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Homework #7

Homework #7. Solutions. 1. a). Relationship is curved between mercury and alkalinity, and mercury and calcium. It is roughly linear between mercury and pH. 1. b).

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Homework #7

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  1. Homework #7 Solutions

  2. 1. a) Relationship is curved between mercury and alkalinity, and mercury and calcium. It is roughly linear between mercury and pH.

  3. 1. b) Residuals versus alkalinity and calcium both indicate non-constant variance, and slight lack of fit. Residuals are fairly evenly distributed in the pH plot. Normal plot of residuals (not shown), seems linear.

  4. 1. c) Relationship is now linear for all predictors.

  5. 1. d) Correlations: pH, Logalka, Logmerc, Logcalc pH Logalka Logmerc Logalka 0.795 Logmerc -0.644 -0.761 Logcalc 0.705 0.827 -0.549 Cell Contents: Pearson correlation Yes, the correlations match the scatter-plots, for example logmerc and logalka appear to have a strong negative correlation which is supported by the -.761 r value.

  6. 1. e)

  7. 1. e) • Non-constant variance and lack of fit is no longer seen in any residual plot. Log transformation improved regression. • Normal plot of residuals is roughly linear. Assumption of normality seems reasonable.

  8. 1. f) Analysis of Variance Source DF SS MS F P Regression 3 9.8908 3.2969 27.11 0.000 Residual Error 33 4.0129 0.1216 Total 36 13.9037 Ho: B1=B2=B3=0 Ha: At least one B not equal to zero. The p-value (<.001) is the probability of getting a test statistic as extreme as F= 27.11 if the null hypothesis is true. Since the p-value is <.05 we reject the null hypothesis and conclude that at least one slope is not equal to zero.

  9. 1. g) Est. SE Logalka -0.4575 0.1005 n-p = 37-4 = 33 95% CI = Est +/- t(n-p,.975)*SE = -.4575 +/- 2.035*.1005 = (-.253, -.662 ) We are 95% confident that B1 lies between -.253 and -.662. 1. h) R-Sq(adj) = 68.5% 68.5% of the variability in log(mercury) is reduced by including all three predictors.

  10. 1. i) Using Minitab: Values of Predictors for New Observations New Obs Logalka Logcalc pH 1 4.09 3.91 7.00 New Obs Fit SE Fit 95.0% CI 95.0% PI 1 5.5688 0.106 ( 5.3530, 5.7846) ( 4.8272, 6.3104) So 95% PI is (e^ 4.8272,e^ 6.3104) = (124.86, 550.265)

  11. 2. a, b, c) • SSE (X1) = 16197.5 • SSR (X2|X1) = SSE(X1) - SSE(X1,X2) = 16197 - 13321 = 2875.6 • SSR(X3|X1,X2) = SSE(X1,X2) – SSE(X1,X2,X3) = 13321-13321 = 0

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