Question

The following data represent the results of an independent-measures study comparing two treatment conditions.

Treatment One	Treatment Two
3.3	2.6
5	4.4
3.6	3.3
5.4	6
6.2	2.8
5.4	5.1
4.3	2.6
4.6	2.5

Run the single-factor ANOVA for this data:
F-ratio:   
p-value:  

Now, run the t test on the same data:
t-statistic:   
p-value:   

Answer 1

Using R code we can have the answers.

CODE:

trt1 <- c(3.3,5,3.6,5.4,6.2,5.4,4.3,4.6)
trt2 <- c(2.6,4.4,3.3,6,2.8,5.1,2.6,2.5)
var <- c(trt1,trt2)
trt <- as.factor(c(rep("trt1",8),rep("trt2",8)))

#ANOVA
summary(aov(var~trt))

#t test
t.test(trt1,trt2)

OUTPUT:

> trt1 <- c(3.3,5,3.6,5.4,6.2,5.4,4.3,4.6)
> trt2 <- c(2.6,4.4,3.3,6,2.8,5.1,2.6,2.5)
> var <- c(trt1,trt2)
> trt <- as.factor(c(rep("trt1",8),rep("trt2",8)))
>
> #ANOVA
> summary(aov(var~trt))
Df Sum Sq Mean Sq F value Pr(>F)
trt 1 4.516 4.516 3.29 0.0912 .
Residuals 14 19.214 1.372   
---
Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
>
> #t test
> t.test(trt1,trt2)

   Welch Two Sample t-test

data: trt1 and trt2
t = 1.8139, df = 12.792, p-value = 0.09321
alternative hypothesis: true difference in means is not equal to 0
95 percent confidence interval:
-0.2050275 2.3300275
sample estimates:
mean of x mean of y
4.7250 3.6625

For ANOVA:

F-ratio = 3.2.

P-value = 0.0912.

For t-test:

t-statistic = 1.8139.

P-value = 0.09321.