Subject: NUMERICAL METHODSDirections: In example no. 2, the problem and solution are both given. Please answer letter a.Thank you in advance tutors! Example 2:Use least-squares regression to fit a straight line to x and y values given below.Along with the slope and intercept, compute the standard error of the estimateand the correlation coefficient. Plot the data and the regression line.4.11121517199 8 7y57101212nEx¡yt-Ex; £ yiη Σx2-Σ x)210(911)-(95)(82)%3Dn = 10Σy82a, == 0.352469959910(1277)-(95)zEx¡yi = 91195= 9.510ao = ỹ – ajã = 8.2 – 0.3524699599(9.5) = 4.851535381Σχ 1277== 8.2ΣΧ 95Therefore, the least square fit is:y = 4.851535381 + 0.3524699599xLeast-Squares Fit of a Straight Line14Standard error of the estimate12SrVn-2Evi-ao-a;x;)²%3Dп-2109.0739652878Sy == 1.065009710-26Correlation coefficient4EGi-y)²-E(y;-ao-a,x;)?r2 = St-Sr -St55.60-9.073965287r2 =55.60101520r2 = 0.8367991855r = 0.9147672849a.)Repeat the problem Example 2, but regress x versus y - that is, switch thevariables. Interpret your results. Plot the data and the regression line.Interpret your results.49.1112151719y67698 7101212

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Example 2: Use least-squares regression to fit a straight line to x and y values given below. Along with the slope and intercept, compute the standard error of the estimate and the correlation coefficient. Plot the data and the regression line. 15 17 4. 11 12 19 9 8 7 6. 7 6 10 12 12 Σy 82 nExiy-ExEy a = nEx1²-Ex1)² n = 10 10(911)-(95)(82) = 0.3524699599 10(1277)-(95)z Ex;yi = 911 95 = 9.5 10 ao = ỹ – aqã = 8.2 – 0.3524699599(9.5) = 4.851535381 Ex;? = 1277 y =- = 8.2 10 Σχ 95 Therefore, the least square fit is: y = 4.851535381 + 0.3524699599x Least-Squares Fit of a Straight Line 14 Standard error of the estimate 12 Sr Evi-ao-a‚x¡)² Vn-2 n-2 10 Sy 9.073965287 %3D = 1.0650097 10-2 6 Correlation coefficient r2 = St-Sr - EGi-y)²-E(yi-ao-a,x;)² St 2 55.60-9.073965287 r2 = 55,60 10 15 20 r2 = 0.8367991855 r = 0.9147672849 a.) Repeat the problem Example 2, but regress x versus y variables. Interpret your results. Plot the data and the regression line. Interpret your results. that is, switch the 4. 6. 11 12 15 17 19 y 7 8 7 10 12 12

Example 2: Use least-squares regression to fit a straight line to x and y values given below. Along with the slope and intercept, compute the standard error of the estimate and the correlation coefficient. Plot the data and the regression line. 15 17 4. 11 12 19 9 8 7 6. 7 6 10 12 12 Σy 82 nExiy-ExEy a = nEx1²-Ex1)² n = 10 10(911)-(95)(82) = 0.3524699599 10(1277)-(95)z Ex;yi = 911 95 = 9.5 10 ao = ỹ – aqã = 8.2 – 0.3524699599(9.5) = 4.851535381 Ex;? = 1277 y =- = 8.2 10 Σχ 95 Therefore, the least square fit is: y = 4.851535381 + 0.3524699599x Least-Squares Fit of a Straight Line 14 Standard error of the estimate 12 Sr Evi-ao-a‚x¡)² Vn-2 n-2 10 Sy 9.073965287 %3D = 1.0650097 10-2 6 Correlation coefficient r2 = St-Sr - EGi-y)²-E(yi-ao-a,x;)² St 2 55.60-9.073965287 r2 = 55,60 10 15 20 r2 = 0.8367991855 r = 0.9147672849 a.) Repeat the problem Example 2, but regress x versus y variables. Interpret your results. Plot the data and the regression line. Interpret your results. that is, switch the 4. 6. 11 12 15 17 19 y 7 8 7 10 12 12

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?
A sixth-grade teacher believes that there is a relationship between his students’ IQscores (y) and the numbers of hours (x) they spend watching television each week. Thefollowing table shows a random sample of 7 sixth-grade students.y 125 116 97 114 85 107 105x 5 10 30 16 41 28 21 Does the data provide sufficient evidence to indicate that the simple linear regressionmodel is appropriate to describe the relationship between x and y? Perform a model utilitytest at α = 0.05. (Give H0, Ha, rejection region, observed test statistic, P-value, decisionand conclusion.)Find the Pearson sample correlation coefficient between x and y. Then interpretthe result.
The least-squares regression equation is y=761.7x+13,208 where y is the median income and x is the percentage of 25 years and older with at least a bachelor's degree in the region. The scatter diagram indicates a linear relation between the two variables with a correlation coefficient of 0.7483. Predict the median income of a region in which 20% of adults 25 years and older have at least a bachelor's degree.
The least-squares regression equation is y=647.8x+17,858 where y is the median income and x is the percentage of 25 years and older with at least a bachelor's degree in the region. The scatter diagram indicates a linear relation between the two variables with a correlation coefficient of 0.7507. predict the median income of a region in which 20% of adults 25 years and older have at least a bachelor's degree. Round to the nearest dollar as needed.
The least-squares regression equation is y=761.7x+13,208 where y is the median income and x is the percentage of 25 years and older with at least a bachelor's degree in the region. The scatter diagram indicates a linear relation between the two variables with a correlation coefficient of 0.7483. In a particular region, 26.5 percent of adults 25 years and older have at least a bachelor's degree. The median income in this region is $30,074. Is this income higher than what you would expect? Why?
You have obtained a sub-sample of 1744 individuals from the Current Population Survey (CPS) and are interested in the relationship between weekly earnings and age. The regression, using heteroskedasticity-robust standard errors, yielded the following result: = 239.16 + 3.75× Age, R2 = 0.15, SER = 287.21., where Earn and Age are measured in dollars and years respectively. Interpret the intercept? Interpret the slope coefficient b) Is the effect of age on earnings large? The average age in this sample is 37.5 years. What is annual income in the sample? (e) Interpret the measures of fit.
The first step of the Durbin-Watson test for the presence of autocorrelation is to estimate the model and determine Select one: a. the residuals lagged one period. b. the current period residuals and the residuals lagged one period. c. the current period residuals. d. the current period residuals, the residuals lagged one period, and the residuals lagged two periods.
The results for the regression are as follows Coefficients Standard Error T stat Intercept 14.63 0.5226 27.995 financial crisis 13.71 1.0610 12.922 COVID-19 crisis 15.71 1.6643 9.439 transition 5.40 1.1835 4.563 In estimating the regression, you are also concerned that the t-statistics may be inflated because of the presence of conditional heteroscedasticity. You conduct a regression of the squared residuals against the dummy variables X1, X2, and X3 and find that for the squared residuals regression: Multiple R 0.4145 R Square 0.1718 Adjusted R Square 0.1600 SEE 92.3760 Conduct a test at the level to see if conditional heteroskedasticity is present In view of your answer for a), what needs to be done?
Consider the following correlations -0.9 , -0.5 , -0.2 , 0 , 0.2 , 0.5 and 0.9. For each give the fraction of the variation in y that is explained by the least-squares regression of y on x.
A random sample of nonindustrialized countries was selected, and the life expectancy in years is listed for both men and women. Men 61.7 42.8 72.6 57.9 55.4 62.1 Women 50.4 74.2 75.3 75.4 49.1 65.2 The correlation coefficient for the data is =r−0.003 and =α0.05. Should regression analysis be done? Now Find the equation of the regression line. Round the coefficients to at least three decimal places. =y′+abx =a =b Now Find women's life expectancy in a country where men's life expectancy =60 years. Round your answer to at least three decimal places. Women's life expectancy is ____ years.
The results for the regression are as follows Coefficients Standard Error T stat Intercept 14.63 0.5226 27.995 financial crisis 13.71 1.0610 12.922 COVID-19 crisis 15.71 1.6643 9.439 transition 5.40 1.1835 4.563 In estimating the regression found above, you are concerned that the t-statistics may be inflated because of serial correlation. You compute the DW statistic at 0.724 for the regression Based on the DW, what can you say about serial correlation between the residuals? Are they positively or negatively correlated? Or not correlated? Compute the sample correlation between the regression residuals from one period and those from the previous period. Perform a statistical test at the level to see if there is serial correlation. If you are using the table in the textbook, assume that the critical values of the DW statistic for 214 observations are about 0.11 higher than the critical…
You run a regression analysis on a bivariate set of data (n=87n=87). With ¯x=75.8x¯=75.8 and ¯y=26.4y¯=26.4, you obtain the regression equationy=−3.905x−60.301y=-3.905x-60.301with a correlation coefficient of r=−0.15r=-0.15. You want to predict what value (on average) for the response variable will be obtained from a value of 70 as the explanatory variable.What is the predicted response value? y =