Exam 11: Simple Linear Regression

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Consider the following pairs of observations: x 2 3 5 5 6 y 1.3 1.6 2.1 2.2 2.7 Find and interpret the value of the coefficient of correlation.

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SSxy=45.1(21)(9.9)5=3.52;SSxx=99(21)25=10.8;SSyy=20.79(9.9)25=1.188;r=3.5210.81.188.9827;\begin{array} { l } S S _ { x y } = 45.1 - \frac { ( 21 ) ( 9.9 ) } { 5 } = 3.52 ; S S _ { x x } = 99 - \frac { ( 21 ) ^ { 2 } } { 5 } = 10.8 ; \\S S _ { y y } = 20.79 - \frac { ( 9.9 ) ^ { 2 } } { 5 } = 1.188 ; r = \frac { 3.52 } { \sqrt { 10.8 \cdot 1.188 } } \approx .9827 ;\end{array}
There is a strong linear relationship between xx and yy .

Consider the following pairs of observations: x 2 3 5 5 6 y 1.3 1.6 2.1 2.2 2.7 Find and interpret the value of the coefficient of determination.

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SSyy=1.188;SSE=.040741;r2=1.0407411.188.9657S S _ { y y } = 1.188 ; S S E = .040741 ; r ^ { 2 } = 1 - \frac { .040741 } { 1.188 } \approx .9657 \text {; }
96.57%96.57 \% of the sample variation in yy values can be attributed to the linear relationship between xx and yy .

A realtor collected the following data for a random sample of ten homes that recently sold in her area. House Asking Price Days on Market \ 114,500 29 \ 149,900 16 \ 154,700 59 \ 159,900 42 \ 160,000 72 \ 165,900 45 \ 169,700 12 \ 171,900 39 \ 175,000 81 \ 289,900 121 a. Construct a scattergram for the data. b. Find the least squares line for the data and plot the line on your scattergram. c. Test whether the number of days on the market, y, is positively linearly related to the  asking price, x. Use α=.05\text { asking price, } x \text {. Use } \alpha = .05 \text {. }

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a.
 a.    b.  S S _ { x y } = 10,232,660 ; S S _ { x x } \approx 1.833 \times 1010 ; \hat { \beta } _ { 1 } = .0005583 ; \bar { x } = 171,140 ;  \bar { y } = 51.6 ; \hat { \beta } _ { 0 } = 51.6 - .0005583 ( 171,140 ) = - 43.94   \hat { y } = 43.94 + .0005583 x  c. The test statistic is  \mathrm { t } = \frac { .0005583 } { 22.58 / \sqrt { 1.833 \times 10 ^ { 10 } } } \approx 3.35 . Based on 8 degrees of freedom, the rejection region is  t > 1.860 . Since the test statistic falls in the rejection region, we reject the null hypothesis and conclude that the number of days on the market,  y , is positively linearly related to the asking price,  x .

b. SSxy=10,232,660;SSxx1.833×1010;β^1=.0005583;xˉ=171,140S S _ { x y } = 10,232,660 ; S S _ { x x } \approx 1.833 \times 1010 ; \hat { \beta } _ { 1 } = .0005583 ; \bar { x } = 171,140 ; yˉ=51.6;β^0=51.6.0005583(171,140)=43.94\bar { y } = 51.6 ; \hat { \beta } _ { 0 } = 51.6 - .0005583 ( 171,140 ) = - 43.94
y^=43.94+.0005583x\hat { y } = 43.94 + .0005583 x
c. The test statistic is t=.000558322.58/1.833×10103.35\mathrm { t } = \frac { .0005583 } { 22.58 / \sqrt { 1.833 \times 10 ^ { 10 } } } \approx 3.35 .
Based on 8 degrees of freedom, the rejection region is t>1.860t > 1.860 . Since the test statistic falls in the rejection region, we reject the null hypothesis and conclude that the number of days on the market, yy , is positively linearly related to the asking price, xx .

A study of the top 75 MBA programs attempted to predict the average starting salary (in $1000's) of graduates of the program based on the amount of tuition (in $1000's)charged by the program. The results of a simple linear regression analysis are shown below: Least Squares Linear Regression of Salary Predictor Variables Coefficient Std Error T P Constant 18.1849 10.3336 1.76 0.0826 Size 1.47494 0.14017 10.52 0.0000  A study of the top 75 MBA programs attempted to predict the average starting salary (in $1000's) of graduates of the program based on the amount of tuition (in $1000's)charged by the program. The results of a simple linear regression analysis are shown below: Least Squares Linear Regression of Salary Predictor  \begin{array} { l r c c l } \text { Variables } & \text { Coefficient } & \text { Std Error } & \text { T } & \text { P } \\ \text { Constant } & 18.1849 & 10.3336 & 1.76 & 0.0826 \\ \text { Size } & 1.47494 & 0.14017 & 10.52 & 0.0000 \end{array}        In addition, we are told that the coefficient of correlation was calculated to be  r = 0.7763 . Interpret this result.  In addition, we are told that the coefficient of correlation was calculated to be r=0.7763r = 0.7763 . Interpret this result.

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 Calculate SSE and s2 for n=30,SSyy=100,SSxy=60, and β^1=.8\text { Calculate SSE and } s ^ { 2 } \text { for } n = 30 , \mathrm { SS } _ { \mathrm { yy } } = 100 , \mathrm { SS } _ { \mathrm { xy } } = 60 \text {, and } \hat { \beta } 1 = .8 \text {. }

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State the four basic assumptions about the general form of the probability distribution of the random error ε.

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The dean of the Business School at a small Florida college wishes to determine whether the grade-point average (GPA)of a graduating student can be used to predict the graduateʹs starting Salary. More specifically, the dean wants to know whether higher GPAs lead to higher starting Salaries. Records for 23 of last yearʹs Business School graduates are selected at random, and data On GPA (x)and starting salary (y, in $thousands)for each graduate were used to fit the model E(y)=β0+β1xE ( y ) = \beta _ { 0 } + \beta _ { 1 } x The results of the simple linear regression are provided below. =4.25+2.75x, S=5.15,S=1.87 SSyy=15.17,SSE=1.0075 Calculate the value of r2r ^ { 2 } , the coefficient of determination.

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For the situation above, give a practical interpretation of s=3213\mathrm { s } = 3213 .

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For the situation above, give a practical interpretation of r=.81r = .81 .

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What is the relationship between diamond price and carat size? 307 diamonds were sampled and a straight-line relationship was hypothesized between y = diamond price (in dollars)and x = size of The diamond (in carats). The simple linear regression for the analysis is shown below: Least Squares Linear Regression of PRICE Predictor What is the relationship between diamond price and carat size? 307 diamonds were sampled and a straight-line relationship was hypothesized between y = diamond price (in dollars)and x = size of The diamond (in carats). The simple linear regression for the analysis is shown below: Least Squares Linear Regression of PRICE Predictor    Interpret the standard deviation of the regression model. Interpret the standard deviation of the regression model.

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What is the relationship between diamond price and carat size? 307 diamonds were sampled and a straight-line relationship was hypothesized between y = diamond price (in dollars)and x = size of The diamond (in carats). The simple linear regression for the analysis is shown below: Least Squares Linear Regression of PRICE Predictor What is the relationship between diamond price and carat size? 307 diamonds were sampled and a straight-line relationship was hypothesized between y = diamond price (in dollars)and x = size of The diamond (in carats). The simple linear regression for the analysis is shown below: Least Squares Linear Regression of PRICE Predictor    Interpret the coefficient of determination for the regression model. Interpret the coefficient of determination for the regression model.

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A county real estate appraiser wants to develop a statistical model to predict the appraised value of houses in a section of the county called East Meadow. One of the many variables thought to be An important predictor of appraised value is the number of rooms in the house. Consequently, the Appraiser decided to fit the simple linear regression model: E(y)=β0+β1xE ( y ) = \beta _ { 0 } + \beta _ { 1 } x where y=y = appraised value of the house (in thousands of dollars) and x=x = number of rooms. Using data collected for a sample of n=74n = 74 houses in East Meadow, the following results were obtained: y=74.80+23.95xy = 74.80 + 23.95 x Give a practical interpretation of the estimate of the slope of the least squares line.

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In a study of feeding behavior, zoologists recorded the number of grunts of a warthog feeding by a lake in the 15 minute period following the addition of food. The data showing the number of grunts and and the age of the warthog (in days)are listed below: Number of Grunts Age (days) 83 118 61 134 32 148 37 153 56 160 33 167 55 176 10 182 13 188 Find and interpret the value of .

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a. Complete the table.  a. Complete the table.     b. Find  S S _ { x y } , S S _ { x x } , \beta 1 , \bar { x } , \bar { y } , and  \hat { \beta } _ { 0 } . c. Write the equation of the least squares line. b. Find SSxy,SSxx,β1,xˉ,yˉS S _ { x y } , S S _ { x x } , \beta 1 , \bar { x } , \bar { y } , and β^0\hat { \beta } _ { 0 } . c. Write the equation of the least squares line.

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A company keeps extensive records on its new salespeople on the premise that sales should increase with experience. A random sample of seven new salespeople produced the data on experience and sales shown in the table. Months on Job Monthly Sales y (\ thousands) 2 2.4 4 7.0 8 11.3 12 15.0 1 .8 5 3.7 9 12.0 Summary statistics yield SSxx=94.8571,SSxy=124.7571,SSyy=176.5171,xˉ=5.8571S S _ { x x } = 94.8571 , S S _ { x y } = 124.7571 , S S _ { y y } = 176.5171 , \bar { x } = 5.8571 , and yˉ=7.4571\bar { y } = 7.4571 . Using SSE=12.435S S E = 12.435 , find and interpret the coefficient of determination.

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What is the relationship between diamond price and carat size? 307 diamonds were sampled and a straight-line relationship was hypothesized between y = diamond price (in dollars)and x = size of The diamond (in carats). The simple linear regression for the analysis is shown below: Least Squares Linear Regression of PRICE  Predictor \text { Predictor }  What is the relationship between diamond price and carat size? 307 diamonds were sampled and a straight-line relationship was hypothesized between y = diamond price (in dollars)and x = size of The diamond (in carats). The simple linear regression for the analysis is shown below: Least Squares Linear Regression of PRICE  \text { Predictor }      The model was then used to create  95 \%  confidence and prediction intervals for  \mathrm { y }  and for E(Y) when the carat size of the diamond was 1 carat. The results are shown here:  95 \%  confidence interval for  \mathrm { E } ( \mathrm { Y } ) : ( \$ 9091.60 , \$ 9509.40 )   95 \%  prediction interval for  Y : ( \$ 7091.50 , \$ 11,510.00 )  Which of the following interpretations is correct if you want to use the model to estimate  E ( Y )  for all 1 -carat diamonds? The model was then used to create 95%95 \% confidence and prediction intervals for y\mathrm { y } and for E(Y) when the carat size of the diamond was 1 carat. The results are shown here: 95%95 \% confidence interval for E(Y):($9091.60,$9509.40)\mathrm { E } ( \mathrm { Y } ) : ( \$ 9091.60 , \$ 9509.40 ) 95%95 \% prediction interval for Y:($7091.50,$11,510.00)Y : ( \$ 7091.50 , \$ 11,510.00 ) Which of the following interpretations is correct if you want to use the model to estimate E(Y)E ( Y ) for all 1 -carat diamonds?

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In team-teaching, two or more teachers lead a class. A researcher tested the use of 74) team-teaching in mathematics education. Two of the variables measured on each teacher in a sample of 159 mathematics teachers were years of teaching experience x and reported success rate y (measured as a percentage)of team-teaching mathematics classes. The correlation coefficient for the sample data was reported as r = -0.3. Interpret this result.

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An academic advisor wants to predict the typical starting salary of a graduate at a top business school using the GMAT score of the school as a predictor variable. A simple linear regression of SALARY versus GMAT using 25 data points is shown below. β0^=92040β^1=228s=3213r2=.66r=.81df=23t=6.67\hat { \beta _ { 0 } } = - 92040 \hat { \beta } 1 = 228 s = 3213 r ^ { 2 } = .66 r = .81 \quad \mathrm { df } = 23 \quad t = 6.67 Give a practical interpretation of r2=.66r ^ { 2 } = .66 .

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Is the number of games won by a major league baseball team in a season related to the teamʹs batting average? Data from 14 teams were collected and the summary statistics yield: y=1,134,x=3.642,y2=93,110,x2=.948622\sum y = 1,134 , \sum ^ { x } = 3.642 , \sum y ^ { 2 } = 93,110 , \sum ^ { x ^ { 2 } } = .948622 , and xy=295.54\sum ^ { x y } = 295.54 Assume β^1=455.27\hat { \beta } 1 = 455.27 and σ^=9.18\hat { \sigma } = 9.18 . Conduct a test of hypothesis to determine if a positive linear relationship exists between team batting average and number of wins. Use α=.05\alpha = .05 .

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Consider the following pairs of observations: x 2 3 5 5 6 y 1.3 1.6 2.1 2.2 2.7 a. Construct a scattergram for the data. Does the scattergram suggest that yy is positively linearly related to xx ? b. Find the slope of the least squares line for the data and test whether the data provide sufficient evidence that yy is positively linearly related to xx . Use α=.05\alpha = .05 .

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