REGRESSION REVIEW (PROFESSOR GIVEN)

2. Another variable the enthusiast recorded was the weight of the car in pounds. The regression with horsepower is given below. Simple linear regression results: Dependent Variable: Weight Independent Variable: HP Weight=2084.29 + 6.938 HP Sample size: 425 R (correlation coefficient) = 0.647 R-sq = 0.418 Estimate of error standard deviation: 581.01 Parameter estimates: Parameter Estimate Std. Err. Alternative Intercept 2084.29 90.14 +0 Slope 6.938 0.398 #0 DF T-Stat P-value 423 23.12 <0.0001 XXXX 423 XXXX a. Interpret the slope of this regression line if possible. If not, explain. b. Interpret the intercept of this regression line if possible. If not, explain. c. If the amount of power produced by the engine is 190 HP, what is the expected weight for this car? d. If the weight is 3,400 pounds for a car with engine power of 190 HP, what is the residual of weight? e. What percentage of the variability in weight can be explained by the regression line relationship with engine power? f. If…

Brain size Does your IQ depend on the size of yourbrain? A group of female college students took a test thatmeasured their verbal IQs and also underwent an MRI scan to measure the size of their brains (in 1000s of pix-els). The scatterplot and regression analysis are shown, and the assumptions for inference were satisfied.Dependent variable is: IQ_VerbalR-squared = 6.5% s = 21.5291 df = 18Variable Coefficient SE(Coeff)Intercept 24.1835 76.38Size 0.098842 0.0884a) Test an appropriate hypothesis about the associationbetween brain size and IQ.b) State your conclusion about the strength of thisassociation.

You may need to use the appropriate technology to answer this question. The commercial division of a real estate firm is conducting a regression analysis of the relationship between x, annual gross rents (in thousands of dollars), and y, selling price (in thousands of dollars) for apartment buildings. Data were collected on several properties recently sold and the following computer output was obtained. Analysis of Variance SOURCE DF Adj SS Regression 1 41587.3 Error 7   Total 8 51984.1 Predictor Coef SE Coef T-Value Constant 20.000 3.2213 6.21 X 7.210 1.3626 5.29 Regression Equation Y = 20.0 + 7.21 X (a) How many apartment buildings were in the sample?   (b) Write the estimated regression equation. ŷ =        (c) What is the value of  sb1?   (d) Use the F statistic to test the significance of the relationship at a 0.05 level of significance. State the null and alternative hypotheses. H0: ?1 ≠ 0Ha: ?1 = 0H0: ?0 = 0Ha: ?0 ≠ 0    H0: ?0 ≠…

Shown below is a portion of a computer output for regression analysis relating Y (dependent variable)X(independent variable). ANOVA                   df.                  Ss Regression 1.             24.011 Residual.     8.            67.989              Coefficients. Standard error Intercept.   11.065.     2.043 X.                 -0.511.      0.304   What is the sample size?

The estimated regression equation relating university GPA to the student's SAT mathematics score and high-school GPA is developed. A portion of Excel output shows the following. ANOVA df MS Significance F Regression 1.7621 ? ? 0.0001 Residual ? Total 1.88 Coefficient Standard Error t Stat Intercept H_GPA SAT_math -1.4053 0.4848 0.0235 0.0087 0.0049 0.0011 where H_GPA = high-school grade point average, SAT_math = SAT %3D mathematics score. (a) (4 points) Complete the eight missing entries (indicated by ?) in the above table. (b) (3 points) At 5% significance level, is the model significant? (c) (3 points) At 5% significance level, is each independent variable significant? (d) (2 points) Did the estimated regression equation provide a good fit to the data? Explain in one or two sentences.

A study on how the time of exercise affects heart rate had the following outputSimple Linear regression results:Dependent Variable: Heart RateIndependent Variable: Exercise timeSample size: 81R (correlation coefficient) = 0.4188R-sq = 0.1754Estimate of error standard deviation: 0.32416Parameter estimates: Parameter Estimate Std. Err. DF T-Stat P-Value Intercept 1.80938 1.9352 79 0.935 0.3524 Slope 0.4752395 0.11594 79 4.099 <0.0001 According to the output, if I exercise for time=163, what should be my heart rate?Use 4 decimal placesAfter exercising everybody has different heart rates, which means there is a lot of variability in heart rates. How much of that variability is explained by exercise time?Use 4 decimal places

The value obtained for the test statistic, z, in a one-mean z-test is given. Also given is whether the test is two tailed, left tailed, or right tailed. Also is given the P-value.A left-tailed test: z = -1.17 P-value: 0.1210 Use technology to create a scatter plot of the data above. Include the regression line. (Hand drawn graphs will not be accepted.) The explanatory (input) variable and the response (output) variable must be clearly labeled, within the context of this problem.

Both arm circumference and BMI measurements have been used as screening tools for being underweight, overweight, or obese. We want to determine if there is a significant correlation between arm circumference (in centimeters or cm) and body mass index or BMI (in kg.m2) among 10 participants. The results of a correlation and regression analysis are indicated in the Excel output below. The mean arm circumference (the independent variable) was 35.2 cm, and the mean BMI (the dependent variable) was 30.7 kg.m2.   SUMMARY OUTPUT                         Regression Statistics           Multiple R 0.855646           R Square 0.732129           Adjusted R Square 0.698646           Standard Error 3.806088           Observations 10                         ANOVA               df SS MS F Significance F   Regression 1 316.7456 316.7456 21.86518 0.001590054   Residual 8 115.8904 14.4863       Total 9 432.636…

Consider a regression that uses X = # of 60 pound bench press repetitions to predict Y = maximum bench press amount in pounds…..for the ith student among a sample of high school athletes.  The estimated predicted sample regression is as follows Maxi= 68.91 + .6885Repsi            *the ei is left off for simplicity   The table below lists the actual data from which the regression above was estimated.  For example, a student in the study who trained by doing 12 “60 pound reps” accomplished a maximum bench press amount of 76 pounds.   Xi Yi 12 76 15 82 17 79 3 71   Predict the average maximum bench press for a student that does 17 “60 pound reps”      a)  a.  around 68.85 pounds   b)  b.  around 85.00 pounds   c)  c.  around 80.61 pounds   d)  d.  around 68.91 pounds

Whta kind of plot is useful for deciding whether it is reasonable to find a regression plane for a set of data points involving several predictor variables?

To investigate the relationship between the milage and sales price for a popular car model the pictured scatterplot was used. a) Based on the excel output that's pictured, what is the estimated regression equation that could be used to predict the price given the miles? b) Does the model fit the data? (ie whether the regression relationship is statistically significant) Did the estimated regression equation provide a good fit? (ie use the coefficient of determination to explain variability independent variable) c) Suppose you are considering purchasing a car of this model with 60000 miles. Using the estimated regression equation, predict the price.

What is the degree of the correlation between sales and display area in square feet? The sales of 14 store branches and display area(in square feet) were obtained. The objective is to predict sales in terms of area of dispslay. Data were encoded into MS Excel. Output is shown below. REGRESSION STATISTICS Multiple R 0.954 R Square 0.910 Adjusted R square 0.902 Standard Error 936.85 Observations 14 Coefficients Standard t stat Error Intercept 901.25 513.02 1.76 Size 1.69 0.15 11.00 CHOICES A. 0.15, low positive B. 0.910, high positive C. 0.954, high positive D. 0.902, high positive

The average height of a large group of children is 43 inches, and the SD is 1.2inches. The average weight of these children is 40 pounds, and the SD is 2pounds. The correlation between the two variables is r = 0.65.A scatter diagram is drawn, with height on the horizontal axis and weight on thevertical axis. The scatter diagram is football shaped. The regression line forpredicting weight based on height is drawn through the scatter.(a) Predict the weights and the typical size of the error for those predictions ineach of the following case:A child who is 43 inches tall is predicted to weigh _____________ pounds, give ortake _____________ pounds.A child who is 41.8 inches tall is predicted to weigh ____________ pounds, give ortake _____________ pounds. 37 pounds and is 41.8 inches tall. Relative to allchildren with the same height, this child’s weight is (pick one)(i)    smaller than average(ii) about average(iii) larger than average(iv) impossible to determineShow your work and justify…

A study was done to compare tree height with trunk thickness. The following output was generated from the regression model.Simple linear regression results:Dependent Variable: Tree HeightIndependent Variable: Trunk SizeHeight = 39.041668 + 8.668677 TrunkSample size: 25R (correlation coefficient) = 0.4418R-sq = 0.1952Estimate of error standard deviation: 10.09977Parameter estimates: Parameter Estimate Std. Err. Intercept 39.041668 15.1736 Trunk 8.668677 3.670058 Assuming the conditions are met test if trunk size is a good predictor of tree height

In a study of the performance of a new engine design, the weight of 12 cars (in pounds) and the top speed (in mph) were recorded. A regression line was generated and shown to be an appropriate description of the relationship. The results of the regression analysis are below. Depend Variable: Top Speed Variable Constant Coefficient 107.58 s.e. of Coeff t-ratio prob Weight 0.8710 11.12 0.4146 9.67 0.000 2.10 0.062 R squared = 30.6% R squared (adjusted) = 23.7% s = 10.42 with 12-2 = 10 degrees of freedom Part A: Provide the regression equation based off the analysis provided & explain it in context. Part B: List the conditions for inference that need to be verified. Assuming these conditions have been met, does the data provide convincing evidence of a relationship between weight and top speed? Part C: Assuming all conditions for inference have been verified, determine a 95% confidence interval estimate for the slope of the regression line.

What kind of plot is useful for deciding whether finding a regression line for a set of data points is reasonable?

The coefficient of determination of a set of data points is 0.862 and the slope of the regression line is - 3.27. Determine the linear correlation coefficient of the data. What is the linear correlation coefficient? r= (Round to three decimal places as needed.)

The coefficient of determination of a set of data points is 0.837 and the slope of the regression line is 3.26. Determine the linear correlation coefficient of the data. What is the linear correlation coefficient? (Round to three decimal places as needed.)

Based on the Descriptive Statistcs table below,  1. Use this formula to calculate Adjusted R2 = 1.0 – (Residual MS/Total MS) 2.  What is the intercept coefficient value (b0)?3. What is the slope coefficient value (b1| in the ANOVA Table and what         does this value mean? 4. Modify the generic wording below to explain the Coefficient of Determination: Positive Coefficient (Direct Relationship) The regression coefficient (XX), or the slope of the linear regression equation, indicates a XX (unit)increase (decrease) in the independent variable (X) is associated with a XX (unit) increase (decrease) in the dependent variable (Y).

A student collected concentration versus absorbance data for a series of standards and produced a standard curve. Which value of 2 would reflect the linear regression curve the student produced? Absorbance 1 0.9 0.8 0.7 0.6 0.5 0.4 0.3 0.2 0.1 0 0 ² = 0.10 ² = 0.76 ² = 0.98 2 = 0.45 1 Absorbance vs. Concentration 2 3 Concentration (ppm) 4 5 6

i)Test individually whether the slope coefficients are significant at 10% significance level. ii)Test the overall significance of the estimated regression at 1% significance level.

Simple linear regression results:   Dependent Variable: Heart. DiseaseIndependent Variable: BikingHeart. Disease = 18.115809 - 0.20845321 BikingSample size: 68R (correlation coefficient) = -0.94616452R-sq = 0.8952273Estimate of error standard deviation: 1.5273175 Correlation between Biking and Heart. Disease is: -0.94616452. Correlation between Smoking and Heart. Disease is: 0.32421.   1) State r 2 (i.e., the coefficient of determination) for “Biking” and “Heart.Disease” and explain what this value means in context of the data set.

A multiple regression analysis produced the following tables. Predictor Coefficients StandardErrort Statistic p-valueIntercept -139.609 2548.989 -0.05477 0.957154x 24.24619 22.25267 1.089586 0.295682x 32.10171 17.44559 1.840105 0.08869Source df SS MS F p-valueRegression 2 302689 151344.5 1.705942 0.219838Residual 13 1153309 88716.07Total 15 1455998Using = 0.01 to test the null hypothesis H :?1 = ?2 = 0, the critical F value is ____.6.701.964.845.995.70

We are interested in the relationship between mid-term exam scores and final exam scores. The Final Exam score is the dependent variable and Midterm score is the independent variable. Use the simple regression output on below to answer the question below. Use a significance level of 0.05 for all hypothesis tests and intervals. What was the sample size for this simple regression analysis? Analysis of Variance Sum of Source DF Squares Mean Square F Ratio Model 1 2632.8012 2632.80 23.2115 Error 54 6125.0381 113.43 Prob > F C. Total 55 8757.8393 |t| Lower 95% Upper 95% Intercept 31.738559 10.26099 Midterm 3.09 0.0031* 11.166522 52.310596 0.8768157 0.61916 0.128514 4.82 <.0001* 0.3615044

Emma’s parents recorded her height at various ages up to 66 months. They entered the data into StatCrunch and found the following statistics: Simple linear regression results: Dependent Variable: Height Independent Variable: Age Height = 22.324324 + 0.34234234 Age Sample size: 5 R (correlation coefficient) = 0.99399497 Check all statements that are correct: The correlation is positive At birth, Emma was approximately 22.3 inches tall. In a month, Emma grew approximately 0.34 inches. The correlation is negative. When Emma is 12 years (144 months) old, the parents can expect her to be about 71.6 inches tall. There is no correlation.