Run a regression analysis on the following bivariate set of data with y as the response variable. x y 90.5 -46 55 38.4 42.9 70.8 60.7 110.2 54.8 76.3 56.6 26.9 83.1 17.8 64.5 67.6 61.2 82.2 90.7 -55.7 89.1 -24.8 Verify that the
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Run a
| x | y |
|---|---|
| 90.5 | -46 |
| 55 | 38.4 |
| 42.9 | 70.8 |
| 60.7 | 110.2 |
| 54.8 | 76.3 |
| 56.6 | 26.9 |
| 83.1 | 17.8 |
| 64.5 | 67.6 |
| 61.2 | 82.2 |
| 90.7 | -55.7 |
| 89.1 | -24.8 |
Verify that the
What is the predicted explanatory value?
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- Run a regression analysis on the following bivariate set of data with y as the response variable. x y 58.3 81.3 77.7 48.5 49.3 67.2 68.5 58.9 46.4 93.1 65.5 75.8 48.9 90.5 65.2 66.9 63.2 79.4 56.5 89.3 65.6 75.5 40.4 96.2 Find the correlation coefficient and report it accurate to three decimal places. r = What proportion of the variation in y can be explained by the variation in the values of x? Report answer as a percentage accurate to one decimal place. (If the answer is 0.84471, then it would be 84.5%...you would enter 84.5 without the percent symbol.)r² = %Based on the data, calculate the regression line (each value to three decimal places)y = x + Predict what value (on average) for the response variable will be obtained from a value of 77.8 as the explanatory variable. Use a significance level of α=0.05 to assess the strength of the linear correlation.What is the predicted response value? (Report answer accurate to one decimal place.)y =Caffeine (mg) Sleep (hours) 60 8 0 8 220 5.5 100 7 80 6.5 For caffeine intake, the mean is 92 and the standard deviation is 80.75. For sleep, the mean is 7 and the standard deviation is 1.06. Calculate the linear correlation coefficient r by hand for this set of paired data values using r=(z_x-z_y)/(n-1). Show your work. Is the value of r statistically significant? Use Pearson's r table to justify your answer. Does the value for r and the statistical significance indicate a positive linear correlation, a negative linear correlation, or no linear correlation between the variables caffeine intake and hours of sleep? What does this mean in context?Run a regression analysis on the following bivariate set of data with y as the response variable. X y 9.4 24.2 36.9 66.5 42.6 140.4 -16 -26.1 27.4 52 27.5 71.9 24.8 66.8 21 41.2 33.1 2.4 23.9 34.4 69.2 34.3 71.9 66 Find the correlation coefficient and report it accurate to three decimal places. r = What proportion of the variation in y can be explained by the variation in the values of x? Report answer as a percentage accurate to one decimal place. (If the answer is 0.84471, then it would be 84.5%...you would enter 84.5 without the percent symbol.) r² = % Based on the data, calculate the regression line (each value to three decimal places) y = X + Predict what value (on average) for the response variable will be obtained from a value of -13.4 as the explanatory variable. Use a significance level of a = 0.05 to assess the strength of the linear correlation. What is the predicted response value? (Report answer accurate to one decimal place.) y =
- Use the given data set to complete parts (a) through (c) below. (Use a = 0.05.) 10 8 13 9. 11 14 4 12 7 y 7.45 6.77 12.73 7.11 7.81 8.84 6.08 5.39 8.16 6.41 5.72 Click here to view a table of critical values for the correlation coefficient. Using the linear correlation coefficient found in the previous step, determine whether there is sufficient evidence to support the claim of a linear correlation between the two variables. Choose the correct answer below. O A. There is sufficient evidence to support the claim of a nonlinear correlation between the two variables. O B. There is insufficient evidence to support the claim of a nonlinear correlation between the two variables. O C. There is sufficient evidence to support the claim of a linear correlation between the two variables. O D. There is insufficient evidence to support the claim of a linear correlation between the two variables. c. Identify the feature of the data that would be missed if part (b) was completed without constructing…Run a regression analysis on the following bivariate set of data with y as the response variable. y 62.8 75.2 60 70.4 59.8 63 59.3 68.5 59 70.3 52.4 50.8 63.4 61.9 70.3 68.1 65.6 69.3 56.8 60.9 67.5 77.5 55.5 57.2 Find the correlation coefficient and report it accurate to three decimal places. What proportion of the variation in y can be explained by the variation in the values of x? Report answer as a percentage accurate to one decimal place. (If the answer is 0.84471, then it would be 84.5%...you would enter 84.5 without the percent symbol.) r2 = Based on the data, calculate the regression line (each value to three decimal places) y = Predict what value (on average) for the response variable will be obtained from a value of 52.5 as the explanatory variable. Use a significance level of a = 0.05 to assess the strength of the linear correlation. What is the predicted response value? (Report answer accurate to one decimal place.) y = Submit Question IO 01 100 O DetailsRun a regression analysis on the following bivariate set of data with y as the response variable. X y 9.8 75.6 17.1 50.7 62 71.6 41.4 60.4 44.8 37.7 52.6 40.6 60.1 10.4 55.6 38.9 74.2 -6.6 49.7 33.6 43.9 54.8 Verify that the correlation is significant at an a = 0.05. If the correlation is indeed significant, predict what value (on average) for the explanatory variable will give you a value of 45.9 on the response variable. What is the predicted explanatory value? X =
- Run a regression analysis on the following bivariate set of data with y as the response variable. X 109.4 69.8 90 87.5 64.1 59.6 63.1 92.2 74.7 61.5 83.3 80.3 58.9. 57.3 73 y 61.4 51.6 38.6. 83.7 71.5 69.5 35.4 79.9 40.5 Find the correlation coefficient and report it accurate to three decimal places. r= Based on the data, calculate the regression line (each value to three decimal places) y' = X +Run a regression analysis on the following bivariate set of data with y as the response variable. x y 59.6 57.3 20.1 61.9 8 87.2 0.1 91.7 31.7 67.6 31.9 62 19.9 74.7 33.1 85.4 0.8 82.8 56.3 59.4 3.7 90.7 Verify that the correlation is significant at an α=0.05α=0.05. If the correlation is indeed significant, predict what value (on average) for the explanatory variable will give you a value of 89.2 on the response variable.What is the predicted explanatory value?x =Life Expectancies A random sample of nonindustrialized countries was selected, and the life expectancy in years is listed for both men and women. Men 71.8 68.0 57.2 68.8 69.1 72.0 Women 65.2 67.3 45.0 66.3 68.1 60.2 The correlation coefficient for the data is r=0.829 and =α0.05. Should regression analysis be done? The correlation coefficient for the data is =r0.829 and =α0.05 . Should regression analysis be done? Find the equation of the regression line. Round the coefficients to at least three decimal places. y=′a+bx =a =b Find women's life expectancy in a country where men's life expectancy = 57 years. Round your answer to at least three decimal places. Women's life expectancy in years.
- Run a regression analysis on the following bivariate set of data with y as the response variable. x y 57.9 -34.4 16.6 92 40.1 89 55.8 54.5 57.9 27.4 17.3 164.1 65.2 62.4 48.3 51.9 66.2 14.5 70.9 -39.2 57 -90.1 58.5 63 Find the correlation coefficient and report it accurate to three decimal places.r = What proportion of the variation in y can be explained by the variation in the values of x? Report answer as a percentage accurate to one decimal place. (If the answer is 0.84471, then it would be 84.5%...you would enter 84.5 without the percent symbol.)r² = %Run a regression analysis on the following bivariate set of data with y as the response variable. x y 60.5 38.5 81.4 13.6 3.5 132.1 52.8 69.8 51.1 30 81.1 11.4 47.1 50 23 119.2 57.9 45.2 58.6 31.8 55.9 98.5 47.6 52 Verify that the correlation is significant at an α=0.05.If the correlation is indeed significant, predict what value (on average) for the explanatory variable will give you a value of 126.3 on the response variable.What is the predicted explanatory value? x=ont Paragraph Styles Editing -A professor from a Quantitative Methods in Business course decided to research the relationship between the Mid Term Exam, the average number of study hours spent per week during the semester and the final course grade for a given student. A sample of containing data from a previous semester was provided by the instructor and summarized in the following table: # of Students Mid-Term exam Study hours per Final course grade week grade 1 50.0 2.0 65.0 60.0 4.0 85.0 55.0 3.5 75.0 4 85.0 6.0 90.0 55.0 5.0 70.0 6. 72.0 4.5 89.0 7 75.0 6.5 91.0 8 45.0 3.0 65.0 6. 88.0 5.5 89.0 10 90.0 7.5 96.0 Develop a linear regression model, where Y is the Final Grade Course and X is the Mid-Term Exam. Calculate the coefficient of determination (r 2 , the coefficient of correlation (r), the variance (o 2) for Text Predictions: On
