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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?Jimmy tested a sample with of n=4 pairs of X and Y scores and found SSY = 48 and a Pearson correlation between X and Y of r = 0.4 Calculate whether the Fobserved in this regression experiment is significant at the ∞ = o.01 level) In estimating the regression in the previous problem (#2), 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 values for 100 observations.
- A set of n = 15 pairs of X and Y values has a correlation of r = +0.80 with SSY = 75, and the regression equation for predicting Y is computed. Find the standard error of estimate for the regression equation. How big would the standard error be if the sample size were n = 30.2. The following data, adapted from Montgomery, Peck, and Vining (2001), present the number of certified mental defectives per 10,000 of estimated population in the United Kingdom ( y) and the number of radio receiver licenses issued (x) by the BBC (in millions) for the years 1924 through 1937. Fit a regression model relating y and x. Comment on the model. Specifically, does the existence of a strong correlation imply a cause-and-effect relationship?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 average midterm score in a large statistics class was 70 with an SD of 10. The average final score in the same class was 65 with an SD of 15. The correlation coefficient between midterm and final scores was r=0.6. Using the regression line, we predict the final score of a student with a midterm score of 60 to be , but this prediction is likely to be off by about . Fill in the blanks, rounding each answer to one decimal point.I see that you ran a correlation for Pearson's r, can you run a linear regression?1) Indicate whether the following statements are true or false. Explain why and show your work. c) In the regression Y= B1+ B2X + B3Z+u , if there is a strong linear correlation between X and Z, then it is more likely you fail to reject the null hypotheses that individual slope parameters are insignificant.
- Using the regression line attached. Based on only the above plot, one can conclude: a) height causes an increase in weight b) weight causes an increase in height c) taller people are more likely to weigh more than shorter people, at least in the sample on which this data is based d) a statistically significant predictive relationship between height and weight e) c and dIf other factors are held constant and the Pearson correlation value between X and Y is r = 0.80, then the regression equation will tend to produce more accurate predictions than would be obtained if the Pearson correlation value was r = 0.60. Group of answer choices True FalseIf a sample of 25 pairs of data yields a correlation coefficient, r, of 0.390 and the scatterplot displays a linear trend, can you use the regression equation to make predictions, assuming your x-values are within the domain of the data set? Choose your answer from the multiple choice answers below A.) Yes, because rcrit = 0.396 and the regression coefficient, r, is less than this value. B.) Yes, because rcrit = 0.381 and the regression coefficient, r, is greater than this value. C.) No, because rcrit = 0.381 and the regression coefficient, r, is greater than this value. D.) No, because rcrit = 0.396 and the regression coefficient, r, is less than this value.