Question

In a study of people first getting acquainted with each other, researchers examined the relationship between the amount of self-disclosure and how much that person likes their partner. Here are the results:

Person	Partner's Self Disclosure	Liking for Partner
1	8	7
2	7	9
3	10	6
4	3	7
5	1	4

First, state the null and alternative hypothesis.

1. How much of the variance in liking for a partner explained by the amount of self-disclosure?
2. Should a person talk about themselves more or less when getting acquainted?

Answer 1

Here is a scatter diagram of the scores, with the explanatory variable self disclosure on the horizontal axis and the response variable degree of liking on the vertical axis. Partner's Self-Disclosure and Liking for Partner 10 10 12 Self-Disclosure of Partner The relationship appears to be positive and weak, or it is possibly curvilinear. It is difficult to give much credence to the strength of the relationship because of the small number of data points.

We calculate the value of the correlation coefficient, using Excel to perform the basic arithmetic. The basic approach is to convert each data value into a z score using the SD then use the formula for the correlation coefficient, which is, where N is the number of pairs of data. Squared Squared z Score Z Score Score Deviation Score Deviation 0.16 0.25 5.76 1.48 0.36 -0.37 0.16 0.25 0.4 2.4 0.6 0.4 2.6 6.761.60 0.66 0.36 4.84 1.44 17.641.27 784 -0.85 23.04 2.2 4.2 2.8 4.8 10 1.45 13.2 SUM SUM .6 equals 264 5.8equals 10.96 3.31 1.62 N e N e Figure 1

Cross Products 2,2. 0.16 0.54 0.47 0.21 2.32 SUM 2.342 gure 2 he calculated value of ris 0.47 d) e perform the hypothesis test to test the significance of the computed correlation oefficient of .47, starting with the statement of the hypotheses.

The null hypothesis states that the population correlation coefficient ρ (rho) is 0, and the alternative indicates a two-tailed test

We determine the cutoff value, which is found using the table of Cutoff Scores for the t Distribution, with a significance level of.05 and N-2 3 degrees of freedom. The cutoff scores are t 3.182. We will reject the null hypothesis if our calculated test statistic is less than-3.182 or greater than +3.182. We next calculate the ttest statistic. (1-r2) N-2 47 (1-47) 5-2 47 /2597 -0.92 The test statistic is 92