Part B

During the summer, three species of penguins gather in the Antarctic Peninsula for breeding: Adlie, Chinstrap and Gentoo. Biologists of the Arctowski station measured the weight (in kg) of 90 adult penguins: 15 males and 15 females for each of the three species.

Below you can find the R output of a regression model for penguin weight, with two categorical predictors for species and sex, as well as their interaction.

> Species=factor(species,levels=c('Gentoo','Adlie','Chinstrap')) > Sex=factor(sex,levels=c('F','M')) > reg_inter=lm(weight~Species+Sex+Species:Sex) > summary(reg_inter)  Call: lm(formula = weight ~ Species + Sex + Species:Sex)  Residuals:  Min 1Q Median 3Q Max -1.25200 -0.31067 -0.01533 0.27300 1.04133  Coefficients:  Estimate Std. Error t value Pr(>|t|) (Intercept) 6.94867 0.12728 54.592 <2e-16 *** SpeciesAdlie -2.02733 0.18001 -11.263 <2e-16 *** SpeciesChinstrap[cSK1] -2.32000 0.18001 -12.888 <2e-16 *** SexM -0.14867 0.18001 -0.826 0.411 [cSK2] SpeciesAdlie:SexM 0.25933 0.25457 1.019 0.311 SpeciesChinstrap:SexM -0.04933 0.25457 -0.194 0.847 --- Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1  Residual standard error: ???? on ???? degrees of freedom Multiple R-squared: 0.8209, Adjusted R-squared: 0.8102 F-statistic: 76.98 on 5 and 84 DF, p-value: < 2.2e-16[cSK3] 

1.Write six separate fitted regression models, one for each combination of species and sex. What is the interpretation of the coefficient (be specific in your answer)?

2.Is the average weight of Gentoo male penguins significantly different from the average weight of Gentoo female penguins? Answer the question referring to the null hypothesis and alternative hypothesis, the value of the test statistics employed, its distribution under the null and the test p-value.

3.Consider the following output from the R anova function.

> anova(reg_spec,reg_inter) Analysis of Variance Table  Model 1: weight ~ Species Model 2: weight ~ Species + Sex + Species:Sex  Res.Df RSS Df Sum of Sq F Pr(>F) 1 ???? 20.965 2 ???? 20.414 ???? ???? ???? ???? 

What are the null and alternative hypotheses? Compute the test statistic, report its distribution under the null hypothesis and the rejection region at level 5%. Interpret the test result. You may use quantiles from an F distribution that are summarized as follows.

 > qf(0.95,df1=1,df2=84) [1] 3.954568 > qf(0.975,df1=1,df2=84) [1] 5.208847 > qf(0.95,df1=2,df2=84) [1] 3.105157 > qf(0.975,df1=2,df2=84) [1] 3.855726 > qf(0.95,df1=3,df2=84) [1] 2.713227 > qf(0.975,df1=3,df2=84) [1] 3.275835 > qf(0.95,df1=1,df2=86) [1] 3.951882 > qf(0.975,df1=1,df2=86) [1] 5.204434 > qf(0.95,df1=2,df2=86) [1] 3.102552 > qf(0.975,df1=2,df2=86) [1] 3.851734 > qf(0.95,df1=3,df2=86) [1] 2.710647 > qf(0.975,df1=3,df2=86) [1] 3.272009 

 > qf(0.95,df1=1,df2=87) [1] 3.950587 > qf(0.975,df1=1,df2=87) [1] 5.202305 > qf(0.95,df1=2,df2=87) [1] 3.101296 > qf(0.975,df1=2,df2=87) [1] 3.849808 > qf(0.95,df1=3,df2=87) [1] 2.709402 > qf(0.975,df1=3,df2=87) [1] 3.270163 

[cSK1]Species$Gentoo is our baseline

[cSK2]SexF is out baseline

[cSK3]Weight = beta0 +beta1 * Species2 + beta2* species3+ beta3*sexM+ beta4*species2*sexM + beta5*species3*sexM

Species2= difference between the so 1 or 0

sexM = female->0, male->1