Homework #7

Solutions

1. a)

Relationship is curved between mercury and alkalinity, and mercury and calcium. It is roughly linear between mercury and pH.

1035

445

89.625

30.675

68.525

24.175

1035445

7.6

5.4

89.625

30.67568.525

24.175 7.65.4

mercury

alka

calc

pH

1. b)

100500

500

0

-500

alka

Res

idua

lResiduals Versus alka

(response is mercury)

9080706050403020100

500

0

-500

calc

Res

idua

l

Residuals Versus calc(response is mercury)

987654

500

0

-500

pH

Res

idua

l

Residuals Versus pH(response is mercury)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.

1. c)

6.64736

5.55621

3.63055

1.33173

3.55395

1.64675

6.64736

5.55621

7.6

5.4

3.63055

1.33173

3.55395

1.64675 7.65.4

Logmerc

Logalka

Logcalc

pH

Relationship is now linear for all predictors.

1. d)

Correlations: pH, Logalka, Logmerc, Logcalc

pH Logalka LogmercLogalka 0.795Logmerc -0.644 -0.761Logcalc 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.

1. e)

543210

0.5

0.0

-0.5

Logalka

Res

idua

l

Residuals Versus Logalka(response is Logmerc)

4.53.52.51.50.5

0.5

0.0

-0.5

Logcalc

Res

idua

l

Residuals Versus Logcalc(response is Logmerc)

987654

0.5

0.0

-0.5

pH

Res

idua

l

Residuals Versus pH(response is Logmerc)

0.50.0-0.5

2

1

0

-1

-2

Nor

mal

Sco

re

Residual

Normal Probability Plot of the Residuals(response is Logmerc)

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.

1. f)

Analysis of Variance

Source DF SS MS F PRegression 3 9.8908 3.2969 27.11 0.000Residual Error 33 4.0129 0.1216Total 36 13.9037

Ho: B1=B2=B3=0Ha: At least one B not equal to zero.

The p-value (<.001) is the probability of getting a test statistic as extreme asF= 27.11 if the null hypothesis is true. Since the p-value is <.05 we reject the null hypothesis and conclude that atleast one slope is not equal to zero.

1. g) Est. SELogalka -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.

R-Sq(adj) = 68.5%

68.5% of the variability in log(mercury) is reduced by including allthree predictors.

1. h)

1. i)

Using Minitab:

Values of Predictors for New Observations

New Obs Logalka Logcalc pH1 4.09 3.91 7.00

New Obs Fit SE Fit 95.0% CI 95.0% PI1 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)

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