ECON 346, NEIU (Scott W. Hegerty)
A statistical distribution shows the possible outcomes of some event on the x-axis and the probability of each outcome on the y-axis. The Normal distribution is best-known: The likeliest outcomes are those closest to the average. In Economics, the t-distibution is very common; the Chi-Square and F are used often, particularly for tests of joint hypotheses. A large critical value shows that a particular outcome is not “average” or attributable to chance.
Here, we show the t- and F-distributions. Both depend on the degrees of freedom, which are based on the sample size.
We begin by generating some random data. We make 25 observations of each of x and y, each with a different mean and standard deviation. We will conduct formal tests for these differences, which may or may not be significant.
set.seed(81)
x<-rnorm(25,10,5)
y<-rnorm(25,12,6)
mean(x)## [1] 9.316308
mean(y)## [1] 13.93007
sd(x)## [1] 4.447163
sd(y)## [1] 6.188187
The means and standard deviations look different, but because of variation in the data these difference might not be significant.
Now, we conduct a two-sample test for differences in the variables means. We can use t.test() or use the textbook formula:
tt<-t.test(x,y) # store the results
tt # show the results##
## Welch Two Sample t-test
##
## data: x and y
## t = -3.0272, df = 43.57, p-value = 0.004134
## alternative hypothesis: true difference in means is not equal to 0
## 95 percent confidence interval:
## -7.686214 -1.541317
## sample estimates:
## mean of x mean of y
## 9.316308 13.930074
(mean(x)-mean(y))/(sqrt(var(x)/length(x)+var(y)/length(y)))## [1] -3.027236
These give the exact same answer; we reject the null hypothesis of no difference in these two variables’ means.
We can also show where this t-statistic falls on the distribution. We create an x-axis and then use dt() to get the probability at each point along the range. We add critical values (using the correct degrees of freedom) before showing the calculated t-statistic in red.
df=2*length(x)-2
x1<-seq(-4,4,length=1000)
y1<-dt(x1,df)
q5<-qt(0.975,df) # critical value at 5%
plot(x1,y1,type = "l",lwd=4,ylab="Prob.",xlab="t",main="T distribution") # basic curve
abline(v=qt(0.975,df),lty=2,lwd=4,col="dark gray") # +0.025
abline(v=qt(0.025,df),lty=2,lwd=4,col="dark gray") # -0.025
abline(v=tt$statistic,col="red",lwd=2)
We see that it falls “outside” the critical value.
Recall that the F-distribution depends on two values for the degrees of freedom, and the curve is different for different values:
Here, we are using the F-test to see whether the variances of x and y are significantly different. We can make a ratio of the larger to the smaller. Note that here, we try both and take the one that is greater than one.
varrat<-max(var(y)/var(x),var(x)/var(y))
varrat## [1] 1.936247
Now, we find the corresponding F-statistic for this ratio.
df<-length(x)-1
cvf<-qf(.95,df,df)
cvf## [1] 1.98376
Looking at the p-value, we see that this ratio is not significantly large. One variance is not greater than the other.
df(cvf,df,df)## [1] 0.1225267
We can plot an F-distribution like we did for the t-distribution, using n-1 for each degree of freedom:
x2<-seq(0,2,length=1000)
y2<-df(x2,df,df)
plot(x2,y2,type = "l",lwd=4,ylab="Prob.",xlab="F",main="F distribution") # basic curve
abline(v=qf(0.95,df,df),lty=2,lwd=4,col="dark gray") # +0.05
abline(v=varrat,col="red",lwd=2)
The calculated statistic is not larger than the critical value. We cannot reject the null and instead conclude that, even though var(x) = 19.8 and var(y) = 38.3, they are not significantly different.
R also has the command vartest(), but you need to know which variable has the larger variance. If you don’t, a large difference will make a small statistic!
var.test(y,x)##
## F test to compare two variances
##
## data: y and x
## F = 1.9362, num df = 24, denom df = 24, p-value = 0.1124
## alternative hypothesis: true ratio of variances is not equal to 1
## 95 percent confidence interval:
## 0.853244 4.393882
## sample estimates:
## ratio of variances
## 1.936247
Note that the results match here using both methods.