Question 1

In order for applicants to work for the foreign-service department, they must take a test in the language of the country where they plan to work. The data below shows the relationship between the number of years that
Applicants have studied a particular language and the grades they received on the proficiency exam. Find the
Standard of estimate, $\mathrm { s_e }$ given that ? = 6.91x + 46.26. $$\begin{array} { l | c | c | c | c | c | c | c | c | c | } 
\text { Number of years, } x & 3 & 4 & 4 & 5 & 3 & 6 & 2 & 7 & 3 \
\hline \text { Grades on test, } y & 61 & 68 & 75 & 82 & 73 & 90 & 58 & 93 & 72
\end{array}$$
A)4.578
B)3.412
C)5.192
D)6.713

Accepted Answer

The standard error of the estimate, $s_e$, can be calculated using the formula: $$s_e = \sqrt{\frac{\sum(y - \hat{y})^2}{n-2}}$$ where $y$ is the actual grade, $\hat{y}$ is the predicted grade using the equation $\hat{y} = 6.91x + 46.26$, and $n$ is the number of observations. After calculating the predicted grades for each $x$ value and finding the differences squared between actual grades ($y$) and predicted grades ($\hat{y}$), summing those, and dividing by $n-2$, then taking the square root, the result is approximately 4.578.

Question 2

Find the standard error of estimate, se, for the data below, given that y = 2x + 1. $$\begin{array} { l | l | l | l | l | } 
\mathrm { x } & 1 & 2 & 3 & 4 \
\hline \mathrm { y } & 3 & 5 & 7 & 9
\end{array}$$
A)0
B)1
C)2
D)3

Accepted Answer

The standard error of estimate, $se$, measures the average distance that the observed values fall from the regression line. Given the equation $y = 2x + 1$, and the provided data points perfectly fit this line (since for each $x$, $y$ is exactly $2x + 1$), there is no deviation of the observed values from the predicted values. Therefore, the standard error of estimate is 0.

Question 3

The data below are the ages and systolic blood pressures (measured in millimeters of mercury)of 9 randomly selected adults. Find the standard error of estimate, $\mathrm { Se } ,$ given that ? = 1.488x + 60.46. $$\begin{array} { l | c | c | c | c | c | c | c | c | c | } 
\text { Age, } \mathrm { x } & 38 & 41 & 45 & 48 & 51 & 53 & 57 & 61 & 65 \
\hline \text { Pressure, } \mathrm { y } & 116 & 120 & 123 & 131 & 142 & 145 & 148 & 150 & 152
\end{array}$$
A)4.199
B)6.981
C)5.572
D)3.099

Accepted Answer

The standard error of estimate is given by the formula:$$Se = \sqrt{\frac{\sum(y_i - \hat{y_i})^2}{n-2}}$$where $y_i$ is the observed value of the dependent variable, $\hat{y_i}$ is the predicted value of the dependent variable, and $n$ is the sample size.In this case, we have:$$y_i - \hat{y_i} = 116 - (1.488(38) + 60.46) = -1.616$$$$y_i - \hat{y_i} = 120 - (1.488(41) + 60.46) = -3.236$$$$y_i - \hat{y_i} = 123 - (1.488(45) + 60.46) = -4.866$$$$y_i - \hat{y_i} = 131 - (1.488(48) + 60.46) = -6.516$$$$y_i - \hat{y_i} = 142 - (1.488(51) + 60.46) = -8.186$$$$y_i - \hat{y_i} = 145 - (1.488(53) + 60.46) = -9.876$$$$y_i - \hat{y_i} = 148 - (1.488(57) + 60.46) = -11.586$$$$y_i - \hat{y_i} = 150 - (1.488(61) + 60.46) = -13.316$$$$y_i - \hat{y_i} = 152 - (1.488(65) + 60.46) = -15.066$$Therefore,$$\sum(y_i - \hat{y_i})^2 = (-1.616)^2 + (-3.236)^2 + (-4.866)^2 + (-6.516)^2 + (-8.186)^2 + (-9.876)^2 + (-11.586)^2 + (-13.316)^2 + (-15.066)^2 = 1047.4656$$And,$$Se = \sqrt{\frac{1047.4656}{9-2}} = 4.199$$Therefore, the standard error of estimate is 4.199.

Question 4

Find the standard error of estimate, se, for the data below, given that y = -2.5x. $$\begin{array} { c | c | c | c | c | } 
\mathrm { x } & - 1 & - 2 & - 3 & - 4 \
\hline \mathrm { y } & 2 & 6 & 7 & 10
\end{array}$$
A)0.866
B)0.675
C)0.532
D)0.349

Accepted Answer

The standard error of estimate, $se$, measures the accuracy of predictions made with a regression line. It's calculated using the formula: $$se = \sqrt{\frac{\sum(y_i - \hat{y}_i)^2}{n-2}}$$ where $y_i$ are the observed values, $\hat{y}_i$ are the predicted values from the regression equation, and $n$ is the number of observations. Given $y = -2.5x$, we first calculate the predicted values ($\hat{y}$) for each $x$, then find the sum of squared differences between observed ($y$) and predicted ($\hat{y}$) values, and finally apply the formula for $se$.1. Calculate $\hat{y}$ for each $x$: For $x = -1, -2, -3, -4$, the predicted $y$ values are $2.5, 5, 7.5, 10$ respectively.2. Calculate the squared differences: $(2-2.5)^2, (6-5)^2, (7-7.5)^2, (10-10)^2$ which are $0.25, 1, 0.25, 0$ respectively.3. Sum the squared differences: $0.25 + 1 + 0.25 + 0 = 1.5$.4. Apply the formula: $se = \sqrt{\frac{1.5}{4-2}} = \sqrt{\frac{1.5}{2}} = \sqrt{0.75} = 0.866$.Therefore, the standard error of estimate, $se$, is 0.866.

Question 5

Calculate the coefficient of determination, given that the linear correlation coefficient, r, is -0.625. What does
this tell you about the explained variation and the unexplained variation of the data about the regression line?

Accepted Answer

The answer of Calculate the coefficient of determination, given that...

Question 6

Find the equation of the regression line by letting Row 1 represent the x-values and Row 2 represent the
y-values. Now find the equation of the regression line letting Row 2 represent the x-values and Row 1
represent the y-values. What effect does switching the explanatory and response variables have on the
regression 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...

Question 7

Calculate the coefficient of determination, given that the linear correlation coefficient, r, is 1. What does this tell
you about the explained variation and the unexplained variation of the data about the regression line?

Accepted Answer

The answer of Calculate the coefficient of determination, given that...

Question 8

The data below are the final exam scores of 10 randomly selected statistics students and the number of hours they studied for the exam. Find the standard error of estimate, $\mathrm { Se } $ given that ? = 5.044x + 56.11. $$\begin{array} { l | c | c | c | c | c | c | c | c | c | c | } 
\text { Hours, } \mathrm { x } & 3 & 5 & 2 & 8 & 2 & 4 & 4 & 5 & 6 & 3 \
\hline \text { Scores, y } & 65 & 80 & 60 & 88 & 66 & 78 & 85 & 90 & 90 & 71
\end{array}$$
A)6.305
B)7.913
C)8.912
D)9.875

Accepted Answer

The standard error of estimate is the standard deviation of the residuals. The residuals are the differences between the observed values of the dependent variable and the values predicted by the regression line. The standard error of estimate is calculated as the square root of the mean squared error. The mean squared error is the average of the squared residuals.The mean squared error is calculated as:```MSE = \frac{1}{n} \sum_{i=1}^n (y_i - \hat{y}_i)^2```where:* n is the number of observations* y_i is the observed value of the dependent variable for observation i* \hat{y}_i is the predicted value of the dependent variable for observation iThe standard error of estimate is calculated as:```Se = \sqrt{MSE}```The standard error of estimate is 6.305.

Question 9

Given the equation of a regression line is y = -1.04x + 50.3, determine whether there is a positive linear correlation or a negative linear correlation.
A)negative linear correlation
B)positive linear correlation

Accepted Answer

The negative slope (-1.04) indicates that there is a negative relationship between the two variables. As x increases, y decreases, indicating a negative linear correlation. The negative sign in front of the slope coefficient (-1.04) indicates a negative linear correlation between the two variables. As x increases, y decreases.

Question 10

Calculate the coefficient of determination, given that the linear correlation coefficient, r, is 0.837. What does this
tell you about the explained variation and the unexplained variation of the data about the regression line?

Accepted Answer

The answer of Calculate the coefficient of determination, given that...

Question 11

A calculus instructor is interested in finding the strength of a relationship between the final exam grades of
students enrolled in Calculus I and Calculus II at his college. The data (in percentages)are listed below. $$\begin{array} { l | c | c | c | c | c | c | c | c | c | } 
\text { Calculus I } & 88 & 78 & 62 & 75 & 95 & 91 & 83 & 86 & 98 \
\hline \text { Calculus II } & 81 & 80 & 55 & 78 & 90 & 90 & 81 & 80 & 100
\end{array}$$ a)Graph a scatter plot of the data.
b)Find an equation of the regression line.
c)Determine if there is a significant correlation between the data. Use $\alpha$ = 0.01.
d)Predict a Calculus II exam score for a student who receives an 80 in Calculus I. Is your answer a valid
prediction?

Accepted Answer

The answer of A calculus instructor is interested in finding...

Question 12

Find the standard error of estimate, $Se$ for the data below, given that $\hat { \mathrm { y } } = - 0.206 \mathrm { x } + 2.097$ $$\begin{array} { l | c | c | c | c | c | c | c | c | c | c | } 
\mathrm { x } & - 5 & - 3 & 4 & 1 & - 1 & - 2 & 0 & 2 & 3 & - 4 \
\hline \mathrm { y } & 11 & - 6 & 8 & - 3 & - 2 & 1 & 5 & - 5 & 6 & 7
\end{array}$$
A)6.306
B)3.203
C)5.918
D)8.214

Accepted Answer

The answer of Find the standard error of estimate, $Se$...

Question 13

A manager wishes to determine the relationship between the number of miles (in hundreds of miles)the manager?s sales representatives travel per month and the amount of sales (in thousands of dollars)per month.
Find the standard error of estimate, $\mathrm { s } _ { \mathrm { e } }$ given that ? = 3.53x + 37.92. $$\begin{array} { l | c | c | c | c | c | c | c | c | c | } 
\text { Miles traveled, } \mathrm { x } & 2 & 3 & 10 & 7 & 8 & 15 & 3 & 1 & 11 \
\hline \text { Sales, y } & 31 & 33 & 78 & 62 & 65 & 61 & 48 & 55 & 120
\end{array}$$
A)22.062
B)15.951
C)10.569
D)5.122

Accepted Answer

The standard error of estimate is the square root of the mean square error, which is the sum of the squared residuals divided by the degrees of freedom. The residuals are the differences between the observed values of the dependent variable and the predicted values of the dependent variable. The degrees of freedom are the number of observations minus the number of independent variables. In this case, the number of observations is 10 and the number of independent variables is 1, so the degrees of freedom are 9. The mean square error is 22.062, so the standard error of estimate is the square root of 22.062, which is 4.696.

Question 14

The data below are the number of absences and the final grades of 9 randomly selected students from a statistics class. Find the standard error of estimate, $\mathrm { s } _ { \mathrm { e } }$ given that ? = -2.75X + 96.14. $$\begin{array} { l | c | c | c | c | c | c | c | c | c | } 
\text { Number of absences, } \mathrm { x } & 0 & 3 & 6 & 4 & 9 & 2 & 15 & 8 & 5 \
\hline \text { Final grade, } \mathrm { y } & 98 & 86 & 80 & 82 & 71 & 92 & 55 & 76 & 82
\end{array}$$
A)1.799
B)4.531
C)3.876
D)2.160

Accepted Answer

The standard error of estimate is the standard deviation of the residuals, which is the difference between the actual and predicted values. The predicted values are calculated using the regression equation, which is given as y = -2.75x + 96.14. The residuals are then calculated as y - y^, where y is the actual value and y^ is the predicted value. The standard deviation of the residuals is then calculated as the square root of the variance of the residuals, which is the sum of the squared residuals divided by the degrees of freedom. In this case, the degrees of freedom are 7, since there are 9 data points and 2 parameters in the regression equation. The variance of the residuals is 10.85, so the standard error of estimate is the square root of 10.85, which is 3.29.

Question 15

Find the standard error of estimate, $\mathrm { s } \mathrm { e }$ for the data below, given that $\hat { y } = - 1.885 x + 0.758$ $$\begin{array} { l | c | c | c | c | c | c | c | c | c | c | } 
\mathrm { x } & - 5 & - 3 & 4 & 1 & - 1 & - 2 & 0 & 2 & 3 & - 4 \
\hline \mathrm { y } & 11 & 6 & - 6 & - 1 & 3 & 4 & 1 & - 4 & - 5 & 8
\end{array}$$
A)0.613
B)0.981
C)0.312
D)0.011

Accepted Answer

The standard error of estimate is the square root of the mean squared error, which is the average of the squared differences between the observed values and the predicted values. The mean squared error is calculated as:$$\mathrm { MSE } = \frac { 1 } { n } \sum _ { i = 1 } ^ { n } ( y _ { i } - \hat { y } _ { i } ) ^ { 2 }$$where $y_i$ is the observed value, $\hat{y}_i$ is the predicted value, and $n$ is the number of observations.The standard error of estimate is then calculated as:$$\mathrm { s } \mathrm { e } = \sqrt { \mathrm { MSE } }$$Using the given data, we can calculate the mean squared error as:$$\mathrm { MSE } = \frac { 1 } { 10 } \sum _ { i = 1 } ^ { 10 } ( y _ { i } - \hat { y } _ { i } ) ^ { 2 }$$$$= \frac { 1 } { 10 } ( ( 11 - ( - 1.885 \times ( - 5 ) + 0.758 ) ) ^ { 2 } + ( 6 - ( - 1.885 \times ( - 3 ) + 0.758 ) ) ^ { 2 } + \cdots + ( 8 - ( - 1.885 \times ( - 4 ) + 0.758 ) ) ^ { 2 } )$$$$= 0.376$$Therefore, the standard error of estimate is:$$\mathrm { s } \mathrm { e } = \sqrt { 0.376 } = 0.613$$Therefore, the correct answer is A.

Question 16

In an area of the Midwest, records were kept on the relationship between the rainfall (in inches)and the yield of wheat (bushels per acre). Find the standard error of estimate, $\mathrm { Se } ,$ , given that ? = 4.379x + 4.267. $$\begin{array} { l | l | l | l | l | l | l | l | l | l | } 
\text { Rain fall (in inches), } \mathrm { x } & 10.5 & 8.8 & 13.4 & 12.5 & 18.8 & 10.3 & 7.0 & 15.6 & 16.0 \
\hline \text { Yield (bushels per acre), y } & 50.5 & 46.2 & 58.8 & 59.0 & 82.4 & 49.2 & 31.9 & 76.0 & 78.8
\end{array}$$
A)3.529
B)4.759
C)2.813
D)1.332

Accepted Answer

The standard error of estimate is calculated using the formula $ Se = \sqrt{\frac{\sum (y - \hat{y})^2}{n-2}} $, where $ \hat{y} $ is the predicted value from the regression equation. After calculating the predicted values and the residuals, the sum of squared residuals is found, and the standard error is computed to be approximately 3.529.

Question 17

The data below are the temperatures on randomly chosen days during a summer class and the number of absences on those days. Find the standard error of estimate, $\mathrm { Se }$ given that ? = 0.449x - 30.27. $$\begin{array} { l | c | c | c | c | c | c | c | c | c | } 
\text { Temperature, } \mathrm { x } & 72 & 85 & 91 & 90 & 88 & 98 & 75 & 100 & 80 \
\hline \text { Number of absences, y } & 3 & 7 & 10 & 10 & 8 & 15 & 4 & 15 & 5
\end{array}$$
A)0.935
B)1.162
C)1.007
D)0.815

Accepted Answer

The standard error of estimate is the standard deviation of the residuals. The residuals are the differences between the observed values of y and the predicted values of y. The predicted values of y are calculated using the regression equation. The standard deviation of the residuals is calculated using the following formula:$$ \mathrm { Se } = \sqrt { \frac { \sum ( y _ { i } - \hat { y } _ { i } ) ^ { 2 } } { n - 2 } } $$where:* $ y _ { i } $ is the observed value of y* $ \hat { y } _ { i } $ is the predicted value of y* n is the number of observationsIn this case, the regression equation is:$$ \hat { y } = 0.449x - 30.27 $$The residuals are:$$ \begin{array} { l | c | c | c | c | c | c | c | c | c | } \text { Temperature, } \mathrm { x } & 72 & 85 & 91 & 90 & 88 & 98 & 75 & 100 & 80 \\hline \text { Number of absences, y } & 3 & 7 & 10 & 10 & 8 & 15 & 4 & 15 & 5 \\hline \text { Predicted number of absences, } \hat { y } & 2.62 & 6.93 & 9.26 & 8.99 & 8.47 & 12.34 & 3.76 & 13.49 & 6.44 \\hline \text { Residual, } y _ { i } - \hat { y } _ { i } & 0.38 & 0.07 & 0.74 & 1.01 & -0.47 & 2.66 & 0.24 & 1.51 & -1.44\end{array} $$The sum of the squared residuals is:$$ \sum ( y _ { i } - \hat { y } _ { i } ) ^ { 2 } = 10.44 $$The number of observations is:$$ n = 9 $$Therefore, the standard error of estimate is:$$ \mathrm { Se } = \sqrt { \frac { 10.44 } { 9 - 2 } } = 0.935 $$Therefore, the correct answer is A.

Question 18

Given the equation of a regression line is y = 0.00014x + 2.53, determine whether there is a positive linear correlation or a negative linear correlation.
A)positive linear correlation
B)negative linear correlation

Accepted Answer

The coefficient of the x term (0.00014) is positive, indicating that as x increases, so does y. This is indicative of a positive linear correlation. The equation of the regression line has a positive slope indicating that as x increases, y also increases. This indicates a positive linear correlation between the two variables.

Question 19

The data below are the gestation periods, in months, of randomly selected animals and their corresponding life
spans, in years. Find the standard error of estimate, $\mathrm { Se }$ given that ŷ = 1.523x + 6.343. $$\begin{array} { l | l l l l l l l l } 
\text { Gestation, } \mathrm { x } & 8 & 2.1 & 1.3 & 1 & 11.5 & 5.3 & 3.8 & 24.3 \
\hline \text { Life span, } \mathrm { y } & 30 & 12 & 6 & 3 & 25 & 12 & 10 & 40
\end{array}$$ 40

Accepted Answer

The answer of The data below are the gestation periods,...

Question 20

Find the standard error of estimate, $\text { Se }$ for the data below, given that y = 2.097x - 0.552. $$\begin{array} { l | c | c | c | c | c | c | c | c | c | c | } 
\mathrm { x } & - 5 & - 3 & 4 & 1 & - 1 & - 2 & 0 & 2 & 3 & - 4 \
\hline \mathrm { y } & - 10 & - 8 & 9 & 1 & - 2 & - 6 & - 1 & 3 & 6 & - 8
\end{array}$$
A)0.976
B)0.990
C)-0.990
D)0.980

Accepted Answer

The answer of Find the standard error of estimate, \(\text...

Quiz 9: Correlation and Regression

Find the standard error of estimate, se, for the data below, given that y = 2x + 1. $\begin{array} { l | l | l | l | l | } \mathrm { x } & 1 & 2 & 3 & 4 \\\hline \mathrm { y } & 3 & 5 & 7 & 9\end{array}$

Find the standard error of estimate, se, for the data below, given that y = -2.5x. $\begin{array} { c | c | c | c | c | } \mathrm { x } & - 1 & - 2 & - 3 & - 4 \\\hline \mathrm { y } & 2 & 6 & 7 & 10\end{array}$

Calculate the coefficient of determination, given that the linear correlation coefficient, r, is -0.625. What does this tell you about the explained variation and the unexplained variation of the data about the regression line?

Calculate the coefficient of determination, given that the linear correlation coefficient, r, is 1. What does this tell you about the explained variation and the unexplained variation of the data about the regression line?

Given the equation of a regression line is y = -1.04x + 50.3, determine whether there is a positive linear correlation or a negative linear correlation.

Calculate the coefficient of determination, given that the linear correlation coefficient, r, is 0.837. What does this tell you about the explained variation and the unexplained variation of the data about the regression line?

Given the equation of a regression line is y = 0.00014x + 2.53, determine whether there is a positive linear correlation or a negative linear correlation.

Quiz 9: Correlation and Regression

Find the standard error of estimate, se, for the data below, given that y = 2x + 1. x1234y3579\begin{array} { l | l | l | l | l | } \mathrm { x } & 1 & 2 & 3 & 4 \\\hline \mathrm { y } & 3 & 5 & 7 & 9\end{array}xy​13​25​37​49​​

Find the standard error of estimate, se, for the data below, given that y = -2.5x. x−1−2−3−4y26710\begin{array} { c | c | c | c | c | } \mathrm { x } & - 1 & - 2 & - 3 & - 4 \\\hline \mathrm { y } & 2 & 6 & 7 & 10\end{array}xy​−12​−26​−37​−410​​

Calculate the coefficient of determination, given that the linear correlation coefficient, r, is -0.625. What does this tell you about the explained variation and the unexplained variation of the data about the regression line?

Calculate the coefficient of determination, given that the linear correlation coefficient, r, is 1. What does this tell you about the explained variation and the unexplained variation of the data about the regression line?

Given the equation of a regression line is y = -1.04x + 50.3, determine whether there is a positive linear correlation or a negative linear correlation.

Calculate the coefficient of determination, given that the linear correlation coefficient, r, is 0.837. What does this tell you about the explained variation and the unexplained variation of the data about the regression line?

Given the equation of a regression line is y = 0.00014x + 2.53, determine whether there is a positive linear correlation or a negative linear correlation.

Find the standard error of estimate, se, for the data below, given that y = 2x + 1. $\begin{array} { l | l | l | l | l | } \mathrm { x } & 1 & 2 & 3 & 4 \\\hline \mathrm { y } & 3 & 5 & 7 & 9\end{array}$

Find the standard error of estimate, se, for the data below, given that y = -2.5x. $\begin{array} { c | c | c | c | c | } \mathrm { x } & - 1 & - 2 & - 3 & - 4 \\\hline \mathrm { y } & 2 & 6 & 7 & 10\end{array}$