Question 1

A company studying the productivity of their employees on a new information system was interested in determining if the age $( X )$ of data entry operators influenced the number of completed entries made per hour (Y). The regression equation is $\hat { y } = 14.374 - 0.145 x$. If $s _ { x } = 14.04$ and $s _ { y } = 2.61$, then the correlation coefficient between age and productivity is
A) $.779$
B) $- .236$
C) $.575$
D) $- .929$
E) $- .779$

Accepted Answer

The correlation coefficient $r$ can be calculated from the regression slope $b$ using the formula $r = b \cdot \frac{s_x}{s_y}$, where $b = -0.145$, $s_x = 14.04$, and $s_y = 2.61$. Substituting these values gives $r = -0.145 \cdot \frac{14.04}{2.61} = -0.779$.

Question 2

Suppose the correlation, r, between two variables x and y is -0.44.What percentage of the variability in y cannot be explained by x?
A)19%
B)44%
C)81%
D)88%
E)12%

Accepted Answer

A

Question 3

R-sq is a measure of ...
A)the change in the y-variable that corresponds with the change in the x-variable.
B)the probability that the regression line makes a correct prediction.
C)the percentage of the accuracy of the regression equation.
D)the initial predicted starting point of the response variable when x is zero.
E)the proportion of the variability in the response variable that is explained by the explanatory variable.

Accepted Answer

R-squared measures the proportion of the variability in the response variable that can be explained by the explanatory variable(s) in a regression model. It does not directly measure changes in variables, probabilities of correct predictions, the accuracy of the regression equation in a percentage form, or the initial predicted starting point of the response variable.

Question 4

Residuals are …
A)the difference between observed responses and values predicted by the model.
B)data collected from individuals that is not consistent with the rest of the group.
C)none of these
D)possible models not explored by the researcher.
E)variation in the data that is explained by the model.

Accepted Answer

Residuals are the difference between observed responses and values predicted by the model.

Question 5

An 8th grade class develops a linear model that predicts the number of cheerios (a small round cereal) that fit on the circumference of a plate by using the diameter in inches.Their
Model is cheerios = 0.56 + 5.11(diameter).
The slope of this model is best interpreted in context as…
A)For every 5.11 inches of diameter, the circumference is about 1 cheerio bigger.
B)For every 1 inch of diameter, the circumference holds about 0.56 more cheerios.
C)It takes 5.11 cheerios to fill a plate' s circumference.
D)For every 1 inch of diameter, the circumference holds about 5.11 more cheerios.
E)A mistake, because π is about 3.14 and that should be the slope.

Accepted Answer

The slope in a linear model represents the rate of change. In this context, the slope of 5.11 indicates that for every 1 inch increase in diameter, the circumference can hold about 5.11 more cheerios.

Question 6

Data were collected on monthly sales revenues (in $\$ 1,000 \mathrm {~s}$ ) and monthly advertising expenditures $( \$ 100 \mathrm {~s} )$ for a sample of drug stores. The regression line relating revenues $( \mathrm { Y } )$ to advertising expenditure $( X )$ is estimated to be $\hat { y } = - 48.3 + 9.00 x$. The correct interpretation of the slope is that for each additional
A) $\$ 1$ spent on advertising, predicted sales revenue increases by $\$ 9,000$.
B) $\$ 100$ spent on advertising, predicted sales revenue increases by $\$ 9,000$.
C) $\$ 100$ spent on advertising, predicted sales revenue decreases by $\$ 9,000$.
D) $\$ 1,000$ in sales revenue, advertising expenditures decrease by $\$ 48.30$.
E) $\$ 100$ in sales revenue, advertising expenditures decrease by $\$ 48.30$.

Accepted Answer

The slope of the regression line is 9.00, which means for every \$100 increase in advertising expenditure (since advertising expenditure is measured in \$100s), the predicted sales revenue increases by \$9,000 (because sales revenue is measured in \$1,000s, and 9 * 1,000 = 9,000).

Question 7

An 8th grade class develops a linear model that predicts the number of cheerios (a small round cereal) that fit on the circumference of a plate by using the diameter in inches.Their
Model is cheerios = 0.56 + 5.11(diameter).
If the diameter is increased from 4 inches to 14 inches, the predicted number of cheerios will
Increase by about…
A)72
B)51
C)10
D)21
E)none of these above

Accepted Answer

The increase in the number of cheerios is calculated by the change in diameter multiplied by the slope of the linear model: 5.11 * (14 - 4) = 5.11 * 10 = 51.

Question 8

The correlation coefficient between high school grade point average (GPA) and college GPA is 0.560.For a student with a high school GPA that is 2.5 standard deviations above the
Mean, we would expect that student to have a college GPA that is ______ the mean.
A)0.56 SD above
B)1.94 SD above
C)2.5 SD above
D)equal to
E)1.4 SD above

Accepted Answer

The correlation coefficient (0.560) indicates the strength and direction of a linear relationship between two variables. In this case, a high school GPA that is 2.5 standard deviations above the mean would result in a college GPA that is 0.560 * 2.5 = 1.4 standard deviations above the mean, hence option E is correct.

Question 9

If the point in the upper right corner of this scatterplot is removed from the data set, then what will happen to the slope of the line of best fit (b) and to the correlation (r)? 
A)b will increase, and r will decrease.
B)b will decrease, and r will increase.
C)both will decrease.
D)both will increase.
E)both will remain the same.

Accepted Answer

The answer of If the point in the upper right...

Question 10

Suppose the correlation, r, between two variables x and y is -0.44.What would you predict about a y value if the x value is 2 standard deviations above its mean?
A)It will be .88 standard deviations below its mean.
B)It will be .88 standard deviations above its mean.
C)It will be 2 standard deviations below its mean.
D)It will be .44 standard deviations below its mean.
E)It will be .44 standard deviations above its mean.

Accepted Answer

The correlation coefficient, r, indicates the strength and direction of a linear relationship between two variables. A negative correlation (r = -0.44) means that as one variable increases, the other decreases. If x is 2 standard deviations above its mean, y would be 2 * -0.44 = -0.88 standard deviations from its mean, indicating it would decrease, hence option A is correct.

Question 11

A small independent organic food store offers a variety of specialty coffees.To determine whether price has an impact on sales, the managers kept track of how many pounds of each
Variety of coffee were sold last month. $$\begin{array} { | l | l | l | } 
\hline & \text { PRICE PER POUND } & \text { POUNDS SOLD } \
\hline \text { Mean } & \$ 8.75 & 54.50 \
\hline \begin{array} { c } 
\text { Standard } \
\text { Deviation }
\end{array} & \$ 3.63 & 18.33 \
\hline \text { Correlation } & - \mathbf { 0 . 9 2 7 } & \
\hline
\end{array}$$ Based on the summary statistics shown below, what percent of the variability in the number
Of pounds of coffee sold per week can be explained by price?
A)95.47%
B)100%
C)85.9%
D)55.6%
E)4.68%

Accepted Answer

The coefficient of determination, $ R^2 $, is the square of the correlation coefficient. Here, $ R^2 = (-0.927)^2 = 0.859 $, which means 85.9% of the variability in the number of pounds of coffee sold can be explained by price.

Question 12

A small independent organic food store offers a variety of specialty coffees.To determine whether price has an impact on sales, the managers kept track of how many pounds of each
Variety of coffee were sold last month. $$\begin{array} { | l | l | l | } 
\hline & \text { PRICE PER POUND } & \text { POUNDS SOLD } \
\hline & \$ 3.99 & 75 \
\hline & \$ 5.99 & 60 \
\hline & \$ 7.00 & 65 \
\hline & \$ 12.00 & 45 \
\hline & \$ 4.50 & 80 \
\hline & \$ 7.50 & 70 \
\hline & \$ 15.00 & 25 \
\hline & \$ 10.00 & 35 \
\hline & \$ 12.50 & 40 \
\hline \text { Mean } & \$ 8.99 & 50 \
\hline \text { Standard } & & \
\hline \text { Deviation } & \$ 3.63 & 54.50 \
\hline & & 18.33 \
\hline
\end{array}$$ Based on the data and summary statistics shown below, the slope of the estimated regression
Line that relates the response variable (monthly sales) to the predictor variable (price per
Pound) is
A)95.459.
B).858.
C)-4.681.
D)-.858.
E)-8.999.

Accepted Answer

The slope of the estimated regression line in a simple linear regression can be calculated using the formula $slope = \frac{r \cdot s_y}{s_x}$, where $r$ is the correlation coefficient, $s_y$ is the standard deviation of the response variable (pounds sold), and $s_x$ is the standard deviation of the predictor variable (price per pound). Given the negative relationship observed in the data (as price increases, pounds sold tends to decrease), and the provided standard deviations, the correct slope must be negative, making -4.681 the plausible choice given the options and the context of the question.

Question 13

A company studying the productivity of its employees on a new information system was interested in determining if the age $( X )$ of data entry operators influenced the number of completed entries made per hour (Y). The regression equation is $\hat { y } = 14.374 - 0.145 x$. Suppose the actual completed entries per hour for an operator who is 35 years old was 8 . The residual is
A) $- 1.3$
B) $2.6$
C) $- 3.5$
D) $1.3$
E) $- 2.2$

Accepted Answer

The residual is the difference between the actual value and the predicted value. The predicted value for an operator who is 35 years old is $\hat { y } = 14.374 - 0.145 x = 14.374 - 0.145 (35) = 9.624$. The residual is therefore 8 - 9.624 = -1.624, which is closest to -1.3.

Question 14

It' s easy to measure the circumference of a tree' s trunk, but not so easy to measure its height. Foresters developed a model for ponderosa pines that they use to predict the tree' $s$ height (in feet) from the circumference of its trunk (in inches): $\ln \hat { h } = - 1.2 + 1.4 ( \ln C$ ). A lumberjack finds a tree with a circumference of $60 ^ { \prime \prime }$; how tall does this model estimate the tree to be?
A) 11 ,
B) $83 ^ { \prime }$
C) $93 ^ { \prime }$,
D) $19 ^ { \prime }$
E) 5 '

Accepted Answer

To find the estimated height, substitute $C = 60$ into the model: $\ln \hat{h} = -1.2 + 1.4 (\ln 60)$. Calculate $\ln 60$, then multiply by 1.4, add -1.2, and exponentiate the result to find $\hat{h}$. This gives approximately 93 feet.

Question 15

A small independent organic food store offers a variety of specialty coffees.To determine whether price has an impact on sales, the managers kept track of how many pounds of each
Variety of coffee were sold last month. $$\begin{array} { | l | l | l | } 
\hline & \text { PRICE PER POUND } & \text { POUNDS SOLD } \
\hline & \$ 3.99 & 75 \
\hline & \$ 5.99 & 60 \
\hline & \$ 7.00 & 65 \
\hline & \$ 12.00 & 45 \
\hline & \$ 4.50 & 80 \
\hline & \$ 7.50 & 70 \
\hline & \$ 15.00 & 25 \
\hline & \$ 10.00 & 35 \
\hline & \$ 12.50 & 40 \
\hline & \$ 8.99 & 50 \
\hline \text { Mean } & & \
\hline \text { Ctandard } & \$ 8.75 & 54.50 \
\hline \text { Deviation } & & 18.33 \
\hline
\end{array}$$ Based on the data and summary statistics, the intercept of the estimated regression line that
Relates the response variable (monthly sales) to the predictor variable (price per pound) is
A)95.459.
B).858.
C)-4.684.
D)-.858.
E)-8.999.

Accepted Answer

The intercept of the estimated regression line can be calculated using the formula for the intercept in a simple linear regression, which is $\bar{y} - b\bar{x}$, where $\bar{y}$ is the mean of the dependent variable (pounds sold), $\bar{x}$ is the mean of the independent variable (price per pound), and $b$ is the slope of the line. Given the mean price per pound is \$8.75 and the mean pounds sold is 54.5, and without the slope directly provided but knowing the relationship between price and quantity sold is generally negative (as price increases, quantity sold decreases), the only positive intercept option provided that makes sense with the given means is 95.459. This suggests that if the price were \$0, the model predicts sales of 95.459 pounds, assuming the linear relationship holds.

Question 16

Data were collected on monthly sales revenues (in $\$ 1,000 \mathrm {~s}$ ) and monthly advertising expenditures $( \$ 100$ s) for a sample of drug stores. The regression line relating revenues (Y) to advertising expenditure $( \mathrm { X } )$ is estimated to be $\hat { y } = - 48.3 + 9.00 x$. The predicted sales revenue for a month in which $\$ 1,000$ was spent on advertising is
A) $\$ 50,000$.
B) $\$ 851.70$.
C) $\$ 8,951.70$.
D) $\$ 41,700$.
E) $\$ 90,000$.

Accepted Answer

The answer of Data were collected on monthly sales revenues...

Question 17

A regression analysis of students' college grade point averages (GPAs) and their high school GPAs found $R ^ { 2 } = 0.311$. Which of these is true?
I. High school GPA accounts for $31.1 \%$ of college GPA.
II. $31.1 \%$ of college GPAs can be correctly predicted with this model.
III. $31.1 \%$ of the variance in college GPA can be accounted for by the model
A) I only
B) II only
C) III only
D) I and II
E) none of these

Accepted Answer

III is correct because $R^2$ (the coefficient of determination) measures the proportion of the variance in the dependent variable that is predictable from the independent variable(s). In this case, it means that $31.1\%$ of the variance in college GPA can be explained by the model that includes high school GPA.

Question 18

If the point in the upper left corner of the scatterplot is removed, what will happen to the correlation (r) and the slope of the line of best fit (b)? 
A) $r$ will decrease and $b$ will increase.
B) $r$ will increase and $b$ will decrease.
C) Both will decrease.
D) Both will increase.
E) They will not change.

Accepted Answer

The answer of If the point in the upper left...

Quiz 7: Linear Regression

Suppose the correlation, r, between two variables x and y is -0.44.What percentage of the variability in y cannot be explained by x?

R-sq is a measure of ...

Residuals are …

An 8th grade class develops a linear model that predicts the number of cheerios (a small round cereal) that fit on the circumference of a plate by using the diameter in inches.Their Model is cheerios = 0.56 + 5.11(diameter) . The slope of this model is best interpreted in context as…

The correlation coefficient between high school grade point average (GPA) and college GPA is 0.560.For a student with a high school GPA that is 2.5 standard deviations above the Mean, we would expect that student to have a college GPA that is ______ the mean.

If the point in the upper right corner of this scatterplot is removed from the data set, then what will happen to the slope of the line of best fit (b) and to the correlation (r) ?

Suppose the correlation, r, between two variables x and y is -0.44.What would you predict about a y value if the x value is 2 standard deviations above its mean?

If the point in the upper left corner of the scatterplot is removed, what will happen to the correlation (r) and the slope of the line of best fit (b) ?