The output below estimates the following regressionTreat; = Bo + B1 · Female; + e;Based on the results, which statement is most accurate.• reg treat st femalelSourceSdfMSNumber of obs619%3DF(1, 617)0.01%3DModel.0030063511.003006351Prob > F0.9128R-squaredAdj R-squaredResidual154.71428617.2507524790.0000-0.0016Total154.717286618.250351595Root MSE.50075%3DtreatCoefficientStd. err.P>|t|(95% conf. interval]st femalel.0051549.04707890.110.913-.0872993.0976092_cons.5033557.041023212.270.000.4227937.5839177O In the treatment group, the fraction female is slightly lower than in the control groupIn the treatment group, the fraction female is slightly higher than in the control groupIn the treatment group, the fraction female is much lower than in the control groupO In the treatment group, the fraction female is much higher than in the control group

As the regression coefficient corresponding to female is 0.0051549 which much lower. Also from P…

The output below estimates the following regression Treat; = Bo + B1 · Female; + e; Based on the results, which statement is most accurate. • reg treat st femalel Source S df MS Number of obs 619 %3D F(1, 617) 0.01 %3D Model .003006351 1 .003006351 Prob > F 0.9128 R-squared Adj R-squared Residual 154.71428 617 .250752479 0.0000 -0.0016 Total 154.717286 618 .250351595 Root MSE .50075 %3D treat Coefficient Std. err. P>|t| (95% conf. interval] st female1 .0051549 .0470789 0.11 0.913 -.0872993 .0976092 _cons .5033557 .0410232 12.27 0.000 .4227937 .5839177 O In the treatment group, the fraction female is slightly lower than in the control group In the treatment group, the fraction female is slightly higher than in the control group In the treatment group, the fraction female is much lower than in the control group O In the treatment group, the fraction female is much higher than in the control group

The output below estimates the following regression Treat; = Bo + B1 · Female; + e; Based on the results, which statement is most accurate. • reg treat st femalel Source S df MS Number of obs 619 %3D F(1, 617) 0.01 %3D Model .003006351 1 .003006351 Prob > F 0.9128 R-squared Adj R-squared Residual 154.71428 617 .250752479 0.0000 -0.0016 Total 154.717286 618 .250351595 Root MSE .50075 %3D treat Coefficient Std. err. P>|t| (95% conf. interval] st female1 .0051549 .0470789 0.11 0.913 -.0872993 .0976092 _cons .5033557 .0410232 12.27 0.000 .4227937 .5839177 O In the treatment group, the fraction female is slightly lower than in the control group In the treatment group, the fraction female is slightly higher than in the control group In the treatment group, the fraction female is much lower than in the control group O In the treatment group, the fraction female is much higher than in the control group

College Algebra

7th Edition

ISBN:9781305115545

Author:James Stewart, Lothar Redlin, Saleem Watson

Publisher:James Stewart, Lothar Redlin, Saleem Watson

Chapter1: Equations And Graphs

Section: Chapter Questions

Problem 10T: Olympic Pole Vault The graph in Figure 7 indicates that in recent years the winning Olympic men’s...

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Olympic Pole Vault The graph in Figure 7 indicates that in recent years the winning Olympic men’s pole vault height has fallen below the value predicted by the regression line in Example 2. This might have occurred because when the pole vault was a new event there was much room for improvement in vaulters’ performances, whereas now even the best training can produce only incremental advances. Let’s see whether concentrating on more recent results gives a better predictor of future records. (a) Use the data in Table 2 (page 176) to complete the table of winning pole vault heights shown in the margin. (Note that we are using x=0 to correspond to the year 1972, where this restricted data set begins.) (b) Find the regression line for the data in part ‚(a). (c) Plot the data and the regression line on the same axes. Does the regression line seem to provide a good model for the data? (d) What does the regression line predict as the winning pole vault height for the 2012 Olympics? Compare this predicted value to the actual 2012 winning height of 5.97 m, as described on page 177. Has this new regression line provided a better prediction than the line in Example 2?
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Suppose that researchers obtain a random sample of adults ages 18 – 40 and collect data on the following variables: shoe size – in inches age – in years height – in inches forearm length – in inches Suppose further that a multiple linear regression model is fit to the resulting data set using R Studio and that the following output is obtained from it. Use this output to answer the question that follows: > summary(lm(shoesize ~ age + height + forearm, data = measures)) Coefficients: (Intercept)ageheightforearm Estimate10.14882 0.06045 -0.02108 -0.06479 Std. Error 4.49245 0.06838 0.06350 0.06847 t value2.259 0.884 -0.332 -0.946 Pr(>|t|) 0.0264 0.3792 0.7408 0.3467 Residual standard error: 1.719 on 85 degrees of freedomMultiple R-squared: 0.01983, Adjusted R-squared: 0.01477 F-statistic: 0.5731 on 3 and 85 DF, p-value: 0.6342 What is the estimate for the standard deviation of the residuals? 1.719 0.01983 -0.946 0.6342
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Suppose that researchers obtain a random sample of adults ages 18 – 40 and collect data on the following variables: shoe size – in inches age – in years height – in inches forearm length – in inches Suppose further that a multiple linear regression model is fit to the resulting data set using R Studio and that the following output is obtained from it. Use this output to answer the question that follows: > summary(lm(shoesize ~ age + height + forearm, data = measures)) Coefficients: (Intercept)ageheightforearm Estimate10.14882 0.06045 -0.02108 -0.06479 Std. Error 4.49245  0.06838 0.06350 0.06847 t value2.259 0.884  -0.332 -0.946 Pr(>|t|) 0.0264  0.3792 0.7408 0.3467 Residual standard error: 1.719 on 85 degrees of freedomMultiple R-squared: 0.01983, Adjusted R-squared: -0.01477 F-statistic: 0.5731 on 3 and 85 DF, p-value: 0.6342 Using the information from above, fill in the blanks for the least-squares regression equation. Input all values to 5…
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A researcher notes that, in a certain region, a disproportionate number of software millionaires were born around the year 1955. Is this a coincidence, or does birth year matter when gauging whether a software founder will besuccessful? The researcher investigated this question by analyzing the data shown in the accompanying table. Complete parts a through c below. a. Find the coefficient of determination for the simple linear regression model relating number (y) of software millionaire birthdays in a decade to total number (x) of births in the region. Interpret the result. The coefficient of determination is 1.___? (Round to three decimal places as needed.) This value indicates that 2.____ of the sample variation in the number of software millionaire birthdays is explained by the linear relationship with the total number of births in the region. (Round to one decimal place as needed.) b. Find the coefficient of determination for the simple linear regression model…