stat221 sum02 lab5 - 3 L ‘ StatsZZl-Mortlock Lt Lab#5...

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Unformatted text preview: 3 L ‘ StatsZZl-Mortlock Lt» July 29, 2002 Lab #5 Lab #5 — 9-9,10-12, 10-13, 10-16, 11-15 9-9 (a) As the average SAT score increased, the percentage of high school seniors who took the SATs’ decreased. So when almost all the students took the SATs’, the scores were generally lower. There clearly is a negative relationship between the two. The correlation is 40.79 = 0.88. R= -0.88. So, -0.88 of the seniors takes the SATs. It is a strong correlation because it is close to —1. o 10 20 30 40 so so 70 so 90 Mala-o SATAvg = -2.18R:tTalhg + 1147; v2 = 0.79 'I would conclude that there is a cause-and-efi‘ect relationship between these two 'ables. The higher the average SAT scores, the smaller the percent taken is. (a) The proportion of the variability in turnpike tolls is explained by the regression line with mileage as 0.999. .123 + 0.0402(150) = “The toll for a person who needs to drive 150 miles 0 he pike is 1.407. (c) The regression equation predicts that the toll will rise 0.0402 for each additional mile that you drive on the turnpike. (d) 1/0.0402 = 24.88. You would have to drive 24.88 miles in order for the toll to increase by one dollar. 10-13 (a) The least squares line for predicting receipts from attendance is: receipts = 65.4 (attendance) — 85000 r"2 = 0.93. 2000 m 10000 Amnmnoe been: = mAAttandame - m r"2 = 0.93 (b) 65.4 (7 500).— 85000 = 405,500. So, the least squares line predicts for the receipts of a show with a weekly attendance of 7,500 is 405,500. (c) 65. (6531) — 85000 = 342,127.4. So, the fitted value is 342,127.4. The ' residual alu for Death ofa Salesman is 351,082 — 342,127.4 = 8,954.6. ‘ (d) abaret = 65.4 (6647) — 85000 = 3497138 F0 e = 65.4 (8195.67) -— 85000 = 450996.818 show with the largest fitted value is Fosse with 450,996.818. 10-16 (a) The regression line for predicting a player’s weight from his jersey number is: weight = 0.85 (number) + 180 where r"2=0.36. 0 102030405060708090100 Nmaer \Abiglt=0.850Nnber+1&Ir"2=0.1£ (b) 0.85 (50) +180 = 222.5. So, the weight of a player wearing number 50 is 222.5. (c) The slope coeflicient in context is 0.85, where 0.85 is used for determining from the sample data when minimizing the sum of the squared residuals. We use this coefficient expression for calculating the means and standard deviations of the two variables. (d) If Dan Loney, a 300-pound center wearing number 66 were to exchange his jersey in practice one day for number 16, previously worn by l85-pound quarterback, Kevin Cooper, I would expect Loney to neither gain nor lose weight. This is due to the fact that that it would only be for one day. Furthermore, for that one day, no ones weight is going to change just because their jersey number does. For this one-day, the weights and jersey numbers are not calculated into this regression line. (e) Knowing a player’s jersey number is of no real use in predicting his weight. I say this because they are not given their jersey numbers by how much they weigh, they are given their numbers at random, so there really should be no correlation between the two. If there were to be, then the higher the number, the more a player would weigh, or the lower the number the higher they would weigh; however, the randomness makes this hard to find the relationship. 1 1-1 5 (a) The residuals seem to be randomly scattered. When a straight line is a reasonable model, the residual plot should reveal a seemingly random scattering of points. (b) Despite the very high correlation of .999 the residual plot does give you reason to suspect that the relationship between toll and mileage is not linear. This is because the residual plot reveals a pattern of some kind. It appears to start above zero, then a little below, and then a little more above, a little farther below, then even farther above zero. ...
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