Question

An experiment on memory was performed, in which 16 subjects were randomly assigned to one of two groups, called "Sentences" or "Intentional". Each subject was given a list of 50 words. Subjects in the "Sentences" group were told to form multiple sentences, each using at least two words from the list, and to keep forming sentences until all the words were used at least once. Subjects in the "Intentional" group were told to spend five minutes memorizing as many of the 50 words as possible. Subjects from both groups were then asked to write down as many words from their lists as they could recall. We are interested in determining if there is a significant difference in the average number of words recalled for subjects in the "sentences" group vs. subjects in the "intentional" group, using α = 0.05. The data is in the table below.

Number of words recalled

"Sentences" group 37 26 29 27 26 29 28 28

"Intentional" group 32 30 34 31 32 30 33 31

A. Enter the values for the following statistics:

xsentences =

ssentences =

xintentional =

sintentional =

(xsentences - xintentional) =

standard error of (xsentences - xintentional) =

Answer 1

There is no significant difference between the averages.

Explanation:

Let's call

`\large X_(sentences)` the mean of the “sentences” group

`\large S_(sentences)` the standard deviation of the “sentences” group

`\large X_(intentional)` the mean of the “intentional” group

`\large S_(intentional)` the standard deviation of the “intentional” group

Then, we can calculate by using the computer

`\large X_(sentences)=28.75`

`\large S_(sentences)=3.53553`

`\large X_(intentional)=31.625`

`\large S_(intentional)=1.40788`

`\large X_(sentences)-X_(intentional)=28.75-31.625=-2.875`

The standard error of the difference (of the means) for a sample of size 8 is calculated with the formula

`\large \sqrt{(S_(sentences))^2/8+(S_(intentional))^2/8}`

So, the standard error of the difference is

`\large √((3.53553)^2/8+(1.40788)^2/8)=1.34546`

In order to see if there is a significant difference in the averages of the two groups, we compute the interval of confidence of 95% for the difference of the means corresponding to a level of significance of 0.05 (5%).

If this interval contains the zero, we can say there is no significant difference.

Since the sample size is small, we had better use the Student's t-distribution with 7 degrees of freedom (sample size-1), which is an approximation to the normal distribution N(0;1) for small samples.

We get the
`\large t_(0.05)` which is a value of t such that the area under the Student's t distribution outside the interval
`\large [-t_(0.05), +t_(0.05)]` is less than 0.05.

That value can be obtained either by using a table or the computer and is found to be

`\large t_(0.05)=2.365`

Now we can compute our confidence interval

`\large (X_(sentences)-X_(intentional)) \pm t_(0.05)*(standard \;error)=-2.875\pm 2.365*1.34546`

and the confidence interval is

[-6.057, 0.307]

Since the interval does contain the zero, we can say there is no significant difference in these samples.