15.) The age in years of 11 children and the number of words in their vocabulary is given below: 15.) The age in years of 11 children and the number of words in their vocabulary is given below: AGE (x) 1 2 3 4 5 6 3 5 2 4 6 Voc size(y) 25 220 540 1100 2100 2600 730 2365 260 1200 2551 Calculate the equation of the regression line (round to two decimal places) and the correlation coefficient. Now use it to predict the Vocabulary size for a child a.) 7 yrs old b.) 11 yrs old Round your answer to the nearest whole number.

15.) The age in years of 11 children and the number of words in their vocabulary is given below: 15.) The age in years of 11 children and the number of words in their vocabulary is given below: AGE (x) 1 2 3 4 5 6 3 5 2 4 6 Voc size(y) 25 220 540 1100 2100 2600 730 2365 260 1200 2551 Calculate the equation of the regression line (round to two decimal places) and the correlation coefficient. Now use it to predict the Vocabulary size for a child a.) 7 yrs old b.) 11 yrs old Round your answer to the nearest whole number.

Linear Algebra: A Modern Introduction

4th Edition

ISBN:9781285463247

Author:David Poole

Publisher:David Poole

Chapter7: Distance And Approximation

Section7.3: Least Squares Approximation

Problem 31EQ

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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?
Suppose IQ scores were obtained from randomly selected twins. For 20 such pairs of people, the linear correlation coefficient is r = 0.914 and the equation of the regression line is y = 9.87+0.9x, where x represents the IQ score of the twin born first and y represents the IQ score of the twin born second. What is the bestpredicted IQ of the twin born second, given that the twin born first has an IQ of 107?What does the r value tell you about the relationship between the IQ scores of twins?What is the Value for r squared? What does the r squared statiscic mean in this scenario?
If x and y in a regression model are totally unrelated, _______. the coefficient of determination would be 0 the MSE would be 0s the SSE would be 0 the correlation coefficient would be -1 the coefficient of determination would be 1
1. Plot the data points on a scatter diagram. 2. Determine the equation of the regression line and find Pearson product-moment correlation coefficient. 3. Determine the point estimate of y at x = 5.5
A sample of 70mwomen is obtained, and their heights (in inches) and pulse rates (in beats per minute) are measured. The linear correlation coefficient is 0.255and the equation of the regression line is y(caret) 17.8 + 0.950 x where x represents height. The mean of the 70 heights is 62.7in and the mean of the 70 pulse rates is74.5 beats per minute. Find the best predicted pulse rate of a woman who is 68 in tall. Use a significance level of alpha equals 0.01
A sample of 50 women is obtained, and their heights (in inches) and pulse rates (in beats per minute) are measured. The linear correlation coefficient is 0.245 and the equation of the regression line is y=17.6 + 0.910x,where x represents height. The mean of the 50 heights is 62.9 in and the mean of the 50 pulse rates is 72.3 beats per minute. Find the best predicted pulse rate of a woman who is 67 in tall. Use a significance level of (alpha equals) α=0.05.
8)Suppose that Y is normal and we have three explanatory unknowns which are also normal, and we have an independent random sample of 11 members of the population, where for each member, the value of Y as well as the values of the three explanatory unknowns were observed. The data is entered into a computer using linear regression software and the output summary tells us that R-square is 0.86, the linear model coefficient of the first explanatory unknown is 7 with standard error estimate 2.5, the coefficient for the second explanatory unknown is 11 with standard error 2, and the coefficient for the third explanatory unknown is 15 with standard error 4. The regression intercept is reported as 28. The sum of squares in regression (SSR) is reported as 86000 and the sum of squared errors (SSE) is 14000. From this information, what is MSE/MST? .5000 NONE OF THE OTHERS .2000 .3000 .4000
A sample of 7070 women is obtained, and their heights (in inches) and pulse rates (in beats per minute) are measured. The linear correlation coefficient is 0.2210.221 and the equation of the regression line is ModifyingAbove y with caret equals 17.6 plus 0.940 xy=17.6+0.940x, where x represents height. The mean of the 7070 heights is 62.862.8 in and the mean of the 7070 pulse rates is 77.377.3 beats per minute. Find the best predicted pulse rate of a woman who is 7474 in tall. Use a significance level of alpha equals 0.05α=0.05. The best predicted pulse rate of a woman who is 7474 in tall is nothing beats per minute.
uppose IQ scores were obtained from randomly selected couplescouples. For 2020 such pairs of people, the linear correlation coefficient is 0.8900.890 and the equation of the regression line is ModifyingAbove y with caret equals negative 0.74 plus 1.02 xy=−0.74+1.02x, where x represents the IQ score of the husbandhusband. Also, the 2020 x values have a mean of 101.17101.17 and the 2020 y values have a mean of 102.7102.7. What is the best predicted IQ of the wifewife, given that the husbandhusband has an IQ of 9999? Use a significance level of 0.05. The best predicted IQ of the wifewife is nothing. (Round to two decimal places as needed.
9)Suppose that Y is normal and we have three explanatory unknowns which are also normal, and we have an independent random sample of 11 members of the population, where for each member, the value of Y as well as the values of the three explanatory unknowns were observed. The data is entered into a computer using linear regression software and the output summary tells us that R-square is 0.79, the linear model coefficient of the first explanatory unknown is 7 with standard error estimate 2.5, the coefficient for the second explanatory unknown is 11 with standard error 2, and the coefficient for the third explanatory unknown is 15 with standard error 4. The regression intercept is reported as 28. The sum of squares in regression (SSR) is reported as 79000 and the sum of squared errors (SSE) is 21000. From this information, what is the adjusted R-square? .8 .7 NONE OF THE OTHERS .6 .5
Suppose IQ scores were obtained from randomly selected twins. For 20 such pairs of people, the linear correlation coefficient is 0.930 and the equation of the regression line is y=−10.69+1.11x, where x represents the IQ score of the twin born first. Also, the 20 x values have a mean of 98.28 and the 20 y values have a mean of 98.5. What is the best predicted IQ of the twin born second, given that the twin born first has an IQ of 92? Use a significance level of 0.05.LOADING... Click the icon to view the critical values of the Pearson correlation coefficient r.Question content area bottomPart 1The best predicted IQ of the twin born second is enter your response here.(Round to two decimal places as needed.)
(1) For neighborhoods with more than 66% of schools participating in free school lunch program, the estimated regression model is: crime_rate = 142.5737 - 1.9215(Prop_change_income) - 38.8741(0) - 19.20521(0) = 142.5737 - 1.9215(Prop_change_income) (2) For neighborhoods with 33% to 66% of schools participating in free school lunch program, the estimated regression model is: crime_rate = 142.5737 - 1.9215(Prop_change_income) - 38.8741(0) - 19.20521(1) = 123.36849 - 1.9215(Prop_change_income) (3) For neighborhoods with less than 33% of schools participating in free school lunch program, the estimated regression model is: crime_rate = 142.5737 - 1.9215(Prop_change_income) - 38.8741(1) - 19.20521(0) = 103.69959 - 1.9215(Prop_change_income) Questions: 1. What is difference between models in 1,2,and 3? 2. Which estimated model 1,2, or 3 results in the line with the highest estimated crime rates (the highest positioned line)?