Comparing two means (Part 2)
M. Drew LaMar
November 4, 2019
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It all starts with experimental design
We will be comparing the means of a numerical variable between two groups.
Definition: In the
paired design , both treatments are applied to every sampled unit. In thetwo-sample design , each treatment group is composed of an independent, random sample of units.
It all starts with experimental design
Paired vs. unpaired
Two-sample design: Estimation
\[ \bar{Y}_{1}\sim N(\mu_{1},\sigma_{\bar{Y}_{1}}^2) \ \mathrm{and} \ \bar{Y}_{2}\sim N(\mu_{2},\sigma_{\bar{Y}_{2}}^2) \\ \bar{Y}_{1} - \bar{Y}_{2}\sim N(\mu_{1}-\mu_{2}, \sigma_{\bar{Y}_{1}}^2+\sigma_{\bar{Y}_{2}}^2) \]
Definition: The
standard error of the difference of the means between two groups is given by \[ \mathrm{SE}_{\bar{Y_{1}}-\bar{Y_{2}}} = \sqrt{s_{p}^2\left(\frac{1}{n_{1}} + \frac{1}{n_{2}}\right)} \] wherepooled sample variance \( s_{p}^{2} \) is given by
\[ s_{p}^2 = \frac{df_{1}s_{1}^2 + df_{2}s_{2}^2}{df_{1}+df_{2}}. \]
Two-sample design: Estimation
Since sampling distribution of \( \bar{Y}_{1} - \bar{Y}_{2} \) is normal
\[ \bar{Y}_{1} - \bar{Y}_{2}\sim N(\mu_{1}-\mu_{2}, \sigma_{\bar{Y}_{1}}^2+\sigma_{\bar{Y}_{2}}^2) \]
the sampling distribution of the statistic
\[ t = \frac{\left(\bar{Y}_{1} - \bar{Y}_{2}\right) - \left(\mu_{1}-\mu_{2}\right)}{\mathrm{SE}_{\bar{Y}_{1} - \bar{Y}_{2}}} \]
has a Student's \( t \)-distribution with total degrees of freedom given by
\[ df = df_{1} + df_{2} = n_{1} + n_{2} - 2. \]
Two-sample design: Estimation
Confidence intervals
Two-sample design: Estimation Example
Practice Problem #16
A study in West Africa (Lefèvre et al. 2010), working with the mosquito species that carry malaria, wondered whether drinking the local beer influenced attractiveness to mosquitoes. They opened a container holding 50 mosquitoes next to each of 25 alcohol-free participants and measured the proportion of mosquitoes that left the container and flew toward the participants. They repeated this procedure 15 minutes after each of the same participants had consumed a liter of beer, measuring the change in proportion (treatment group).
Two-sample design: Estimation Example
Practice Problem #16
(cont'd) This procedure was also carried out on another 18 human participants who were given water instead of beer (control group).
Two-sample design: Estimation Example
mydata <- read.csv("http://whitlockschluter.zoology.ubc.ca/wp-content/data/chapter12/chap12q16BeerAndMosquitoes.csv")
str(mydata)
'data.frame': 43 obs. of 4 variables:
$ drink : Factor w/ 2 levels "beer","water": 1 1 1 1 1 1 1 1 1 1 ...
$ beforeDrink: num 0.13 0.13 0.21 0.25 0.25 0.32 0.43 0.44 0.46 0.5 ...
$ afterDrink : num 0.49 0.59 0.27 0.43 0.5 0.5 0.37 0.3 0.58 0.89 ...
$ change : num 0.36 0.46 0.06 0.18 0.25 0.18 -0.06 -0.14 0.12 0.39 ...
Two-sample design: Estimation Example
Discuss: What type of plot would be good here?
Two-sample design: Estimation Example
The long way, first computing the means, and then sample sizes:
(meanChange <- tapply(mydata$change,
mydata$drink,
mean))
beer water
0.154400000 0.007777778
(n <- tapply(mydata$change,
mydata$drink,
length))
beer water
25 18
Two-sample design: Estimation Example
Now, degree of freedoms and variances:
dfs <- n-1
(vari <- tapply(mydata$change, mydata$drink, var))
beer water
0.02632567 0.01611242
Pooled sample variance, which is actually a weighted mean (weighted by degrees of freedom):
(sp <- weighted.mean(vari, dfs))
[1] 0.02209091
Two-sample design: Estimation Example
Finally, find 95% confidence interval:
sderr <- sqrt(sp*sum(1/n)) # Standard error
df <- sum(n) - 2 # Degree of freedom
tcrit <- qt(0.025, df=df, lower.tail=FALSE)
diffMeans <- meanChange["beer"] - meanChange["water"]
lower <- diffMeans - tcrit*sderr
upper <- diffMeans + tcrit*sderr
(CI <- unname(c(lower, upper)))
[1] 0.05383517 0.23940928
Two-sample design: Estimation Example
Using dplyr:
mydata %>%
group_by(drink) %>%
summarize(mu = mean(change),
n = n(),
df = n-1,
vari = var(change)) %>%
summarize(sp = weighted.mean(vari, df),
sderr = sqrt(sp*sum(1/n)),
df = sum(df),
tcrit = qt(0.025, df=df, lower.tail=FALSE),
diffMeans = first(mu)-last(mu),
lower = diffMeans - tcrit*sderr,
upper = diffMeans + tcrit*sderr)
# A tibble: 1 x 7
sp sderr df tcrit diffMeans lower upper
<dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl>
1 0.0221 0.0459 41 2.02 0.147 0.0538 0.239
Two-sample design: Estimation Example
Now the fast way using t.test:
t.test(change ~ drink,
data=mydata,
mu=0,
var.equal=TRUE)$conf.int
[1] 0.05383517 0.23940928
attr(,"conf.level")
[1] 0.95
Two-sample design: Hypothesis testing
Two-sample design: Hypothesis testing
Two-sample design: Testing Example
Practice Problem #16
Two-sample \( t \)-test
\[ H_{0}: \mu_{\mathrm{beer}} - \mu_{\mathrm{water}} = 0 \] \[ H_{A}: \mu_{\mathrm{beer}} - \mu_{\mathrm{water}} \neq 0 \]
\[ H_{0}: \mu_{\mathrm{beer}} = \mu_{\mathrm{water}} \] \[ H_{A}: \mu_{\mathrm{beer}} \neq \mu_{\mathrm{water}} \]
Two-sample design: Testing Example
Starting with, as usual, the long way, we need to calculate the test statistic:
\[ t = \frac{\left(\bar{Y}_{1} - \bar{Y}_{2}\right)}{SE_{\bar{Y}_{1} - \bar{Y}_{2}}} \]
(tstat <- unname(meanChange["beer"] - meanChange["water"])/sderr)
[1] 3.191281
tstat > tcrit
[1] TRUE
Two-sample design: Testing Example
Let's calculate \( P \)-value as well:
(pval <- 2*pt(abs(tstat),
df=df,
lower.tail=FALSE))
[1] 0.00271747
Two-sample design: Testing Example
Short way, again using t.test (note the var.equal=TRUE):
t.test(change ~ drink,
mu=0,
var.equal=TRUE,
data=mydata)
Two-sample design: Testing Example
Short way, again using t.test (note the var.equal=TRUE):
Two Sample t-test
data: change by drink
t = 3.1913, df = 41, p-value = 0.002717
alternative hypothesis: true difference in means is not equal to 0
95 percent confidence interval:
0.05383517 0.23940928
sample estimates:
mean in group beer mean in group water
0.154400000 0.007777778
Two-sample design: Assumptions
Two-sample design: Testing Example
- Moderate sample sizes (\( n_{1}, n_{2} > 30 \))
- Balanced: \( n_{1} \approx n_{2} \)
- \( |s_{1} - s_{2}| \leq 3 \)
n
beer water
25 18
sqrt(vari)
beer water
0.1622519 0.1269347
Two-sample design: Testing Example
What to do if can't meet assumptions of two-sample \( t \)-test?
Definition:
Welch’s t-test compares the mean of two groups and can be used even when the variances of the two groups are not equal.
Standard error and degrees of freedom are calculated differently than two-sample \( t \)-test, but otherwise the same (i.e. uses a \( t \)-distribution).
Two-sample design: Welch's t-test
Same as two-sample in R, except var.equal=FALSE (default).
t.test(change ~ drink,
mu=0,
var.equal=FALSE,
data=mydata)
Two-sample design: Welch's t-test
Same as two-sample in R, except var.equal=FALSE (default).
Welch Two Sample t-test
data: change by drink
t = 3.3219, df = 40.663, p-value = 0.001897
alternative hypothesis: true difference in means is not equal to 0
95 percent confidence interval:
0.05746134 0.23578311
sample estimates:
mean in group beer mean in group water
0.154400000 0.007777778