x 2 4 5 6 y 7 11 13 20 STRATEGY Put the data in L1 L2 Use prgm PREDICT in group

# X 2 4 5 6 y 7 11 13 20 strategy put the data in l1 l2

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x 2 4 5 6 y 7 11 13 20 STRATEGY: Put the data in L1 & L2. Use prgm PREDICT in group CH10Vn . Enter the default alpha (.05). (Note: Same data as 13, 15, 16, 17 and 18) 12) The paired data below consists of test scores and hours of preparation for 5 randomly selected students. The equation of the regression line is y ˆ = 44.8447 + 3.52427 x . Find the explained variation. x 5 2 9 6 10 y 64 48 72 73 80 Note: Find the explained variation = Σ ( y ˆ - y ) 2 when y ˆ = 44.8447 + 3.52427 x . STRATEGY: Put the data in L1 & L2. Use prgm VARIATN in group CH10Vn . press 1 (Note: Same data as 12, 15, 16, 17 and 18) 13) The paired data below consists of test scores and hours of preparation for 5 randomly selected students. The equation of the regression line is y^ = 44.8447 + 3.52427x. Find the unexplained variation. x Hours of preparation 5 2 9 6 10 y Test score 64 48 72 73 80 STRATEGY: Put the data in L1 & L2. Use prgm VARIATN in group CH10Vn . press 1 Use program PREDICT Use program VARIATN Answer: 511.724 the explained variation E U = Explained Variation U V = Unexplained Variation T V = Total Variation E U = Explained + Unexplained (used to check T V) Answer: 87.476 the unexplained variation E U = Explained Variation U V = Unexplained Variation T V = Total Variation E U = Explained + Unexplained (used to check T V) Use program VARIATN Answer: .8873
14) Use the computer display to answer the question. A collection of paired data consists of the number of years that students have studied Spanish and their scores on a Spanish language proficiency test. A computer program was used to obtain the least squares linear regression line and the computer output is shown below. Along with the paired sample data, the program was also given an x value of 2 (years of study) to be used for predicting test score. The regression equation is Score = 31.55 + 10.90 Years. Predictor Coef StDev T P Constant 31.55 6.360 4.96 0.000 Years 10.90 1.744 6.25 0.000 S = 5.651 R-Sq = 83.0% R-Sq (Adj) = 82.7% Predicted values Fit StDev Fit 95.0% CI 95.0% PI 53.35 3.168 (42.72, 63.98) (31.61, 75.09) Use the information in the display to find the value of the linear correlation coefficient r . Determine whether there is significant linear correlation between years of study and test scores. Use a significance level of 0.05. There are 10 pairs of data. ANSWER: (No calculator program to do this---do by reasoning) Use the Coef column. The constant is 31.55 and the x coefficient is 10.90. x y 90 . 10 3155 ˆ + = . Now find r and test it. R -Sq = 83.0%, R = 83 = .911, n = 5, α = .05. Look up the critical value in A-6 or use prgmA6PEARSN. The critical value is .632. The test stat | r | = | .911 | > .632 the critical so Reject Ho. There is significant linear correlation. Conclusions: r = 0.91; There is significant linear correlation. Note: P-val = 0 so we have another reason to reject Ho. ((Note: Same data as 12, 13, 16 and 17) 15) Find the total variation for the paired data. The paired data below consists of test scores and hours of preparation for 5 randomly selected students. The equation of the regression line is y^ = 44.8447 + 3.52427x. Find the total variation. x Hours of preparation 5 2 9 6 10 y Test score 64 48 72 73 80 (Note: Same data as 12 and 13) STRATEGY: Put the data in L1 & L2. Use prgm VARIATN in group CH10Vn .

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