Question

A team of researchers examined the brain systems involved in human romantic love. One issue was whether romantic love images a part of the brain called the caudate (a brain structure know to become active when people win money in an experimental tasks, are given cocaine, and other such “reward” situations). Thus, the researchers recruited individuals who had very recently fallen “madly in love.” (For example, to be in the study participants had to think about the partner at least 80% of their waking hours.) Participants brought a picture of their beloved with them, plus a picture of a neutral person (someone of the same age and sex as their beloved with whom they were very familiar, but about whom they did not have any particularly strong feelings). Participants then went in to the functional magnetic resonance imaging (fMRI) machine and their brain was scanned while they looked at the two pictures – 30 seconds at the neutral persons’ picture, 30 seconds at their beloved, 30 seconds at the neutral person, and so forth. The data are below. Student Beloved’s photo Control photo 1 1487.8 1487.2 2 1329.4 1328.1 3 1407.9 1405.9 4 1236.1 1234.0 5 1299.8 1298.2 6 1447.2 1444.7 7 1354.1 1354.3 8 1204.6 1203.7 9 1322.3 1320.8 10 1388.5 1386.8 What do you make of this data? 1. You should probably state the null and alternative hypotheses; 2. You should probably examine the descriptive statistics (esp. the means); 3. You should probably run a paired-t (because you are comparing means of the same people, a paired t is required); 4. You should present a conclusion;

Answer 1

1.)

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The null hypothesis is defined as there is no difference in the test score of two treatment and the alternative hypothesis is that there is a significant difference in test score of two treatment.

2)

The mean and standard deviation for the difference of scores are obtained as follow,

Student	Beloved’s photo	Control photo	Difference
1	1487.8	1487.2	0.6
2	1329.4	1328.1	1.3
3	1407.9	1405.9	2
4	1236.1	1234	2.1
5	1299.8	1298.2	1.6
6	1447.2	1444.7	2.5
7	1354.1	1354.3	-0.2
8	1204.6	1203.7	0.9
9	1322.3	1320.8	1.5
10	1388.5	1386.8	1.7
Average			1.4
SD			0.7930252
n			10

Average difference Xp 1..4

Std. Dev of difference sn = 0.793

3)

The paired sample t test is performed as follow,

The t statistic is obtained using the formula,

XD 1.4 0,793 5.583 10 t = sp n

The P-value for the t-statistic is obtained from t-distribution table for degree of freedom = n -1 = 9.

P-value 0.0003

4)

P-value 0.0003<a0.05

Since the P-value id less than significance level = 0.05 at 5% significance level. The null hypothesis is rejected. Hence we can conclude that there is significant difference in both treatment score