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) =

There is no significant difference between the averages.

Step-by-step explanation:

Let's call

the mean of the “sentences” group

the standard deviation of the “sentences” group

the mean of the “intentional” group

the standard deviation of the “intentional” group

Then, we can calculate by using the computer

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

So, the standard error of the difference is

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 which is a value of t such that the area under the Student's t distribution  outside the interval is less than 0.05.

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

Now we can compute our confidence interval

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.

10/60

step-by-step explanation:

if you divide 10 by 60 the answer is 0.166666 while if you divide or solve the other problems it equals 0.6