Furness and Bryant (1996) measured the metabolic rates of eight male and six female breeding northern fulmars and were interested in testing the null hypothesis that there was no difference in metabolic rate between the sexes (Box 3.2 of Quinn and Keough(2002)).

To start this activity, I reset first the R’s memory using the code below:

rm(list=ls())

Step 1

I import the Furness and Bryant (1996) data set that will be used in this activity using the codes below.

setwd("C://Users/April Mae Tabonda//Documents//MS Marine Science//Biostat//PLP//RMDs//PLP_7 T-Test")
getwd()
## [1] "C:/Users/April Mae Tabonda/Documents/MS Marine Science/Biostat/PLP/RMDs/PLP_7 T-Test"
furness <-read.csv('furness.csv', header = T, sep=',')
head(furness)
##            SEX METRATE BODYMASS
## 1 Male          2950.0      875
## 2 Female        1956.1      635
## 3 Male          2308.7      765
## 4 Male          2135.6      780
## 5 Male          1945.6      790
## 6 Female        1490.5      635
tail(furness)
##             SEX METRATE BODYMASS
## 9  Female        1091.0      645
## 10 Male          1195.5      788
## 11 Female         727.7      635
## 12 Male           843.3      855
## 13 Male           525.8      860
## 14 Male           605.7     1005
str(furness)
## 'data.frame':    14 obs. of  3 variables:
##  $ SEX     : Factor w/ 2 levels "Female      ",..: 2 1 2 2 2 1 1 1 1 2 ...
##  $ METRATE : num  2950 1956 2309 2136 1946 ...
##  $ BODYMASS: int  875 635 765 780 790 635 668 640 645 788 ...

Step 2

In this activity, I assessed assumptions of normality and homogeneity of variance for the null hypothesis that the population mean metabolic rate is the same for male and female breeding northern fulmars.

boxplot(METRATE~SEX, furness)

with(furness, rbind(MEAN=tapply(METRATE, SEX, mean),
                    VAR=tapply(METRATE, SEX, var),
                    SD=tapply(METRATE, SEX, sd)))
##      Female       Male        
## MEAN    1285.5167    1563.7750
## VAR   177209.4177  799902.5250
## SD       420.9625     894.3727

In order to visually check normality, I used qqplot by running the codes below.

qqnorm(furness$METRATE)
qqline(furness$METRATE)

We were instructed to used qqplot to produce both boxplot and quantile-quantile to visually checked normality.

Shapiro Wilks

In this activity, we used Shapiro Wilks to test for normality.

shapiro.test(furness$METRATE)
## 
##  Shapiro-Wilk normality test
## 
## data:  furness$METRATE
## W = 0.94392, p-value = 0.4709

Step2b

Levene’s test can be used to answer the following question:

Is the assumption of equal variances valid?

We used the package .

pacman::p_load(car)
leveneTest(furness$METRATE,furness$SEX)
## Levene's Test for Homogeneity of Variance (center = median)
##       Df F value  Pr(>F)  
## group  1  7.4057 0.01856 *
##       12                  
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1

Conclusion While there is no evidence of non-normality (boxplots not grossly asymmetrical), variances are a little unequal (although perhaps not grossly unequal- one of the boxplots is not more than three times smaller than the other). Hence, a separate variances t-test is more appropriate than a pooled variances t-test.

Step 3

Welch’s T-test

We were instructed to perform a separate variances (Welch’s) t-test to test the null hypothesis that the population mean metabolic rate is the same for both male and female breeding northern fulmars.

t.test(METRATE~SEX, furness, var.equal=FALSE)
## 
##  Welch Two Sample t-test
## 
## data:  METRATE by SEX
## t = -0.77317, df = 10.468, p-value = 0.4565
## alternative hypothesis: true difference in means is not equal to 0
## 95 percent confidence interval:
##  -1075.3208   518.8042
## sample estimates:
## mean in group Female       mean in group Male         
##                   1285.517                   1563.775

Conclusion

Do not reject the null hypothesis. Metabolic rate of male breeding northern fulmars was not found differ significantly (t= -0.773, df= 10.468, P= 0.457) from that of females.