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Calculate the correlation coefficient.

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The following results were obtained from a simple regression analysis: y^\hat { y } = 37.2895 - 1.2024x, r2 = 0.6744, sb1s _ { b _ { 1 } } = 0.2934 -For each unit change in x,the estimated change in the mean of y is equal to:


A) 36.0871
B) 0.6774
C) 37.2895
D) 0.2934
E) -1.2024

F) A) and B)
G) A) and C)

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When using simple regression analysis,if there is a strong positive correlation between the independent and dependent variable,then we can conclude that an increase in the value of the independent variable causes an increase in the value of the dependent variable.

A) True
B) False

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In simple regression analysis,the quantity (yyˉ)2\sum ( y - \bar { y } ) ^ { 2 } is called the total variation.

A) True
B) False

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In a simple linear regression analysis,when the constant variance assumption for the error term holds,a plot of the residual versus x:


A) Fans out
B) Funnels in
C) Fans out,but then funnels in
D) Forms a horizontal band pattern
E) Suggests an increasing error variance

F) A) and D)
G) A) and E)

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Consider the following partial computer output from a simple linear regression analysis:  Predictor  Coef  SE Coef TP Constant 5566.1254.021.910.000 Independent Var 210.3524.19\begin{array} { l l c c c } \text { Predictor } & \text { Coef } & \text { SE Coef } & \mathrm { T } & \mathrm { P } \\\text { Constant } & 5566.1 & 254.0 & 21.91 & 0.000 \\\text { Independent Var } & - 210.35 & 24.19 & - &\end{array} S = _________ R-Sq = Analysis of Variance  Source  DF  SS  MS  F  P  Regression 13963719396371975.590.000 Residual Error 14___52439 Total 15___\begin{array}{lccrcc}\text { Source } & \text { DF } & \text { SS } & \text { MS } & \text { F } & \text { P } \\\text { Regression } & 1 & 3963719 & 3963719 & 75.59 & 0.000 \\\text { Residual Error } & 14 &\_\_\_ & 52439 & &\\\text { Total }&15&\_\_\_\end{array} -Calculate the standard error of the model.

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s = ?MSE =...

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An experiment was performed on a certain metal to determine if the strength is a function of heating time.Results based on 10 metal sheets are given below.Use the simple linear regression model. X\sum X = 30 X2\sum X ^ { 2 } = 104 Y\sum Y = 40 Y2\sum Y ^ { 2 } = 178 XY\sum X Y = 134 -Find the estimated y-intercept.

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A data set with 7 observations yielded the following.Use the simple linear regression model where y is the dependent variable and x is the independent variable. x\sum x = 21.57 X2\sum X ^ { 2 } = 68.31 Y2\sum Y ^ { 2 } = 188.9 Y2\sum Y ^ { 2 } = 5,140.23 XY\sum X Y = 590.83 SSE = 1.06 -Use the least squares regression equation,

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= 12.36 + 4.745x,and determi...

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A local grocery store wants to predict the daily sales in dollars. The manager believes that the amount of newspaper advertising significantly affects the store sales. The manager randomly selects 7 days of data consisting of daily grocery store sales (in thousands of dollars) and advertising expenditures (in thousands of dollars) . The Excel/Mega-Stat output given below summarizes the results of fitting a simple linear regression model using this data. Regression Analysis r20.762n7R0.873k 1  Std. Error 11.547  Dep. Var. Sales \begin{array}{rc}\mathrm{r}^{2} 0.762 & \mathrm{n} 7 \\\mathrm{R} 0.873 & \mathrm{k} \text { 1 } \\\text { Std. Error 11.547 } & \text { Dep. Var. Sales }\end{array} ANOVA table  Source SSdfMSFp-value  Regression 2,133.333312,133.333316.00.0103 Residual 666.66675133.3333 Total 2,800.00006\begin{array}{rrrrrr}\hline \text { Source } & S S & d f & M S & F & p \text {-value } \\\hline \text { Regression } & 2,133.3333 & 1 & 2,133.3333 & 16.00 & .0103 \\\text { Residual } & 666.6667 & 5 & 133.3333 & & \\\hline \text { Total } & 2,800.0000 & 6 & & &\\\hline\end{array}  Regression output \text { Regression output }\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad Confidence interval \text { Confidence interval }  Variables Coefficients  std. error t(df=5)  p-value 95%95% upper  lower  Intercep 63.33337.96827.948.000542.850583.8162 Advertising 6.66671.66674.000.0103\begin{array}{rrrrrrrr}\hline\text { Variables} & \text { Coefficients } & \text { std. error } & t(d f=5) & \text { p-value } & 95 \% &95 \% \text { upper } \\& & & & & \text { lower } & \\\hline \text { Intercep } & 63.3333 & 7.9682 & 7.948 & .0005 & 42.8505 & 83.8162 \\\text { Advertising } & 6.6667 & 1.6667 & 4.000 & .0103 & &\end{array} -In testing the population slope for significance at a significance level of .05,what is the rejection point condition for the two-sided t test?


A) Reject H0if |t| > 2.571
B) Reject H0if t > 2.571
C) Reject H0if |t| < 2.571
D) Reject H0if |t| > 2.051
E) Reject H0if t > 2.051

F) A) and C)
G) B) and D)

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If there is significant autocorrelation present in a data set,the error terms are not ________________.

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The following results were obtained as a part of simple linear regression analysis: R2= 0.9162 F test statistic = 81.87 At α\alpha = 0) 05,the null hypothesis of no linear relationship between the dependent variable and the independent variable _____.


A) is rejected.
B) cannot be tested with the given information.
C) is not rejected.

D) None of the above
E) B) and C)

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The point estimate of the error variance in a regression model is:


A) SSE
B) b0
C) MSE
D) b1
E) r

F) All of the above
G) C) and D)

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Determine the 95% prediction interval for the strength of a metal sheet when the average heating time is 2.5 minutes.

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_____ is a statistical technique in which we use observed data to relate a dependent variable to one or more predictor (independent)variables.

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What is the coefficient of determination?

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r2 = Explained variat...

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The simple linear regression model assumes there is a _____ relationship between the dependent variable and the independent variable.

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linear or ...

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Consider the following partial computer output from a simple linear regression analysis.  Predictor  Coef  SE Coef  T  P  Constant 67.0520.903.210.012 Independent Var 5.81670.7085___0.000\begin{array} { l c c c c } \text { Predictor } & \text { Coef } & \text { SE Coef } & \text { T } & \text { P } \\\text { Constant } & 67.05 & 20.90 & 3.21 & 0.012 \\\text { Independent Var } & 5.8167 & 0.7085 & \_\_\_& 0.000\end{array} S = _________ R-Sq = _______ Analysis of Variance  Source  DF  SS  MS  F  P  Regression 13492067.390.000 Residual Error 8518 Total 939065\begin{array} { l c c c c c } \text { Source } & \text { DF } & \text { SS } & \text { MS } & \text { F } & \text { P } \\\text { Regression } & 1 & - & 34920 & 67.39 & 0.000 \\\text { Residual Error } & 8 & - & 518 & & \\\text { Total } & 9 & 39065 & & &\end{array} -Calculate the SSE.

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SSE = MSE ...

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A local tire dealer wants to predict the number of tires sold each month. The dealer believes that the number of tires sold is a linear function of the amount of money invested in advertising. The dealer randomly selects 6 months of data consisting of tire sales (in thousands of tires) and advertising expenditures (in thousands of dollars) . Based on the data set with 6 observations, the simple linear regression model yielded the following results. x\sum x = 24 X2\sum X ^ { 2 } =124 Y\sum Y =42 Y2\sum Y ^ { 2 } =338 XY\sum X Y =196 -What is the value of SSE?


A) 24
B) 28
C) 44
D) 16
E) 6

F) A) and B)
G) A) and E)

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The dependent variable is the variable that is being described,predicted,or controlled.

A) True
B) False

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Consider the following partial computer output from a simple linear regression analysis.  Predictor  Coef  SE Coef  T  P  Constant 4.86150.52019.350.000 Independent Var 0.346550.05866___S=0.4862RSq=____\begin{array}{l}\begin{array} { l l l l l } \text { Predictor } & \text { Coef } & \text { SE Coef } & \text { T } & \text { P } \\\text { Constant } & 4.8615 & 0.5201 & 9.35 & 0.000 \\\text { Independent Var } - & 0.34655 & 0.05866 &\_\_\_ &\end{array}\\\mathrm { S } = 0.4862 \quad \mathrm { R } - \mathrm { Sq } =\_\_\_\_\end{array} Analysis of Variance  Source  DF  SS  MS  F  P  Regression 1__34.900.000 Residual Error 13____ Total 1411.3240\begin{array}{lccccc}\text { Source } & \text { DF } & \text { SS } & \text { MS } & \text { F } & \text { P } \\\text { Regression } & 1 & \_\_& & 34.90 & 0.000\\\text { Residual Error }&13& \_\_& \_\_\\\text { Total }&14&11.3240\end{array} -Write the equation of the least squares line.

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= 4.8615...

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