Question

A physician measured the relationship between labor pains and infant bonding. The following table lists the ranks for each mother based on the amount of pain she experienced during labor (from lowest to highest) and the amount of time she spent bonding with her infant following birth (from least to most).

Labor Pains	Infant Bonding
X	Y
8	1
7	3
2	7
4	5
3	4
6	2
5	6
1	8

1. State/write out the null hypothesis using appropriate IV and DV terms.

2. Is this a Pearson r or Spearman Rho correlation?

3. Compute and report SPSS results that includes the means and SD of the appropriate measure(s)

4. Report the linear correlation between labor pains and infant bonding.

5. Report the CV at alpha .05 level

6. Report dfs

7. Report the actual p (alpha) level reported by SPSS for the correlation coefficient.

8. Using a two-tailed test at the .05 level of significance state the decision to retain or reject the null hypothesis.

9. Explain all conclusions.

Answer 1

A physician measured the relationship between labor pains and infant bonding. The following table lists the ranks for each mother based on the amount of pain she experienced during labor (from lowest to highest) and the amount of time she spent bonding with her infant following birth (from least to most).

1. State/write out the null hypothesis using appropriate IV and DV terms.

Here we have to test the hypothesis that,

H0 : There is no relationship between X and Y.

H1 : There is some relationship between X and Y.

Assume alpah = level of significance = 0.05

2. Is this a Pearson r or Spearman Rho correlation?

Yes this is Pearson r or Spearman rho correlation.

3. Compute and report SPSS results that includes the means and SD of the appropriate measure(s)

Descriptive Statistics: X, Y

Variable Mean StDev
X 4.500 2.449
Y 4.500 2.449

4. Report the linear correlation between labor pains and infant bonding.

Pearson correlation of X and Y = -0.881

There is negative relationship between X and Y.

5. Report the CV at alpha .05 level

CVx = Cvy = $\frac{sd}{mean}*100$

CVx = CVy = $\frac{2.449}{4.500} = 0.5442$

6. Report dfs

dfs = n-2 = 8-2 = 6

7. Report the actual p (alpha) level reported by SPSS for the correlation coefficient.

P-Value = 0.004

8. Using a two-tailed test at the .05 level of significance state the decision to retain or reject the null hypothesis.

P-value < alpha

Reject H0 at 5% level of significance.

Conclusion : There is some relationship between X and Y.

9. Explain all conclusions.

There is negative relationship between X and Y.

We get significant result about test of correlation.