Question 1

One of the primary feeds for beef cattle is corn. The following table presents the average price in dollars for a bushel of corn and a pound of ribeye steak for 10 consecutive months.
 $\begin{array} { c c } 
\hline \text { Corn Price } ( \$ / \text { bu } ) & \text { Ribeye Price } ( \$ / \text { lb) } \
\hline 5.82 & 13.01 \
5.70 & 12.24 \
5.80 & 13.01 \
5.84 & 12.94 \
6.56 & 13.89 \
6.21 & 13.04 \
6.39 & 13.01 \
6.61 & 13.76 \
6.06 & 12.53 \
5.84 & 12.95 \
\hline
\end{array}$ 
The correlation coefficient between the corn price and the ribeye price is 0.773. Which of the following is the best interpretation of the correlation coefficient?
A)The price of ribeye tends to go down and the price of corn goes up.
B)Increasing corn prices cause ribeye prices to increase.
C)The changes in corn price and ribeye price tend to go up and down together.
D)There is no correlation between the price of corn and the price of ribeye.

Accepted Answer

The correlation coefficient of 0.773 indicates a strong positive relationship between the price of corn and the price of ribeye, meaning they tend to move in the same direction.

Question 2

The following display from a graphing calculator presents the least-squares regression line for predicting the price of a certain commodity (y) from the price of a barrel of oil(x).
 Predict the commodity price when oil costs $107 per barrel.
A)$20
B)$44
C)$485
D)$70

Accepted Answer

A

Question 3

Characterize the relationship shown in the figure. 
A)negative nonlinear
B)positive nonlinear
C)negative linear
D)positive linear

Accepted Answer

B

Question 4

One of the primary feeds for beef cattle is corn. The following table presents the average price in dollars for a bushel of corn and a pound of ribeye steak for 10 consecutive months.
$\begin{array} { c c } 
\hline \text { Corn Price } ( \$ / \text { bu } ) & \text { Ribeye Price } ( \$ / \text { lb) } \
\hline 5.89 & 12.94 \
6.66 & 14.20 \
6.07 & 13.05 \
5.72 & 12.86 \
6.20 & 13.61 \
6.42 & 13.24 \
5.81 & 12.98 \
6.68 & 13.45 \
5.75 & 12.13 \
5.90 & 12.79 \
\hline
\end{array}$ 
Construct a scatter plot of the price of ribeye (y) versus the price of corn (x)
A)

B)

C)

D)

Accepted Answer

The answer of One of the primary feeds for beef...

Question 5

Compute the correlation coefficient.
 $\begin{array} { c | c c c c c c } 
x & 3 & 4 & 5 & 6 & 7 & 8 \
\hline y & 86 & 77 & 113 & 119 & 133 & 128
\end{array}$
A)0.809
B)0.899
C)48.99
D)0.101

Accepted Answer

The correlation coefficient is calculated using the formula for Pearson's correlation, which measures the linear relationship between two variables. The correct calculation yields approximately 0.899, indicating a strong positive correlation.

Question 6

The following table presents the number of police officers (per 100,000 citizens) and the annual murder rate (per 100,000 citizens) for a sample of cities.
 $$\begin{array} { c c } 
\hline \text { Police Officers (per 100,000) } & \text { Murders (per 100,000 per yr) } \
\hline 1377 & 32 \
1324 & 38 \
1068 & 47 \
1443 & 33 \
1289 & 42 \
1206 & 39 \
\hline
\end{array}$$ Compute the correlation coefficient between the per capita number of police officers and the per capita murder rate.
A)-0.900
B)0.810
C)-0.810
D)-0.846

Accepted Answer

The correlation coefficient is a measure of the strength and direction of a linear relationship between two variables. It can range from -1 to 1, with -1 indicating a perfect negative relationship, 0 indicating no relationship, and 1 indicating a perfect positive relationship.In this case, the correlation coefficient between the per capita number of police officers and the per capita murder rate is -0.900. This indicates a strong negative relationship between the two variables, meaning that as the number of police officers per capita increases, the murder rate per capita decreases.

Question 7

Compute the least-squares regression line for predicting y from x given the following summary statistics:
 $$\begin{array} { l l l } 
\bar { x } = 6.7 & s _ { x } = 1.6 & \bar { y } = 23.8 \
s _ { y } = 6.5 & r = 0.93 & \

\end{array}$$
A) $y = - 1.5133 + 3.7781 x$
B) $y = - 1.5133 + 0.2289 x$
C) $y = 0.2289 - 1.5133 x$
D) $y = 3.7781 - 1.5133 x$

Accepted Answer

The slope of the least-squares regression line is calculated as $b = r \frac{s_y}{s_x} = 0.93 \times \frac{6.5}{1.6} = 3.7781$. The intercept is calculated as $a = \bar{y} - b\bar{x} = 23.8 - 3.7781 \times 6.7 = -1.5133$. Thus, the equation is $y = -1.5133 + 3.7781x$.

Question 8

One of the primary feeds for beef cattle is corn. The following table presents the average price in dollars for a bushel of corn and a pound of ribeye steak for 10 consecutive months.
$\begin{array}{cc}
\hline \text { Corn Price }(\$ / \mathrm{bu}) & \text { Ribeye Price }(\$ / \mathrm{lb}) \
\hline 6.34 & 13.22 \
5.93 & 12.06 \
6.45 & 13.72 \
6.19 & 12.75 \
6.56 & 13.27 \
5.94 & 12.17 \
\hline
\end{array}$
The least-squares regression line for predicting the ribeye price from the corn price is $y = - 1.7478 + 2.3437 x$.
If the price of corn differs by $\$ 0.35$ per bushel, by how much would you expect the price , ribeye to differ?
A) $- \$ 0.61$ per lb
B) $\$ 0.61$ per lb
C) $\$ 0.93$ per lb
D) $\$ 0.82$ per $\mathrm { lb }$

Accepted Answer

The answer of One of the primary feeds for beef...

Question 9

For which of the following scatter plots is the correlation coefficient an appropriate summary?
A)

B)

C)

D)

Accepted Answer

The answer of For which of the following scatter plots...

Question 10

One of the primary feeds for beef cattle is corn. The following table presents the average price in dollars for a bushel of corn and a pound of ribeye steak for 10 consecutive months.
$\begin{array}{cc}
\hline \text { Corn Price }(\$ / \mathrm{bu}) & \text { Ribeye Price }(\mathrm{S} / \mathrm{lb}) \
\hline 5.76 & 12.56 \
6.45 & 13.97 \
5.70 & 12.07 \
6.17 & 12.52 \
6.61 & 13.51 \
5.89 & 11.97 \
\hline
\end{array}$
Compute the least-squares regression line for predicting the ribeye price from the corn pri
A) $y = - 1.5651 + 0.5443 x$
B) $y = 1.5651 + 1.8373 x$
C) $y = 1.8373 + 1.5651 x$
D) $y = 1.5651 + 0.5443 x$

Accepted Answer

The answer of One of the primary feeds for beef...

Question 11

The common cricket can be used as a crude thermometer. The colder the temperature, the slower the rate of chirping. The table below shows the average chirp rate of a cricket at various temperatures.
$\begin{array}{cc}
\hline \begin{array}{c}
\text { Chirp Rate } \
\text { (chirps/second) }
\end{array} & \begin{array}{c}
\text { Temperature } \
\left({ }^{\circ} \mathrm{F}\right)
\end{array} \
\hline 2.4 & 61.9 \
3.8 & 83.6 \
3.7 & 71.8 \
2.0 & 47 \
3.4 & 65.1 \
3.9 & 75.8 \
\hline
\end{array}$
Compute the least-squares regression line for predicting the temperature from the chirp ra
A) $y = 14.354 + 21.6004 x$
B) $y = 21.6004 + 19.2632 x$
C) $y = 19.2632 + 21.6004 x$
D) $y = 21.6004 + 14.354 x$

Accepted Answer

The answer of The common cricket can be used as...

Question 12

One of the primary feeds for beef cattle is corn. The following table presents the average price in dollars for a bushel of corn and a pound of ribeye steak for 10 consecutive months.
$\begin{array}{cc}
\hline \text { Corn Price }(\$ / \mathrm{bu}) & \text { Ribeye Price }(\$ / \mathrm{lb}) \
\hline 6.35 & 12.87 \
6.20 & 13.30 \
5.82 & 11.85 \
6.18 & 12.64 \
6.00 & 12.29 \
6.57 & 14.16 \
\hline
\end{array}$
The least-squares regression line for predicting the ribeye price from the corn price is $y = - 4.8904 + 2.8678 x$.
Predict the ribeye price in a month when the corn price was $\$ 6.40$ per bushel.
A) $\$ 12.79$ per lb
B) $\$ 15.30$ per $\mathrm { lb }$
C) $\$ 18.35$ per $\mathrm { lb }$
D) $\$ 13.46$ per lb

Accepted Answer

To predict the ribeye price when the corn price is $6.40 per bushel, substitute $x = 6.40$ into the regression equation $y = -4.8904 + 2.8678x$. This gives $y = -4.8904 + 2.8678(6.40) = -4.8904 + 18.3541 = 13.4637$, which rounds to $\$13.46$ per lb.

Question 13

The following table presents the number of police officers (per 100,000 citizens) and the annual murder rate (per 100,000 citizens) for a sample of cities.
 $$\begin{array} { c c } 
\hline \text { Police Officers (per 100,000) } & \text { Murders (per 100,000 per yr) } \
\hline 1196 & 38 \
1351 & 36 \
1250 & 40 \
1500 & 27 \
1319 & 33 \
1483 & 35 \
\hline
\end{array}$$ Construct a scatter plot of the per capita murder rate (y)versus the per capita number of police officers (x).
A)

B)

C)

D)

Accepted Answer

The answer of The following table presents the number of...

Question 14

The following table presents the number of police officers (per 100,000 citizens) and the annual murder rate (per 100,000 citizens) for a sample of cities.
 $\begin{array} { c c } 
\hline \text { Police Officers (per 100,000) } & \text { Murders (per 100,000 per yr) } \
\hline 1109 & 46 \
1238 & 44 \
1235 & 38 \
1298 & 40 \
1077 & 49 \
1076 & 42 \
\hline
\end{array}$ 
The correlation coefficient between the per capita number of police officers and the per capita murder rates -0.666. Which of the following is the best interpretation of the correlation coefficient?
A)Higher murder rates make it more difficult for cities to hire police officers.
B)The per capita number of police officers and the per capita murder rates arepositively associated.
C)More per capita police officers results in fewer per capita murders.
D)The per capita murder rate tends to go down as the per capita number of policeofficers goes up.

Accepted Answer

The correlation coefficient of -0.666 indicates a negative relationship between the number of police officers per 100,000 citizens and the murder rate per 100,000 citizens. This means as the number of police officers increases, the murder rate tends to decrease.

Question 15

The following display from a graphing calculator presents the least-squares regression line for predicting the price of a certain commodity (y) from the price of a barrel of oil (x)

Write the equation of the least-squares regression line.
A) $y = 3.72 + 0.49 x$
B) $y = 0.49 + 3.72 x$
C) $y = 0.49 + 0.72009 x$
D) $y = 3.72 + 0.72009 x$

Accepted Answer

The answer of The following display from a graphing calculator...

Question 16

Compute the least-squares regression line for the given data set. $\begin{array}{c|cccccc}
x & 2 & 3 & 4 & 5 & 6 & 7 \
\hline y & 17.1 & 18.6 & 22.7 & 23.4 & 25.3 & 29
\end{array}$

A) $y = 2.2943 x + 12.3590$
B) $y = 2.3800 x + 12.3590$
C) $y = 12.3590 x + 2.2943$
D) $y = 2.3800 x + 17.1$

Accepted Answer

To find the least-squares regression line, we first calculate the means of $x$ and $y$, the slope ($m$), and the y-intercept ($b$) using the formulas:$$m = \frac{n(\sum xy) - (\sum x)(\sum y)}{n(\sum x^2) - (\sum x)^2}$$$$b = \bar{y} - m\bar{x}$$where $n$ is the number of data points.Given data:$x = [2, 3, 4, 5, 6, 7]$$y = [17.1, 18.6, 22.7, 23.4, 25.3, 29]$Calculations:$\sum x = 27$, $\sum y = 136.1$, $\sum xy = 736.1$, $\sum x^2 = 147$, $n = 6$Mean of $x$, $\bar{x} = \frac{27}{6} = 4.5$Mean of $y$, $\bar{y} = \frac{136.1}{6} = 22.6833$Slope ($m$):$$m = \frac{6(736.1) - (27)(136.1)}{6(147) - (27)^2} = \frac{4416.6 - 3674.7}{882 - 729} = \frac{741.9}{153} = 2.2943$$Y-intercept ($b$):$$b = 22.6833 - 2.2943(4.5) = 22.6833 - 10.32435 = 12.35895$$Thus, the least-squares regression line is $y = 2.2943x + 12.3590$.

Question 17

The following display from a graphing calculator presents the least-squares regression line for predicting the price of a certain commodity (y) from the price of a barrel of oil (x).
 What is the correlation between the oil price and the commodity price?
A)5.11
B)0.24
C)0.71939
D)0.84817

Accepted Answer

The answer of The following display from a graphing calculator...

Question 18

The common cricket can be used as a crude thermometer. The colder the temperature, the slower the rate of chirping. The table below shows the average chirp rate of a cricket at various temperatures
$\begin{array}{cc}
\hline \begin{array}{c}
\text { Chirp Rate } \
\text { (chirps/second) }
\end{array} & \begin{array}{c}
\text { Temperature } \
\left({ }^{\circ} \mathrm{F}\right)
\end{array} \
\hline 3.4 & 74 \
3.1 & 61.3 \
1.9 & 45.7 \
1.0 & 35.4 \
1.4 & 49 \
2.5 & 66 \
\hline
\end{array}$
The least-squares regression line for predicting the temperature from the chirp rate is $y = 24.3278 + 13.9423 x$. If two chirp rates differ by $1.5$ chirps per second, by how much would the temperature differ?
A) $21 ^ { \circ } \mathrm { F }$
B) $24 ^ { \circ } \mathrm { F }$
C) $36 ^ { \circ } \mathrm { F }$
D) $20 ^ { \circ } \mathrm { F }$

Accepted Answer

The temperature difference for a 1.5 chirps per second difference in chirp rate can be calculated using the slope of the regression line, which is 13.9423. Multiplying this slope by the chirp rate difference gives $13.9423 \times 1.5 = 20.91345$, which rounds to approximately $21 ^{\circ}F$.

Question 19

The common cricket can be used as a crude thermometer. The colder the temperature, the slower the rate of chirping. The table below shows the average chirp rate of a cricket at various temperatures
$\begin{array}{cc}
\hline \begin{array}{c}
\text { Chirp Rate } \
\text { (chirps/second) }
\end{array} & \begin{array}{c}
\text { Temperature } \
\left({ }^{\circ} \mathrm{F}\right)
\end{array} \
\hline 2.9 & 61.4 \
2.0 & 55.5 \
3.8 & 77.5 \
2.5 & 58.8 \
1.7 & 53.4 \
1.6 & 52.6 \
\hline
\end{array}$
The least-squares regression line for predicting the temperature from the chirp rate is $y = 34.0812 + 10.6698 x$. Predict the temperature if the chirp rate is $3.6$ chirps per second.
A) $75 ^ { \circ } \mathrm { F }$
B) $72 { } ^ { \circ } \mathrm { F }$
C) $38 ^ { \circ } \mathrm { F }$
D) $76 ^ { \circ } \mathrm { F }$

Accepted Answer

The answer of The common cricket can be used as...

Question 20

One of the primary feeds for beef cattle is corn. The following table presents the average price in dollars for a bushel of corn and a pound of ribeye steak for 10 consecutive months
$$\begin{array} { c c } 
\hline \text { Corn Price } ( \$ / b u ) & \text { Ribeye Price } ( \$ / \text { b) } \
\hline 6.40 & 13.20 \
6.08 & 13.21 \
6.11 & 13.30 \
6.37 & 13.55 \
5.91 & 12.57 \
6.22 & 13.37 \
6.49 & 13.54 \
6.20 & 13.37 \
6.00 & 12.88 \
6.04 & 12.71 \
\hline
\end{array}$$
Compute the correlation coefficient between the corn price and the ribeye price.
A)0.815
B)0.965
C)0.664
D)0.185

Accepted Answer

The correlation coefficient is a measure of the strength and direction of a linear relationship between two variables. It is calculated by dividing the covariance of the two variables by the product of their standard deviations. The covariance is a measure of how much the two variables vary together, and the standard deviation is a measure of how much each variable varies on its own.In this case, the correlation coefficient between the corn price and the ribeye price is 0.815. This indicates that there is a strong positive linear relationship between the two variables. As the corn price increases, the ribeye price also tends to increase.

Quiz 4: Probability

The following display from a graphing calculator presents the least-squares regression line for predicting the price of a certain commodity (y) from the price of a barrel of oil(x) . Predict the commodity price when oil costs $107 per barrel.

Characterize the relationship shown in the figure.

Compute the correlation coefficient. $\begin{array} { c | c c c c c c } x & 3 & 4 & 5 & 6 & 7 & 8 \\\hline y & 86 & 77 & 113 & 119 & 133 & 128\end{array}$

Compute the least-squares regression line for predicting y from x given the following summary statistics: $\begin{array} { l l l } \bar { x } = 6.7 & s _ { x } = 1.6 & \bar { y } = 23.8 \\s _ { y } = 6.5 & r = 0.93 & \\\end{array}$

For which of the following scatter plots is the correlation coefficient an appropriate summary?

The following display from a graphing calculator presents the least-squares regression line for predicting the price of a certain commodity (y) from the price of a barrel of oil (x) Write the equation of the least-squares regression line.

Compute the least-squares regression line for the given data set. $\begin{array}{c|cccccc}x & 2 & 3 & 4 & 5 & 6 & 7 \\\hline y & 17.1 & 18.6 & 22.7 & 23.4 & 25.3 & 29\end{array}$

The following display from a graphing calculator presents the least-squares regression line for predicting the price of a certain commodity (y) from the price of a barrel of oil (x) . What is the correlation between the oil price and the commodity price?

Quiz 4: Probability

The following display from a graphing calculator presents the least-squares regression line for predicting the price of a certain commodity (y) from the price of a barrel of oil(x) . Predict the commodity price when oil costs $107 per barrel.

Characterize the relationship shown in the figure.

Compute the correlation coefficient. x345678y8677113119133128\begin{array} { c | c c c c c c } x & 3 & 4 & 5 & 6 & 7 & 8 \\\hline y & 86 & 77 & 113 & 119 & 133 & 128\end{array}xy​386​477​5113​6119​7133​8128​​

For which of the following scatter plots is the correlation coefficient an appropriate summary?

The following display from a graphing calculator presents the least-squares regression line for predicting the price of a certain commodity (y) from the price of a barrel of oil (x) Write the equation of the least-squares regression line.

Compute the least-squares regression line for the given data set. x234567y17.118.622.723.425.329\begin{array}{c|cccccc}x & 2 & 3 & 4 & 5 & 6 & 7 \\\hline y & 17.1 & 18.6 & 22.7 & 23.4 & 25.3 & 29\end{array}xy​217.1​318.6​422.7​523.4​625.3​729​​

The following display from a graphing calculator presents the least-squares regression line for predicting the price of a certain commodity (y) from the price of a barrel of oil (x) . What is the correlation between the oil price and the commodity price?

Compute the correlation coefficient. $\begin{array} { c | c c c c c c } x & 3 & 4 & 5 & 6 & 7 & 8 \\\hline y & 86 & 77 & 113 & 119 & 133 & 128\end{array}$

Compute the least-squares regression line for the given data set. $\begin{array}{c|cccccc}x & 2 & 3 & 4 & 5 & 6 & 7 \\\hline y & 17.1 & 18.6 & 22.7 & 23.4 & 25.3 & 29\end{array}$