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Statistics

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Feb 20, 2024

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 11.14 (p.629)  11.26 (p.632, OJUICE.xlsx Download OJUICE.xlsx )  11.106 (p.672, for part f, use the confidence interval in the table , BRNOUT.xlsx Download BRNOUT.xlsx )  11.116 (p.677, all parts excepting part e , ( BTU.xlsx Download BTU.xlsx ),  12.16 (p.701, WELLS.xlsx Download WELLS.xlsx ) 11.14 x i y i x i 2 x i y i 7 2 49 14 4 4 16 16 6 2 36 12 2 5 4 10 1 7 1 7 1 6 1 6 3 5 9 15 Sum: 24 Sum: 31 Sum: 116 Sum: 80 SSxy = 7*2 + 4*4 + 6*2 + 2*5 + 7*1 + 6*1 + 3*5 = 80 80 – sum(x)*sum(y)/7 = 80 – (24*31)/7 = -26.28 SSxx = 49 + 16 + 36 + 4 + 1 + 1 + 9 = 116 SSxx = 116 – 24^2/7 = 33.72 B1 = SSxy/SSxx = -26.28/33.72 = -.779 Xbar = sum(x)/n = 24/7 Ybar = sum(y)/n = 31/7 B0 = ybar – B1*xbar = 4.42 – (-.779)*(3.42) = 4.42 – (-2.66) = 7.08 SSxy = -26.28 SSxx = 33.72 B1 = -.779 Xbar = 3.42 Ybar = 4.42 B0 = 7.08 Least squares line = 7.08 + -.779x

11.26 a. Find the least squares line for the data. b. Interpret B0 and B1 in the words of the problem. c. Predict the sweetness index if amount of pectin in the orange juice is 300 ppm. [ Note: A measure of reliability of such a prediction is discussed in Section 11.6] y = B0 + B1x ybar = 5.65 xbar = 256.95 SSxy = -130.442 SSxx = 56,452.95 B1 = -.00231 B0 = 6.25 Least squares line = 6.26 + -.00231x B0 is the y intercept and represents the estimated sweetness index where there is no detectable pectin B1 is the slope and represents the rate of change in the sweetness index Sweetness index if pectin is 300ppm = 6.26 + -.00231(300) = 5.567 11.106 Yes they appear to be related. As the concentration increases, the emotional exhaustion index increases Correlation coefficient (calculations done in excel) xbar 578.32 ybar 68.56 SSxx 1800417 Ssxy 124348. 5 Ssyy 14026.1 6 R = 124348.5/sqrt(1800417*14026.16) = .7825 This value would imply that concentration does cause an increase in emotional exhaustion Value of test statistic: t = r(sqrt(n-2))/(sqrt(1-r^2)) .7825(sqrt(25-2))/(sqrt(1-.7825^2)) = 6.027 Critical value = 25 -2 = 23. Alpha/2 = .025 0 200 400 600 800 1000 1200 0 20 40 60 80 100 120

T = 2.069 6.027 > 2.069 This result means that we have enough evidenc to prove that the concentration index is useful when predicting burnout Reg est = -29.4967 + 8.86x SSE = 14448.79 (1800417-14448.79)/1800417 = .991 About 99.1% of the sample variation in exhaustion can be explained by using concentration to predict exhaustion in the straight line model 11.116 Calculations done in excel ybar 1357904.54 5 xbar 14170.7272 7 SSyy 2.44773E+1 3 SSxx 1568952926 Ssxy 1.6131E+11 B1 102.814047 8 B0 -99045.2863 Least squares estimate: 5221.85 +.0065x R = 1.6131E+11/sqrt(1568952926*2.44773E+13) = .823. This would suggest there is a positive linear relationship Test statistic = (.823*sqrt(22-2))/sqrt(1-.823^2) = 7.445 Alpha/2 = .05 T = 2.069 7.445 > 2.069 which means there is enough evidence to prove that there is a positive linear relationship between the two variables Coefficient of determination = SSyy – SSE/SSyy = .99999 About 99% of the sample variation in y can be explained by (or Attributed to) using x to predict y in the straight-line model. 5221.85 +.0065(8000) = 5273.85

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S = SSE/22-2 = 628184.69 5273.85 +/- (2.086)(628184.69)*sqrt(1 + 1/22 + ((8000- 14170.72727)^2)/1568952926 5273.85 +/- 1310393 * sqrt(1.069) 5273.85 +/- 1355306.599 = ( 1406083.967, -1395536.27) We can predict with 95% confidence that BTU will be within ( 1406083.967, -1395536.27) for a warehouse of 8000sq ft. Assumption 1: The mean of the probability distribution of e is 0—that is, the Average of the values of e over an infinitely long series of experiments is 0 for each setting of the independent variable x . This assumption implies that the mean value of y, E 1 y 2, for a given value of x is E 1 y 2 = b 0 + b 1 x . Assumption 2: The variance of the probability distribution of e is constant for all settings of the independent variable x . For our straight-line model, this assumption means that the variance of e is equal to a constant, say s 2 , for all values of x . Assumption 3: The probability distribution of e is normal. Assumption 4: The values of e associated with any two observed values of y are independent—that is, the value of e associated with one value of y has no effect on the values of e associated with other y -values. 12.16 SUMMARY OUTPUT Regression Statistics Multiple R 0.357715 R Square 0.12796 Adjusted R Square 0.119861 Standard Error 103.3014 Observations 327 ANOVA df SS MS F Significance F Regression 3 505770.4 168590.1 15.79864 1.31E-09 Residual 323 3446791 10671.18 Total 326 3952562 Coefficients Standard Error t Stat P-value Lower 95% Upper 95% Lower 95.0% Upper 95.0% Intercept -86867.9 31224.27 -2.78206 0.005719 -148297 -25439.3 -148297 -25439.3 LATITUDE (x1) -2218.76 526.8165 -4.21163 3.29E-05 -3255.18 -1182.33 -3255.18 -1182.33 LONGITUDE (x2) 1542.163 373.0721 4.133686 4.55E-05 808.2048 2276.121 808.2048 2276.121

DEPTH-FT (x3) -0.34962 0.156617 -2.23235 0.026277 -0.65774 -0.04151 -0.65774 -0.04151 Write a first-order model for arsenic level ( y ) as a function of latitude, longitude, and depth. b. Fit the model to the data using the method of least squares. c. Give practical interpretations of the b estimates. d. Find the model standard deviation, s , and interpret its value. e. Find and interpret the values of R 2 and R 2a . f. Conduct a test of overall model utility at a = .05. g. Based on the results, parts d–f, would you recommend using the model to predict arsenic level ( y )? Explain. X1 = latitude X2 = longitude X3 = depth Y = arsenic N =327 Significance level = .05 Predictor variables = 3 E(y) = β0+β1Latitude+β2Longitude+ β3Depth Y-hat(arsenic level) = -86867.917 – 2218.757*latitude + 1542.163*longtitude - .350*depth The β coefficients represent how much the arsenic level changes for a one unit change in latitude, longitude, or depth when the other variables are held constant at the same time. R^2 = .128 or 12.8% R^2a = .120 or 12% The model is predicting only 12.8% of the variation in the arsenic level. 87% of the information is remaining to be predicted. The three variables only capture 13% of the information for why arsenic varies from well to well Ho: B1 = B2 = B3 = 0 Ha: At least one Beta not equals zero Significance F < .05 We reject the null hypothesis. Yes, I would recommend using the model to predict arsenic levels because at least one of the variables is relevant in determining the arsenic level. At this point we don’t have a better alternative model.

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