Analysis of Scientific Data - Semester 2, 2020 Week 11 Tutorial Polychlorinated biphenyls (PCBs) were once used in industry but were banned in the 1970s because of concerns about their toxicity. Despite the ban, PCBs can still be detected in most people because they are persistent in the environment. A team of researchers recorded the amount of PCBs detected in maternal milk from mothers who had eaten fish from a particular lake considered to be contaminated with PCBs. They subsequently administered an IQ test to the children when they were 11 years old. The results are shown in the following scatter plot along with the least-squares line fitting a linear relationship between the two variables: O 0 120 O o 110 O Full Scale IQ 0 0 100 0 O o 90 - o 600 800 1000 1200 1400 1600 Maternal milk PCB (ng/g) A regression analysis in R produced the following (edited) summary: Coefficients: Estimate Std. Error ( Intercept) 125.773972 7.008028 PCB -0.021109 0.007538 Residual standard error: 9.314 on 12 degrees of freedom Multiple R-squared: 0. 3952, Adjusted R-squared: 0.3448 F-statistic: 7.842 on 1 and 12 DF a) Based on the degrees of freedom, how many pairs of observations were used in the analysis? 12 + 2 = 14
b) Based on the coefficients for the least-squares line provided by R, estimate the mean IQ of children if their mothers had a maternal milk PCB measurement of 1400 ng/g. ! ! 125.773972 + (-0.021109)*(1400) = 96.221372 ! c) Which assumption of linear regression does the following plot check? Comment on the validity of the assumption here. Assumption of normality. 0 0 10 0 0 0 O 0 0 0 Residuals 0 -10 O 0 -20 O -1 0 1 Z d) Circle the point with a residual of -20 in the original scatter plot. e) Is there evidence of a negative association between maternal milk PCB levels and IQ outcome? T-test: (Estimate-0)/std error = - 0.021109/0.007538 = - 2.8 P value: pt(-2.8, df=12) = 0.008022124 Yes, there is sufficient evidence of a negative association You may find the following output from R useful in answering the questions on this sheet. > pt(-2.800,df=12) [1] 0.008022124 > 2*pt(-2.800,df=12) [1] 0.01604425 > 1-pf(7.842,df1=1,df2=12) [1] 0.01603363