13. What proportion of the variance in Course % is attributable to the regression model? A- .568 B- .322 C- 960.521 D- 22.822 14. What is the regression equation for this analysis? A. Y=55.423x+.370 B. Y=55.423r+6.331 C. Y=.370x+55.423 D. Y=.370x+.077
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13. What proportion of the variance in Course % is attributable to the regression model?
A- .568
B- .322
C- 960.521
D- 22.822
14. What is the regression equation for this analysis?
A. Y=55.423x+.370
B. Y=55.423r+6.331
C. Y=.370x+55.423
D. Y=.370x+.077
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- call: Researchers measured the percent 1m(formula = Symptoms - wear_mask, data - some_states) of people in 25 states who ʻknew someone with COVID symptoms' (ŷ) and regressed this on the percent of the population frequently wearing a mask in public (x). Residuals: Min -7.9167 -2.3306 -0.2469 2.5020 7. 3345 10 Median 30 Маx coefficients: (Intercept) 111.0981 wear_mask Estimate std. Error t value Pr (>|t|) 10. 5423 10. 538 2.82e-10 *** -8. 375 1. 94 e-08 *** -1.0419 0.1244 signif. codes: 0 ****' 0.001 ***' 0.01 **' 0.05 '.' 0.1 ' ' 1 Residual standard error: 3.859 on 23 degrees of freedom Multiple R-squared: 0.7531, F-statistic: 70.15 on 1 and 23 DF, p-value: 1.936e-08 Adjusted R-squared: 0.7423 If 75 percent of people in a state wear masks regularly, what % of people does this model predict will know someone with COVID symptons? 1) 32 2) 33 3) 34 4) 35Researchers measured the percent call: Im(formula - Symptoms - wear_mask, data - some_states) of people in 25 states who 'knew someone with COVID symptoms' (ŷ) Residuals: Min 10 Median 30 мах -7.9167 -2.3306 -0.2469 2. 5020 7.3345 coefficients: and regressed this on the (Intercept) 111.0981 wear_mask Estimate std. Error t value Pr(>|t|) 10. 5423 10. 538 2.82e-10 *** -8. 375 1.94 e-08 *** -1.0419 0.1244 percent of the population frequently wearing a mask in public (x). signif. codes: 0 ***** 0.001 **** 0.01 * 0.05 '.' 0.1''1 Residual standard error: 3. 859 on 23 degrees of freedom Multiple R-squared: 0.7531, F-statistic: 70.15 on 1 and 23 DF, p-value: 1.936e-08 Adjusted R-squared: 0.7423 Which of the following represents the correct regression equation? 1) ŷ = -1.04 + 111x 2) x = 111.10 – 1.04ý 3) ŷ = 111.10 + 10.54x %3D 4) ŷ = 111.10 – 1.04xDefine Ordinary Least Square predicted values and residuals?
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- a residual is the distance between the data value in the line of best fit true or false?What is the expected average height of 8 year old boys, using the regression equation?The Average final score was 78, the average midterm score was 80. The standard deviation for the average final midterm score was 4.5 and the standard deviation for the average midterm score was 5.5. The correlation coefficient is 0.76 Find the least squares regression line. Supposed that wanted to find to predict the final exam score based the midterm score
- The data show the chest sire and weight of several bears Find the regression equation, leting chest size be the independent(x)variable. Then find the best predicted we ght of a bear inith a chest sizo of 40 inches. Is the result close to the actual weight of 272 pounds? Use a signifcance level of 0.05 8 Click the icon to verw the critical values of the Pearson correlation coefficient . What is the regrestion equation? \[ \hat{y}=\text {. K(Round ta one decimal place as needed) } \]The standard error of the mean (Qx) is a(n) estimate of population means which make up the sampling distribution. indication of variability in the distribution of sample means indication of variability in the raw data derived from the sample measure of variability in the populationOn minitab or excel Calculating the coefficient of determination of a set of data, it was obtained that r^2 = 0.845 . Based on this value, it can be concluded that: Options: The least squares line provided a good fit as a large proportion of the variability in "y" has been explained by the least squares line. The least squares line provided a good fit as a small proportion of the variability in "y" has been explained by the least squares line. The least squares line did not provide a good fit since a small proportion of the variability in "y" has been explained by the least squares line. The least squares line did not provide a good fit since a large proportion of the variability in "y" has been explained by the least squares line.