# Statistical Tests

--

# Generate Random Data

`x <- rnorm(1000,1000,20)`

m <- 6

c <- 10

y <- m*x + c

simdf <- data.frame(x = x , y=y)

- This is what the data looks like

`head(simdf)`

## x y

## 1 1005.4237 6042.542

## 2 1009.2766 6065.660

## 3 1001.2252 6017.351

## 4 995.0024 5980.014

## 5 964.0087 5794.052

## 6 979.7846 5888.707

- Sampling few rows from our “Population”

sampSize <- 100

sample1 <- simdf[sample(seq(1 , nrow(simdf) , 1) , sampSize ),]

sample2 <- simdf[sample(seq(1 , nrow(simdf) , 1) , sampSize ),]

head(sample1)## x y

## 440 970.4631 5832.779

## 48 1011.4937 6078.962

## 199 1006.4098 6048.459

## 723 992.0803 5962.482

## 375 999.7168 6008.301

## 705 978.0303 5878.182head(sample2)## x y

## 336 1024.1023 6154.614

## 476 1005.8980 6045.388

## 439 1024.4581 6156.749

## 157 972.8277 5846.966

## 30 993.1578 5968.947

## 233 994.1998 5975.199

# Are these two samples from the same distribution?

• what if we didn’t know already

s1mean <- mean(sample1$y)

s2mean <- mean(sample2$y)

print(s1mean)## [1] 5997.703print(s2mean)## [1] 6009.034

# Is the difference between the means actually statistically significant

• Differences of means will be normally distributed if we repeatedly sample.

• Sampling distribution of difference of means will be normally distributed

• But since we don’t know the distribution parameters

• We assume T-distribution

# Two sample T-test

## [1] “combined sd of the t distribution”## [1] 15.78279## t_stat t_crit degf

## 1 -0.7179117 1.652648 196.4361t.test(sample1$y , sample2$y)##

## Welch Two Sample t-test

##

## data: sample1$y and sample2$y

## t = -0.71791, df = 196.44, p-value = 0.4737

## alternative hypothesis: true difference in means is not equal to 0

## 95 percent confidence interval:

## -42.45612 19.79481

## sample estimates:

## mean of x mean of y

## 5997.703 6009.034data.frame(fstat = sd(sample1$y)²/sd(sample2$y)²

, fcrit = qf(0.95 , df1 = nrow(sample1) , df2 = nrow(sample2)))## fstat fcrit

## 1 1.195933 1.39172var.test(sample1$y , sample2$y)##

## F test to compare two variances

##

## data: sample1$y and sample2$y

## F = 1.1959, num df = 99, denom df = 99, p-value = 0.3749

## alternative hypothesis: true ratio of variances is not equal to 1

## 95 percent confidence interval:

## 0.8046737 1.7774364

## sample estimates:

## ratio of variances

## 1.195933

# Conclusions

• We can see that these two samples have similar means and variances therefore its safe to assume they come from the same distribution/population

summary(aov(sample1$y ~ sample2$y))`## Df Sum Sq Mean Sq F value Pr(>F)

## sample2$y 1 139428 139428 11.35 0.00108 **

## Residuals 98 1203617 12282

## — -

## Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ‘ 1