2. A researcher records the number of words recalled by students presented with

Question

2. A researcher records the number of words recalled by students presented with a list of words for one minute. In one group, students were presented with the list of words in color and for the other group, the words were presented in black and white. Test to see if there are differences in the number of words recalled, using alpha -.05 and four steps. Also calculate and interpret effect size and explained variance if appropriate. Words recalled for B/W condition 4 Word recalled in color condition 7 12

Explanation / Answer

This is the r code you need.

x=c(4,5,1,3)

y=c(7,12,8,11)

t.test(x,y)

Explanation:

Let X be the no of words recalled for B/W condition. We assume X~N(µ1,?12 )

Let Y be the no of words recalled for color condition We assume Y~N(µ2,?22)

so here we want to test H0:µ1=µ2 ag H1:µ1=!µ2 .

The test I have written code for , that is Welch Two Sample t-test. Here By default R assumes ?1 =!?2( This is called Behrens-Fisher Problem)

For homoscedastic assumption just add var.equal=TRUE in which we get Fisher’s t-test

So then the modified code will be as follows

x=c(4,5,1,3)

y=c(7,12,8,11)

t.test(x,y var.equal=TRUE)

The resullts of the firs code is

x=c(4,5,1,3)
>
> y=c(7,12,8,11)
>
> t.test(x,y)

Welch Two Sample t-test

data: x and y
t = -4.2666, df = 5.4414, p-value = 0.006589
alternative hypothesis: true difference in means is not equal to 0
95 percent confidence interval:
-9.925505 -2.574495
sample estimates:
mean of x mean of y
3.25 9.50

And the result of the second test is

> x=c(4,5,1,3)
>
> y=c(7,12,8,11)
>
> t.test(x,y, var.equal=TRUE)

Two Sample t-test

data: x and y
t = -4.2666, df = 6, p-value = 0.005283
alternative hypothesis: true difference in means is not equal to 0
95 percent confidence interval:
-9.834399 -2.665601
sample estimates:
mean of x mean of y
3.25 9.50

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