Question

The Mozart Effect pertains to the hypothesis that listening to Mozart might induce a short-term improvement on the performance of certain kinds of mental tasks. A team of researchers were interested in seeing if this would apply to a general intelligence test. They do not know if this will improve or lower scores for this particular task but they collected data from two groups:

No Music	Music
1	2
10	12
7	6
3	5
2	3
0	1
3	2
13	16
0	0
5	8

a. Treating these two groups as independent, answer the following:
* State the hypotheses for this analysis and if this is a one- or two-tailed test.
* State your alpha value and the critical values.
* Test this hypothesis (showing your calculations).
* State your decision regarding the null hypothesis.

b. Now, conduct those four steps again but treat these two groups as dependent (e.g., each row now belongs to the same person in a repeated-measures design).

c. Do the two tests lead to different conclusions? Comment on why or why not.

Answer 1

A) If two groups are considered as independent then, we would apply a 2-sample t-test. The 2-sample t-test takes your sample data from two groups and boils it down to the t-value. It requires independent groups for each sample.

• Null Hypothesis: The default null hypothesis for a 2-sample t-test is that the two groups are equal. In other words, we can construct a hypothesis that the difference between the two groups is zero.

For one-tailed test -       against ( Mozart might induce a short-term improvement on the performance of certain kinds of mental tasks.)

For two-tailed test--   against   ( do not know if this will improve or lower scores)

• Here we would like to choose  the level of significance at 5%  

    For one-tailed test - At this level of significance, the critical value would be 1.734(df=18)

For two-tailed test- At this level of significance, the critical value would be 2.101(df=18)

• Testing of hypothesis:

For one-tailed test-

  

S = pooled variance and given by

  

Here n1=10, n2=10,   AND  

so we get  

hence  

• Since the p-value is greater than 0.05 in both cases. so we failed to reject the null hypothesis based on the given sample and hence conclude that there is no difference found in both groups (Music and Non=Music)

B) To compare the means of the two paired sets of data, the differences(d) between all pairs must be, first, calculated.  

• Null Hypothesis ( mean of the difference is equal to zero).

Alternate Hypothesis      For one-tailed test

For two-tailed test

• Here we would like to choose  the level of significance at 5%  

•     For one-tailed test - At this level of significance, the critical value would be 1.833(df=9) = n-1

  For two-tailed test- At this level of significance, the critical value would be 2.262(df=9)= n-1

• Testing of hypothesis:  m and s are the mean and the standard deviation of the difference (d), respectively. n is the size of d.

• This test is significant as the p-value is less than 0.05 there we have enough evidence to reject the null hypothesis and conclude that there is a significant difference between both samples.

C) Of course, both tests lead to a different conclusion because in an independent sample test the test allows greater variation to both groups as both groups are independent and may have some variation because of random chance.

Whereas in paired t-test, we assumed that both samples were taken from same group so both are dependent. In this case, smaller variation may lead to significant results. Therefore a small difference made a significant t-statistic which in turn lead to rejecting the null hypothesis.