1. Download the dataframe pirate_survey_noerrors.txt from http://nathanieldphillips.com/wp-content/uploads/2015/05/pirate_survey_noerrors.txt. The data are stored in a tab-separated text file with headers. Load the dataframe into an object called pirates. Because it’s tab-separated, use sep = “”.
pirates <- read.table("http://nathanieldphillips.com/wp-content/uploads/2015/05/pirate_survey_noerrors.txt", sep = "\t", header = T, stringsAsFactors = F)
  1. The function pairs() can create a matrix of scatterplots of different ratio or interval variables in a dataset. Enter the following code to see a matrix of scatterplots for the pirate dataset
pairs(~ age + tattoos + tchests.found + parrots.lifetime + sword.speed, data = pirates)

  1. What variables reliably predict the number of treasure chests a pirate has found? Conduct a simple linear regression analysis with treasure chests found as the dependent variable and 3 independent variables: parrots.lifetime, age, and tattoos. Save the model as the object model.1. Then, use the summary() function to see the coefficients. What are your conclusions?
model.1 <- lm(tchests.found ~ parrots.lifetime + age + tattoos,
                 data = pirates)
summary(model.1)
## 
## Call:
## lm(formula = tchests.found ~ parrots.lifetime + age + tattoos, 
##     data = pirates)
## 
## Residuals:
##    Min     1Q Median     3Q    Max 
## -9.566 -5.225 -2.271  2.636 46.003 
## 
## Coefficients:
##                   Estimate Std. Error t value Pr(>|t|)    
## (Intercept)       1.266327   1.407182   0.900 0.368389    
## parrots.lifetime -0.007083   0.088349  -0.080 0.936115    
## age               0.123838   0.044491   2.783 0.005480 ** 
## tattoos           0.274414   0.074297   3.693 0.000233 ***
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## Residual standard error: 7.785 on 996 degrees of freedom
## Multiple R-squared:  0.02135,    Adjusted R-squared:  0.0184 
## F-statistic: 7.241 on 3 and 996 DF,  p-value: 8.283e-05
# F (3, 996) = 7.241, p < 0.01, R2 = .018.

# for parrots, the results was non-significant (t(996) = -0.08, p = .936)
# for age, the results were significant (t(996) = 2.783, p < .01)
# for tattoos, the results were significant (t(996) = 3.693, p < .01 )

Conclusion: The independent variables age and tattoos seem to reliably predict the number of treasure chests a pirate has found.

  1. Using the results from the previous question, create a scatterplot with the true values of the dependent variable (treasure chests found) on the x-axis and the model fits on the y-axis. Make the plot look nice with appropriate labels.
plot(x = pirates$tchests.found,
     y = model.1$fitted.values,
     xlab = "True Amount of Treasure Chests Found",
     ylab = "Model Amount of Chests Found",
     main = "Treasure Chests Found\n True versus Model", 
    pch = 16, 
    col = gray(.05, .15),
    abline(a = 0, b = 1) #add diagonal line
)

  1. Repeat your analysis from question 2, but only include pirates who are female and have owned less than 5 parrots in their lives. Do your conclusions change?
female.parrots <- subset(pirates, subset = sex == "female" & parrots.lifetime < 5)

model.1 <- lm(tchests.found ~ parrots.lifetime + age + tattoos,
                 data = female.parrots)

summary(model.1)
## 
## Call:
## lm(formula = tchests.found ~ parrots.lifetime + age + tattoos, 
##     data = female.parrots)
## 
## Residuals:
##     Min      1Q  Median      3Q     Max 
## -10.426  -4.722  -2.113   2.851  44.026 
## 
## Coefficients:
##                  Estimate Std. Error t value Pr(>|t|)   
## (Intercept)      -2.72969    2.61996  -1.042  0.29821   
## parrots.lifetime -0.18443    0.29642  -0.622  0.53423   
## age               0.25240    0.08075   3.126  0.00193 **
## tattoos           0.28510    0.11336   2.515  0.01237 * 
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## Residual standard error: 7.258 on 338 degrees of freedom
## Multiple R-squared:  0.04689,    Adjusted R-squared:  0.03843 
## F-statistic: 5.542 on 3 and 338 DF,  p-value: 0.001006
# F (3, 338) = 5.542, p < 0.01, R2 = .038.

# for parrots, the results was non-significant (t(338) = -1.042, p = .30)
# for age, the results were significant (t(338) = 3.126, p < .01)
# for tattoos, the results were significant (t(338) = 2.515, p = .012)
  1. Is there a relationship between whether or not a pirate wears a headband and his/her sword speed? Test this using linear regression. What is your conclusion?
model.3 <- lm(sword.speed ~ headband, data = pirates)
summary(model.3)
## 
## Call:
## lm(formula = sword.speed ~ headband, data = pirates)
## 
## Residuals:
##    Min     1Q Median     3Q    Max 
## -1.658 -0.895 -0.576  0.063 43.483 
## 
## Coefficients:
##             Estimate Std. Error t value Pr(>|t|)    
## (Intercept)   1.6576     0.2553   6.494 1.32e-10 ***
## headbandyes  -0.5449     0.2686  -2.029   0.0428 *  
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## Residual standard error: 2.514 on 998 degrees of freedom
## Multiple R-squared:  0.004107,   Adjusted R-squared:  0.003109 
## F-statistic: 4.115 on 1 and 998 DF,  p-value: 0.04276
anova(model.3)
## Analysis of Variance Table
## 
## Response: sword.speed
##            Df Sum Sq Mean Sq F value  Pr(>F)  
## headband    1   26.0 26.0079  4.1152 0.04276 *
## Residuals 998 6307.3  6.3199                  
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
# Wether or not a pirate wears a headband does have a significant influence on the swordspeed. Wearing a headband leads to a greater sword speed.
  1. Now, repeat the analysis from question 4, but this time add sword.type as a second independent variable. What is your conclusion now?
model.4 <- lm(sword.speed ~ headband + sword.type,
              data = pirates)

summary(model.4)
## 
## Call:
## lm(formula = sword.speed ~ headband + sword.type, data = pirates)
## 
## Residuals:
##    Min     1Q Median     3Q    Max 
## -6.304 -0.564 -0.261  0.249 36.805 
## 
## Coefficients:
##                    Estimate Std. Error t value Pr(>|t|)    
## (Intercept)          3.8331     0.3433  11.164  < 2e-16 ***
## headbandyes          3.9581     0.3044  13.003  < 2e-16 ***
## sword.typecutlass   -7.0595     0.3967 -17.796  < 2e-16 ***
## sword.typesabre     -3.3909     0.4250  -7.978 4.06e-15 ***
## sword.typescimitar  -1.5348     0.4317  -3.556 0.000395 ***
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## Residual standard error: 2.063 on 995 degrees of freedom
## Multiple R-squared:  0.3314, Adjusted R-squared:  0.3287 
## F-statistic: 123.3 on 4 and 995 DF,  p-value: < 2.2e-16
anova(model.4)
## Analysis of Variance Table
## 
## Response: sword.speed
##             Df Sum Sq Mean Sq  F value Pr(>F)    
## headband     1   26.0   26.01   6.1116 0.0136 *  
## sword.type   3 2073.1  691.02 162.3837 <2e-16 ***
## Residuals  995 4234.2    4.26                    
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
# Sword type and headband both have a significant influence on the sword speed. But there also is an interaction between sword type and headband.
  1. Is there an interaction between sex and headband use when predicting a pirate’s sword speed? Test this only using pirates whose sex is male or female
model.5 <- lm(sword.speed ~ headband * sex, 
              data = pirates)

summary(model.5)
## 
## Call:
## lm(formula = sword.speed ~ headband * sex, data = pirates)
## 
## Residuals:
##    Min     1Q Median     3Q    Max 
## -1.721 -0.894 -0.582  0.065 43.452 
## 
## Coefficients:
##                      Estimate Std. Error t value Pr(>|t|)    
## (Intercept)           1.72108    0.35613   4.833 1.56e-06 ***
## headbandyes          -0.62761    0.37687  -1.665   0.0962 .  
## sexmale              -0.04571    0.53419  -0.086   0.9318    
## sexother             -0.61876    1.01623  -0.609   0.5427    
## headbandyes:sexmale   0.09614    0.56089   0.171   0.8639    
## headbandyes:sexother  0.45839    1.11105   0.413   0.6800    
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## Residual standard error: 2.518 on 994 degrees of freedom
## Multiple R-squared:  0.004748,   Adjusted R-squared:  -0.0002582 
## F-statistic: 0.9484 on 5 and 994 DF,  p-value: 0.4487
anova(model.5)
## Analysis of Variance Table
## 
## Response: sword.speed
##               Df Sum Sq Mean Sq F value  Pr(>F)  
## headband       1   26.0 26.0079  4.1014 0.04312 *
## sex            2    2.9  1.4719  0.2321 0.79290  
## headband:sex   2    1.1  0.5597  0.0883 0.91553  
## Residuals    994 6303.2  6.3413                  
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
# When predicting a pirate's sword speed, the interaction between sex and headband does not have predictive value (p-values for sexmale:headbandyes and sexother:headbandyes are p > .05)

9.Is there an effect of a pirate’s favorite pirate on the number of tattoos they have? Test this once using an ANOVA (the aov() function) and once using linear regression. How do the two p-values compare?

#using the aov-function
model.6 <- aov(tattoos ~ favorite.pirate, 
               data = pirates) 

summary(model.6)
##                  Df Sum Sq Mean Sq F value Pr(>F)
## favorite.pirate   5     51   10.11   0.919  0.468
## Residuals       994  10939   11.01
#No - there are no significant results meaning that the favorite pirate does not have an effect on the number of tattoos. 


#using the lm-function

model.7 <- lm(tattoos ~ favorite.pirate, 
               data = pirates) 
summary(model.7)
## 
## Call:
## lm(formula = tattoos ~ favorite.pirate, data = pirates)
## 
## Residuals:
##    Min     1Q Median     3Q    Max 
## -9.713 -1.713  0.287  2.380  9.393 
## 
## Coefficients:
##                             Estimate Std. Error t value Pr(>|t|)    
## (Intercept)                  9.10000    0.30283  30.050   <2e-16 ***
## favorite.pirateBlackbeard    0.52000    0.44917   1.158    0.247    
## favorite.pirateEdward Low    0.24211    0.43387   0.558    0.577    
## favorite.pirateHook          0.61304    0.43290   1.416    0.157    
## favorite.pirateJack Sparrow  0.50706    0.34059   1.489    0.137    
## favorite.pirateLewis Scot   -0.01837    0.45167  -0.041    0.968    
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## Residual standard error: 3.317 on 994 degrees of freedom
## Multiple R-squared:  0.004601,   Adjusted R-squared:  -0.000406 
## F-statistic: 0.9189 on 5 and 994 DF,  p-value: 0.4678
#The favorite pirate does not have an effect on the number of tattoos.
  1. Create a scatterplot showing the relationship between a pirate’s age and the number of parrots they’ve had in their life. Include different colored points for pirates who went to Jack Sparrow’s School of Fashion and Piratry and those who went to Captain Chunk’s Cannon Crew. Then, add a regression line for each dataset showing the linear relationship between the two variables for each college.

Note that the echo = FALSE parameter was added to the code chunk to prevent printing of the R code that generated the plot.