# problem-11.25

problem-11.25 The data can be entered and models fit as follows:

    > likability = c(-1,4,0,-1,4,1,6,2,7,1,2,2,7,5,2,3,6,1)
    > web = factor(rep(c("N","Y"), c(9,9)))
    > tv = factor(rep(c(0,"1-2","3+"), c(6,6,6)))
    > res.web = lm(likability ~ web)
    > res.tv = lm(likability ~ tv)
    > res.both = lm(likability ~ tv + web)

To see whether the web itself has an effect we can look at the output of summary():

    > summary(res.web)
    ...
    Coefficients:
                Estimate Std. Error t value Pr(>|t|)
    (Intercept)    2.444      0.873    2.80    0.013 *
    webY           0.778      1.235    0.63    0.538
    ---
    Signif. codes:  0 `***' 0.001 `**' 0.01 `*' 0.05 `.' 0.1 ` ' 1

    Residual standard error: 2.62 on 16 degrees of freedom
    Multiple R-Squared: 0.0242,     Adjusted R-squared: -0.0368
    F-statistic: 0.397 on 1 and 16 DF,  p-value: 0.538

Web advertising is not effective in this assessment. However, controlling first for television exposure we have

    > anova(res.tv, res.both)
    Analysis of Variance Table

    Model 1: likability ~ tv
    Model 2: likability ~ tv + web
      Res.Df  RSS Df Sum of Sq    F Pr(>F)
    1     15 86.2
    2     14 69.5  1      16.7 3.36  0.088 .
    ---
    Signif. codes:  0 `***' 0.001 `**' 0.01 `*' 0.05 `.' 0.1 ` ' 1

That is, the effect of web advertising seems more likely. It may be worthwhile to perform this test on a bigger sample to try to detect any differences. It is difficult to detect differences with so few observations per category.