options(digits=5,width=70) # R console display parameters
options(contrasts=c("contr.sum","contr.poly") )
# install.packages("BSDA")
library(BSDA) # use this for z.test()
Answer the questions in Section 2 and submit your answers as a script file on Avenue 2 Learn. Make sure to begin your script file with the following lines:
# PSYCH 710 Lab 2 Homework
# Date: 15-SEP-2022
# Your Name: <<Your name here>>
# Student ID: <<Your ID here>>
# Collaborators: <<Names of your collaborators here>>
Also, make sure that text that is not an R command is preceded by a comment symbol (#). For example, you can insert questions or comments among your commands like this:
# The following command doesn't work... not sure why...
# ttest(x=g1,y=g2) # was trying to do a t test
set.seed(210980) # set random number seed
dscores <- rnorm(n=20,5,10) # get the values
dscores
. Use a boxplot
to display the
distribution of values.c(mean(dscores),var(dscores),sd(dscores))
## [1] 4.5103 103.7138 10.1840
boxplot(dscores,ylab="dscore")
dscores
is distributed normally. Do the data
look approximately normal? Use
shapiro.test
to address this question with a quantitative
test.# the data fall along a straight line and therefore look normal
qqnorm(dscores)
qqline(dscores)
shapiro.test(dscores) # we do not reject the null hypothesis of normality
##
## Shapiro-Wilk normality test
##
## data: dscores
## W = 0.95, p-value = 0.37
dscores
. Use a \(t\) to
evaluate the hypothesis that memory was affected (positively
or negatively) by the drug. State the null and
alternative hypotheses evaluated by your test, and whether or not you
reject the null hypothesis.# H0: mu = 0 H1: mu ≠ 0
t.test(dscores,mu=0,alternative="two.sided")
##
## One Sample t-test
##
## data: dscores
## t = 1.98, df = 19, p-value = 0.062
## alternative hypothesis: true mean is not equal to 0
## 95 percent confidence interval:
## -0.25597 9.27655
## sample estimates:
## mean of x
## 4.5103
Answer: The null hypothesis is that the population mean of difference scores is zero and the alternative is that the mean is not equal to zero. This test is two-sided because we are examining if the drug affected memory positively OR negatively. The test was not significant (\(t=1.98\), \(p=0.06\)) and therefore I do not reject the null hypothesis.
z.test
to repeat the analysis performed in question
3 with a \(z\) test. To use
z.test
, you need to set sigma.x
to a value
that equals the population standard deviation (\(\sigma\)) of dscores
. Although
we do not know \(\sigma\), we
can estimate it with the sample standard deviation,
sigma.x = sd(dscores)
. Perform the \(z\) test this way. How do the results
differ from the \(t\) test in question
4? Why do they differ? Which test do you think is more valid?# H0: mu = 0 H1: mu ≠ 0
z.test(x=dscores,alternative="two.sided",mu=0,sigma.x=sd(dscores),conf.level=0.95)
##
## One-sample z-Test
##
## data: dscores
## z = 1.98, p-value = 0.048
## alternative hypothesis: true mean is not equal to 0
## 95 percent confidence interval:
## 0.047038 8.973539
## sample estimates:
## mean of x
## 4.5103
# Unlike the t test, the z test is significant p<.05 and therefore we reject H0 in favor of H1
# The z and t test differ because the ESTIMATE of pop SD is biased (too small) and therefore
# inflates z and results in more Type I errors. I trust the t test more.
# H0: mu ≤ 0 H1: mu > 0
# I reject the null hypothesis, t(19)=1.98, p=0.031
t.test(dscores,mu=0,alternative="greater")
##
## One Sample t-test
##
## data: dscores
## t = 1.98, df = 19, p-value = 0.031
## alternative hypothesis: true mean is greater than 0
## 95 percent confidence interval:
## 0.57269 Inf
## sample estimates:
## mean of x
## 4.5103