frac{1.536-0}{0.424}\\ =3.6226\end{array}$

Thus, the test statistic is 3.6226.

Decision criteria for the classical approach:

If $\left|t\right|>t_{\frac{\alpha }{2}}\left(\text{test}\text{statistic}\text{> critical}\text{value}\right)$, then reject the null hypothesis $\left(H_0\right)$.

Conclusion:

Here, the test statistic is 3.6226 and critical value is 2.131.

The t statistic is greater than the critical value.

That is, $3.6226\left(\text{=test statistic}\right)>2.131\left(=\text{critical value}\right)$

Based on the decision rule, the null hypothesis is rejected.

Hence, there is a linear relationship between the predictor variable % reported nausea and the response variable motion sickness dose.

Therefore, there is sufficient evidence to conclude that the predictor variable motion sickness dose is useful for predicting the value of the response variable % reported nausea.

To determine

Check whether it is plausible to estimate the expected % reported nausea when the motion sickness dose is 5.0 using the obtained regression line.

Answer to Problem 35E

No, it is not plausible to estimate the expected % reported nausea when the motion sickness dose is 5.0 using the obtained regression line.

Explanation of Solution

Calculation:

Linear regression model:

A linear regression model is given as $\widehat{y}={\widehat{\beta }}_0+{\widehat{\beta }}_{1}x$ where $\widehat{y}$ be the predicted values of response variable and x be the predictor variable. The ${\widehat{\beta }}_1$ be the estimate of slope and ${\widehat{\beta }}_0$ be the estimate of intercept of the line.

Y-intercept:

In a linear equation $\widehat{y}={\widehat{\beta }}_0+{\widehat{\beta }}_{1}x$ the constant $b_1$ be the slope and $b_0$ be the y-intercept form and x is the independent variable and y is the independent variable.

The general formula to obtain y-intercept is,

$\begin{array}{c}{\widehat{\beta }}_0=\overline{y}-{\widehat{\beta }}_1\overline{x}\\ =\frac{{\displaystyle \sum _i{y}_i}-{\widehat{\beta }}_1{\displaystyle \sum _i{x}_i}}{n}\end{array}$

The y-intercept of the regression model is obtained as follows:

$\begin{array}{c}{\widehat{\beta }}_0=\frac{{\displaystyle \sum _i{y}_i}-{\widehat{\beta }}_1{\displaystyle \sum _i{x}_i}}{n}\\ =\frac{193-1.536\times 222.1}{17}\\ =-8.715\end{array}$

Thus, the y-intercept of the regression model is ${\widehat{\beta }}_0=-8.715$.

From part (a), the slope coefficient of the regression line is ${\widehat{\beta }}_1=1.536$.

Therefore, the regression equation of the variables motion sickness dose $\left(x\right)$ and % reported nausea $\left(y\right)$ is $\stackrel{⌢}{y}=-8.715+1.536x$.

Predicted value of % reported nausea when the motion sickness dose is 5.0:

The predicted value of % reported nausea when the motion sickness dose is 5.0 is obtained as follows:

$\begin{array}{c}\stackrel{⌢}{y}=-8.715+1.536x\\ =-8.715+1.536\times 0.5\\ =-7.947\end{array}$

Thus, the predicted value of % reported nausea for 5.0 motion sickness dose is –7.947.

Here, the % reported nausea is resulted as a negative value, which is not possible in reality.

Thus, the predicted value is a flaw.

Moreover, it is given that the range of the values of the variable motion sickness dose is 6.0 to 17.6.

The value 5.0 is outside the range of the variable motion sickness dose. That is, the observation 5.0 is not available.

Hence, the regression line may not give good estimate of expected % reported nausea when the motion sickness dose is 5.0.

Therefore, it is not plausible to estimate the expected % reported nausea when the motion sickness dose is 5.0 using the obtained regression line.

To determine

Find the interval estimate for the slope of the population regression after eliminating the observation $\left(6.0,2.50\right)$.

Comment whether the observation $\left(6.0,2.50\right)$ have a substantial impact on the regression model

Answer to Problem 35E

The 95% confidence interval for the slope of the population regression after eliminating the observation $\left(6.0,2.50\right)$ is $\underset{\_}{0.3719\le {\beta }_1\le 2.7301}$.

Yes, the observation $\left(6.0,2.50\right)$ has a substantial impact on the regression model

Explanation of Solution

Calculation:

Linear regression model:

In a linear equation $y=b_0+b_1{x}_i$ the constant $b_1$ be the slope and $b_0$ be the y-intercept and x is the independent variable and y is the independent variable.

A linear regression model is given as $\widehat{y}={\widehat{\beta }}_0+{\widehat{\beta }}_{1}x$ where $\widehat{y}$ be the predicted values of response variable and x be the predictor variable. The ${\widehat{\beta }}_1$ be the estimate of slope and ${\widehat{\beta }}_0$ be the estimate of intercept of the line.

Here, the observation $\left(6.0,2.50\right)$ has to be removed from the data set.

That is, the value 6.0 has to be removed from the variable motion sickness dose $\left(x\right)$ and 2.50 has to be removed from the variable % reported nausea $\left(y\right)$.

The results of the summary statistics after eliminating the observation $\left(6.0,2.50\right)$ from the data set are as follows:

Sample size:

$n=17-1=16$.

Sum of the variable:

${\displaystyle \sum _{i=1}^n{x}_i}=222.1-6=216.1$,${\displaystyle \sum _{i=1}^n{y}_i}=193-2.50=191.5$.

Sum of squares of the variable:

${\displaystyle \sum _{i=1}^n{x}_i^2}=3,056.69-6^2=3,020.69$,${\displaystyle \sum _{i=1}^n{y}_i^2}=2,975-{2.5}^2=2,968.75$,and ${\displaystyle \sum _{i=1}^n{x}_i{y}_i}=2,759.6-6\times 2.5=2,7444.6$.

Y-intercept:

In a linear equation $\widehat{y}={\widehat{\beta }}_0+{\widehat{\beta }}_{1}x$ the constant $b_1$ be the slope and $b_0$ be the y-intercept form and x is the independent variable and y is the independent variable.

The general formula to obtain y-intercept is,

$\begin{array}{c}{\widehat{\beta }}_0=\overline{y}-{\widehat{\beta }}_1\overline{x}\\ =\frac{{\displaystyle \sum _i{y}_i}-{\widehat{\beta }}_1{\displaystyle \sum _i{x}_i}}{n}\end{array}$

Slope:

In a linear equation $\widehat{y}={\widehat{\beta }}_0+{\widehat{\beta }}_{1}x$ the constant $b_1$ be the slope and $b_0$ be the y-intercept form and x is the independent variable and y is the independent variable.

The general formula to obtain slope is,

$\begin{array}{c}{\widehat{\beta }}_1=\frac{S_{xy}}{S_{xx}}\\ =\frac{\left[{\displaystyle \sum _i{x}_i{y}_i}-\frac{\left({\displaystyle \sum _i{x}_i}\right)\left({\displaystyle \sum _i{y}_i}\right)}{n}\right]}{{\displaystyle \sum _i{x}_i^2}-\frac{{\left({\displaystyle \sum _i{x}_i}\right)}^2}{n}}\end{array}$

The slope coefficient of the simple linear regression is,

$\begin{array}{c}{\widehat{\beta }}_1=\frac{\left[{\displaystyle \sum _i{x}_i{y}_i}-\frac{\left({\displaystyle \sum _i{x}_i}\right)\left({\displaystyle \sum _i{y}_i}\right)}{n}\right]}{{\displaystyle \sum _i{x}_i^2}-\frac{{\left({\displaystyle \sum _i{x}_i}\right)}^2}{n}}\\ =\frac{2,744.6-\frac{216.1\times 191.5}{16}}{3,020.69-\frac{216.1\times 216.1}{16}}\\ =\frac{2,744.6-2,586.447}{3,020.69-2,918.701}\\ =1.551\end{array}$

Thus, the point estimate of the slope is ${\widehat{\beta }}_1=1.551$.

Total sum of square: (SST)

The total variation in the observed values of the response variable is defined as the total sum of squares. The formula for total sum of square is $SST={{\displaystyle \sum _i\left(y_i-\overline{y}\right)}}^2$ where $y_i$ be the i^th observation value and $\overline{y}$ be the sample mean.

The total sum of square is obtained as ,

$\begin{array}{c}SST=S_{yy}\\ ={\displaystyle \sum _i{y}_i^2}-\frac{{\left({\displaystyle \sum _i{y}_i}\right)}^2}{n}\\ =2,968.75-\frac{\left(191.5\times 191.5\right)}{16}\\ =676.7344\end{array}$

Therefore, the total sum of squares is $SST=S_{yy}=676.7344$

Regression sum of square: (SSR)

The variation in the observed values of the response variable explained by the regression is defined as the regression sum of squares. The formula for regression sum of square is $SSR={{\displaystyle \sum _i\left({\widehat{y}}_i-\overline{y}\right)}}^2$ where ${\widehat{y}}_i$ be the predicted value of the i^th observation and $\overline{y}$ be the sample mean.

The regression sum of squares is obtained as is,

$\begin{array}{c}SSR=\frac{S_{xy}^2}{S_{xx}}\\ =\frac{{\left[{\displaystyle \sum _i{x}_i{y}_i}-\frac{\left({\displaystyle \sum _i{x}_i}\right)\left({\displaystyle \sum _i{y}_i}\right)}{n}\right]}^2}{{\displaystyle \sum _i{x}_i^2}-\frac{{\left({\displaystyle \sum _i{x}_i}\right)}^2}{n}}\\ =\frac{{\left(2,744.6-\frac{216.1\times 191.5}{16}\right)}^2}{3,020.69-\frac{216.1\times 216.1}{16}}\\ =\frac{{\left(2,744.6-2,586.447\right)}^2}{3,020.69-2,918.701}\\ =\frac{2,5012.41}{101.9894}\end{array}$.

$=245.2453$

Error sum of square: (SSE)

The variation in the observed values of the response variable which is not explained by the regression is defined as the error sum of squares. The formula for error sum of square is $SSE={{\displaystyle \sum _i\left(y_i-\widehat{y}\right)}}^2$ where $y_i$ be the predicted value of the i^th observation and $\overline{y}$ be the predicted value for the i^th observation.

The general formula to obtain error sum of square is,

$SSE=SST-SSR$.

The error sum of squares is obtained as,

$\begin{array}{c}SSE=SST-SSR\\ =676.7344-245.2453\\ =431.4891\end{array}$

Therefore, the error sum of squares is $SSE=431.4891$

Estimate of error standard deviation:

The general formula for the estimate of error standard deviation is,

$\sigma =s=\sqrt{\frac{\text{SSE}}{n-2}}$.

The estimate of error standard deviation is obtained as,

$\begin{array}{c}s=\sqrt{\frac{\text{SSE}}{n-2}}\\ =\sqrt{\frac{431.4891}{16-2}}\\ =5.552\end{array}$

Thus, the estimate of error standard deviation is $\underset{\_}{s=5.552}$.

Error sum of square: (SSE)

The variation in the observed values of the response variable that is not explained by the regression is defined as the regression sum of squares. The formula for error sum of square is $SSE={{\displaystyle \sum _i\left(y_i-\widehat{y}\right)}}^2$ where $y_i$ be the predicted value of the i^th observation and $\overline{y}$ be the predicted value for the i^th observation.

Estimate of error standard deviation of slope coefficient:

The general formula for the estimate of error standard deviation of slope coefficient is,

${\sigma }_{{\widehat{\beta }}_1}=\frac{\sigma }{\sqrt{S_{xx}}}$,

The defining formula for $S_{\text{xx}}$ is,

$S_{xx}={\displaystyle \sum _i{x}_i^2}-\frac{{\left({\displaystyle \sum _i{x}_i}\right)}^2}{n}$

The estimate of error standard deviation of slope coefficient is,

$\begin{array}{c}{s}_{{\widehat{\beta }}_1}=\frac{s}{\sqrt{{\displaystyle \sum _i{x}_i^2}-\frac{{\left({\displaystyle \sum _i{x}_i}\right)}^2}{n}}}\\ =\frac{5.552}{\sqrt{3,020.69-\frac{216.1\times 216.1}{16}}}\\ =0.5497\end{array}$

Thus, the estimate of error standard deviation of slope coefficient is $\underset{\_}{s_{{\widehat{\beta }}_1}=0.5497}$.

Confidence interval:

The general formula for the confidence interval for the slope of the regression line is,

$CI={\widehat{\beta }}_1\pm t_{a/2,\left(n-2\right)}\times s_{{\widehat{\beta }}_1}$

Where, ${\widehat{\beta }}_1$ be the slope of the sample regression line, $s_{{\widehat{\beta }}_1}$ be the estimate of error standard deviation of slope coefficient.

Since, the level of confidence is not specified. The prior confidence level 95% can be used.

Critical value:

For 95% confidence level,

$\begin{array}{c}1-\alpha =1-0.95\\ \alpha =0.05\\ \frac{\alpha }{2}=\frac{0.05}{2}\\ =0.025\end{array}$

Degrees of freedom:

The sample size is $n=16$

The degrees of freedom is,

$\begin{array}{c}d.f=n-2\\ =16-2\\ =14\end{array}$

From Table A.5 of the t-distribution in Appendix A, the critical value corresponding to the right tail area 0.025 and 14 degrees of freedom is 2.145.

Thus, the critical value is $\left(t_{0.025,14}\right)=2.145$.

The 95% confidence interval is,

$\begin{array}{c}C.I={\widehat{\beta }}_1-\left(t_{a/2}\times s_{{\widehat{\beta }}_1}\right)\le {\beta }_1\le {\widehat{\beta }}_1+\left(t_{a/2}\times s_{{\widehat{\beta }}_1}\right)\\ =\left(1.551-\left(2.145\times 0.5497\right)\le {\beta }_1\le 1.551+\left(2.145\times 0.5497\right)\right)\\ =\left(1.551\pm 1.1791\right)\\ \simeq \left(0.3719,2.7301\right)\end{array}$

Thus, the 95% confidence interval for the slope of the population regression is $\underset{\_}{0.3719\le {\beta }_1\le 2.7301}$.

Interpretation:

There is 95% confident, that the expected change in % reported nausea associated with 1 unit increase in motion sickness dose lies between 0.3719 and 2..7301.

Comparison:

The 95% confidence interval for the slope of the population regression with the observation $\left(6.0,2.50\right)$ is $\underset{\_}{0.632\le {\beta }_1\le 2.440}$.

The 95% confidence interval for the slope of the population regression after eliminating the observation $\left(6.0,2.50\right)$ is $\underset{\_}{0.3719\le {\beta }_1\le 2.7301}$.

Here, by observing both the intervals it is clear that the $\left(6.0,2.50\right)$ has an impact on the slope coefficient of the regression line.