Question 1

The data below are the number of absences and the final grades of 9 randomly selected students from aliterature class. Find the equation of the regression line for the given data. What would be the predicted finalgrade if a student was absent 14 times? Round the regression line values to the nearest hundredth. Round thepredicted grade to the nearest whole number. $\begin{array}{l|c|c|c|c|c|c|c|c|c|}
\text { Number of absences, } \mathrm{x} & 0 & 3 & 6 & 4 & 9 & 2 & 15 & 8 & 5 \
\hline \text { Final grade, } \mathrm{y} & 98 & 86 & 80 & 82 & 71 & 92 & 55 & 76 & 82
\end{array}$
A) $\hat { y } = - 2.75 x + 96.14 ; 58$
B) $\hat { y } = 96.14 x - 2.75 ; 1343$
C) $\hat{ y } = - 2.75 \mathrm { x } - 96.14 ; 134.64$
D) $\hat { y } = - 96.14 \mathrm { x } + 2.75 ; 1343$

Accepted Answer

The correct equation of the regression line is $\hat { y } = - 2.75 x + 96.14$, which is used to predict the final grade based on the number of absences. Using this equation, if a student was absent 14 times, the predicted final grade would be 58 when rounded to the nearest whole number.

Question 2

Find the equation of the regression line for the given data. Round values to the nearest thousandth. $\begin{array}{l|c|c|c|c|c|c|c|c|c|c|}
\mathrm{x} & -5 & -3 & 4 & 1 & -1 & -2 & 0 & 2 & 3 & -4 \
\hline \mathrm{y} & 11 & -6 & 8 & -3 & -2 & 1 & 5 & -5 & 6 & 7
\end{array}$
A) $\hat { y }  = - 0.206 \mathrm { x } + 2.097$
B) $\hat { y } = 2.097 \mathrm { x } - 0.206$
C) $\hat { y } = 0.206 x - 2.097$
D) $\hat { y } = - 2.097 x + 0.206$

Accepted Answer

To find the equation of the regression line, we first calculate the slope ($m$) and the y-intercept ($b$) of the line. The formula for the slope is $m = \frac{n(\sum xy) - (\sum x)(\sum y)}{n(\sum x^2) - (\sum x)^2}$ and for the y-intercept is $b = \frac{\sum y - m(\sum x)}{n}$, where $n$ is the number of data points. After calculating these using the given data, we find the equation of the regression line to be $\hat { y }  = - 0.206 \mathrm { x } + 2.097$, rounding the coefficients to the nearest thousandth.

Question 3

A manager wishes to determine the relationship between the number of years her sales representatives havebeen employed by the firm and their amount of sales (in thousands of dollars) per month. Find the equation ofthe regression line for the given data. What would be the predicted sales if the sales representative wasemployed by the firm for 30 years Is this reasonable? Why or why not? Round the regression line values to thenearest hundredth. $\begin{array}{l|c|c|c|c|c|c|c|c|c|}
\text { Years employed, } \mathrm{x} & 2 & 3 & 10 & 7 & 8 & 15 & 3 & 1 & 11 \
\hline \text { Sales, } \mathrm{y} & 31 & 33 & 78 & 62 & 65 & 61 & 48 & 55 & 120
\end{array}$
A) $\hat  { y } = 3.53 \mathrm { x } + 37.92 ; \$ 143,820$; No; it is not reasonable. 30 years of employment is well outside the scope of the model.
B) $\hat  { y } = 3.53 \mathrm { x } + 37.92 ; \$ 143,820 ;$ Yes, it is reasonable.
C) $\hat  { y } = 3.53 \mathrm { x } - 37.92 ; \$ 67,980 ; \mathrm { No }$; it is not reasonable. 30 years of employment is well outside the scope of the model.
D) $\hat { y } = 3.53 \mathrm { x } - 37.92 ; \$ 67,980$; Yes; it is reasonable.

Accepted Answer

The correct regression equation is $\hat  { y } = 3.53 \mathrm { x } + 37.92$, and the predicted sales for 30 years of employment would be \$143,820. However, predicting for 30 years is not reasonable because it extrapolates far beyond the range of the data provided, which could lead to inaccurate predictions.

Question 4

The data below are the final exam scores of 10 randomly selected history students and the number of hoursthey slept the night before the exam. Find the equation of the regression line for the given data. What would bethe predicted score for a history student who slept 15 hours the previous night? Is this a reasonable question?Round your predicted score to the nearest whole number. Round the regression line values to the nearesthundredth. $\begin{array}{l|c|c|c|c|c|c|c|c|c|c|}
\text { Hours, } \mathbf{x} & 3 & 5 & 2 & 8 & 2 & 4 & 4 & 5 & 6 & 3 \
\hline \text { Scores, } \mathrm{y} & 65 & 80 & 60 & 88 & 66 & 78 & 85 & 90 & 90 & 71
\end{array}$
A) $\hat { y } = 5.04 \mathrm { x } + 56.11 ; 132$; No, it is not reasonable. 15 hours is well outside the scope of the model.
B)  $\hat { y }  = 5.04 \mathrm { x } + 56.11 ; 132$; Yes, it is reasonable,
C) $\hat { y } = - 5.04 x + 56.11 ; - 20$; No, it is not reasonable.
D)  $\hat { y } = - 5.04 \mathrm { x } + 56.11 ; - 20$; Yes, it is reasonable.

Accepted Answer

The regression equation is calculated based on the given data points, and extrapolating to 15 hours of sleep, which is beyond the range of the provided data, makes the prediction unreliable.

Question 5

In an area of the Great Plains, records were kept on the relationship between the rainfall (in inches) and theyield of wheat (bushels per acre). Find the equation of the regression line for the given data. Round theregression line values to the nearest thousandth. $\begin{array}{|l|c|c|c|c|c|c|c|c|c|}
\text { Rainfall (in inches), } \mathrm{x} & 10.5 & 8.8 & 13.4 & 12.5 & 18.8 & 10.3 & 7.0 & 15.6 & 16.0 \
\hline \text { Vield (hushels ner acre) y } & 505 & 462 & 588 & 590 & 824 & 492 & 319 & 760 & 788
\end{array}$
A) $\hat { y } = 4.379 x + 4.267$
B) $\hat { y } = - 4.379 \mathrm { x } + 4.267$
C) $\hat { y } = 4.267 \mathrm { x } + 4.379$
D) $\hat { y } = 4.267 x - 4.379$

Accepted Answer

To find the equation of the regression line, you typically use the formula $\hat{y} = mx + b$, where $m$ is the slope of the line and $b$ is the y-intercept. The slope $m$ is calculated based on the change in $y$ over the change in $x$, and the y-intercept $b$ is calculated such that the line best fits the data points in terms of minimizing the sum of squared differences between the observed values and the values predicted by the line. Without performing the actual calculations here, since the correct answer requires the use of statistical software or manual calculations involving the sum of the products of deviations of x and y from their means, divided by the sum of squared deviations of x from its mean (for the slope), and then using the mean values of x and y to solve for $b$, we cannot directly compute these values here. However, the correct choice is provided based on the assumption that the calculations were done correctly to arrive at the given options.

Question 6

A manager wishes to determine the relationship between the number of miles traveled (in hundreds of miles)by her sales representatives and their amount of sales (in thousands of dollars) per month. Find the equation ofthe regression line for the given data. What would be the predicted sales if the sales representative traveled 0miles? Is this reasonable? Why or why not? Round the regression line values to the nearest hundredth. $\begin{array}{|l|c|c|c|c|c|c|c|c|c|}
\text { Miles traveled, } \mathrm{x} & 2 & 3 & 10 & 7 & 8 & 15 & 3 & 1 & 11 \
\hline \text { Sales, } \mathrm{y} & 31 & 33 & 78 & 62 & 65 & 61 & 48 & 55 & 120 \
\hline
\end{array}$
A) $\hat { \mathrm { y } } = 3.53 \mathrm { x } + 37.92 ; \$ 37,920$; No; it is not reasonable for a representative to travel 0 miles and have a positive amount of sales.

B) $\hat { y } = 3.53 x + 37.92 ; \$ 3792 ;$ No; it is not reasonable for a representative to travel 0 miles and have a positive amount of sales.
C) $\hat { y } = 3.53 \mathrm { x } + 37.92 ; \$ 37,920$; Yes, it is reasonable.
D)  $\hat { y }= 37.92 \mathrm { x } + 3.53 ; \$ 3792 ; \mathrm { Yes }$, it is reasonable.

Accepted Answer

The equation of the regression line is given as $\hat { y } = 3.53x + 37.92$, which means if a sales representative traveled 0 miles, the predicted sales would be \$37,920. This prediction may not be reasonable because it suggests that sales can occur without any travel, which might not be practical for a sales role that typically requires travel to generate sales.

Question 7

The data below are ages and systolic blood pressures (measured in millimeters of mercury) of 9 randomlyselected adults. Find the equation of the regression line for the given data. What would be the predictedpressure if the age was 60? Round the predicted pressure to the nearest whole number. Round the regressionline values to the nearest hundredth. $\begin{array}{l|c|c|c|c|c|c|c|c|c|}
\text { Age, } \mathrm{x} & 38 & 41 & 45 & 48 & 51 & 53 & 57 & 61 & 65 \
\hline \text { Pressure, } \mathrm{y} & 116 & 120 & 123 & 131 & 142 & 145 & 148 & 150 & 152
\end{array}$
A) $\hat { y } = 1.49 x + 60.46 ; 150 \mathrm {~mm}$
B) $\hat { y } = 60.46 \mathrm { x } - 1.49 ; 3626 \mathrm {~mm}$
C) $\hat { y } = 1.49 \mathrm { x } - 60.46 ; 29 \mathrm {~mm}$
D) $\hat{ y } = 60.46 \mathrm { x } + 1.49 ; 3629 \mathrm {~mm}$

Accepted Answer

The correct equation of the regression line is found by calculating the slope and intercept from the given data, which results in $\hat { y } = 1.49 x + 60.46$. Using this equation to predict the pressure for an age of 60 gives a value of approximately 150 mmHg when rounded to the nearest whole number.

Question 8

The data below are the final exam scores of 10 randomly selected history students and the number of hoursthey slept the night before the exam. Find the equation of the regression line for the given data. What would bethe predicted score for a history student who slept 7 hours the previous night? Is this a reasonable question?Round the regression line values to the nearest hundredth, and round the predicted score to the nearest wholenumber. $\begin{array}{l|c|c|c|c|c|c|c|c|c|c|}
\text { Hours, } \mathrm{x} & 3 & 5 & 2 & 8 & 2 & 4 & 4 & 5 & 6 & 3 \
\hline \text { Scores, } \mathrm{y} & 65 & 80 & 60 & 88 & 66 & 78 & 85 & 90 & 90 & 71
\end{array}$
A) $\hat { y }  = 5.04 \mathrm { x } + 56.11 ; 91$; Yes, it is reasonable.
B) $\hat { y } = 5.04 \mathrm { x } + 56.11 ; 91$; No, it is not reasonable. 7 hours is well outside the scope of the model.
C) $\hat { y }  = - 5.04 \mathrm { x } + 56.11 ; 21$; No, it is not reasonable. 7 hours is well outside the scope of the model.
D) $\hat { y }  = - 5.04 \mathrm { x } + 56.11 ; 21$; Yes, it is reasonable.

Accepted Answer

Without the specific calculations provided in the question, it's not possible to directly verify the accuracy of the regression equation or the predicted score. However, assuming the regression equation and the predicted score are calculated correctly based on the given data, the choice that includes a positive relationship between hours slept and exam score (indicating that more hours slept is associated with a higher exam score) and considers predicting a score for 7 hours of sleep as reasonable (since 7 hours is within the range of the provided data) would be the correct choice.

Question 9

The data below are the average one-way commute times (in minutes) for selected students and the number ofabsences for those students during the term. Find the equation of the regression line for the given data. Whatwould be the predicted number of absences if the commute time was 95 minutes? Is this a reasonable question?Round the predicted number of absences to the nearest whole number. Round the regression line values to thenearest hundredth. $\begin{array} { | l | c | c | c | c | c | c | c | c | c | } 
\text { Commute time } ( \mathrm { min } ) , \mathrm { x } & 72 & 85 & 91 & 90 & 88 & 98 & 75 & 100 & 80 \
\hline \text { Number of absences, } \mathrm { y } & 3 & 7 & 10 & 10 & 8 & 15 & 4 & 15 & 5
\end{array}$
A) $\hat { y } = 0.45 \mathrm { x } - 30.27 ; 12$ absences; $Y e s$, it is reasonable.
B)$\hat { y }  = 0.45 \mathrm { x } - 30.27 ; 12$ absences; No, it is not reasonable. 95 minutes is well outside the scope of the model.
C) $\hat { y } = 0.45 \mathrm { x } + 30.27 ; 73$ absences; Yes, it is reasonable.
D) $\hat { y }  = 0.45 \mathrm { x } + 30.27 ; 73$ absences; No, it is not reasonable. 95 minutes is well outside the scope of the model.

Accepted Answer

The regression line is $\hat { y } = 0.45 \mathrm { x } - 30.27$. Substituting $x = 95$ into the regression line, we get $\hat { y } = 0.45 ( 95 ) - 30.27 = 12.38$. Rounding to the nearest whole number, the predicted number of absences is 12. Since 95 minutes is within the range of the commute times in the data, it is reasonable to use the regression line to predict the number of absences.

Question 10

In order for a companyʹs employees to work in a foreign office, they must take a test in the language of thecountry where they plan to work. The data below shows the relationship between the number of years thatemployees have studied a particular language and the grades they received on the proficiency exam. Find theequation of the regression line for the given data. Round the regression line values to the nearest hundredth. $\begin{array}{|l|c|c|c|c|c|c|c|c|c|}
\text { Number of years, } \mathrm{x} & 3 & 4 & 4 & 5 & 3 & 6 & 2 & 7 & 3 \
\hline \text { Grades on test, } \mathrm{y} & 61 & 68 & 75 & 82 & 73 & 90 & 58 & 93 & 72
\end{array}$
A) $\hat {y }= 6.91 x + 46.26$
B) $\hat  { y } = 6.91 \mathrm { x } - 46.26$
C) $\hat  { y } = 46.26 \mathrm { x } - 6.91$
D) $\hat { y } = 46.26 x + 6.91$

Accepted Answer

To find the equation of the regression line, you typically use the formula $\hat{y} = mx + b$, where $m$ is the slope of the line and $b$ is the y-intercept. The calculation of the slope ($m$) and intercept ($b$) requires statistical analysis of the given data points, often involving the use of statistical software or a calculator capable of regression analysis. The correct option, A, $\hat{y} = 6.91x + 46.26$, likely results from such an analysis, providing a linear model that best fits the given data points by minimizing the sum of the squares of the vertical distances of the points from the line.

Question 11

The data below are the average one-way commute times (in minutes) for selected students and the number ofabsences for those students during the term. Find the equation of the regression line for the given data. Whatwould be the predicted number of absences if the commute time was 40 minutes? Is this a reasonable question?Round the predicted number of absences to the nearest whole number. Round the regression line values to thenearest hundredth. $\begin{array}{l|c|c|c|c|c|c|c|c|c|}
\text { Commute time (min), } \mathrm{x} & 72 & 85 & 91 & 90 & 88 & 98 & 75 & 100 & 80 \
\hline \text { Number of absences. } \mathrm{y} & 3 & 7 & 10 & 10 & 8 & 15 & 4 & 15 & 5
\end{array}$
A) $hat {y }= 0.45 x - 30.27 ; - 12$ absences; No, it is not reasonable. 40 minutes is well outside the scope of the model.
B) $\hat  { y } = 0.45 \mathrm { x } - 30.27 ; - 12$ absences; $\mathrm { Yes }$, it is reasonable.
C) $\hat  { y } = 0.45 \mathrm { x } + 30.27 ; 48$ absences; Yes, it is reasonable.
D) $\hat  { y } = 0.45 \mathrm { x } + 30.27 ; 48$ absences; No, it is not reasonable. 40 minutes is well outside the scope of the model.

Accepted Answer

Without the actual calculation of the regression line and the predicted value for a 40-minute commute, it's impossible to directly verify the correct equation and prediction. However, predicting -12 absences is not reasonable because the number of absences cannot be negative. The statement about the 40-minute commute being outside the scope of the model is valid because the given data do not include commute times as low as 40 minutes, making extrapolation to this value potentially inaccurate.

Question 12

Find the equation of the regression line by letting Row 1 represent the x-values and Row 2 represent they-values. Now find the equation of the regression line letting Row 2 represent the x-values and Row 1represent the y-values. What effect does switching the explanatory and response variables have on theregression line? $$\begin{array} { l | c | c | c | c | c | c | c | c | c | c | } 
\text { Row 1 } & - 5 & - 3 & 4 & 1 & - 1 & - 2 & 0 & 2 & 3 & - 4 \
\hline \text { Row 2 } & - 10 & - 8 & 9 & 1 & - 2 & - 6 & - 1 & 3 & 6 & - 8
\end{array}$$

Accepted Answer

The answer of Find the equation of the regression line...

Quiz 4: Describing the Relation Between Two Variables

Find the equation of the regression line for the given data. Round values to the nearest thousandth. $\begin{array}{l|c|c|c|c|c|c|c|c|c|c|}\mathrm{x} & -5 & -3 & 4 & 1 & -1 & -2 & 0 & 2 & 3 & -4 \\\hline \mathrm{y} & 11 & -6 & 8 & -3 & -2 & 1 & 5 & -5 & 6 & 7\end{array}$