1. **State the problem:** We want to determine if the correlation coefficient $r=0.582$ between tension in partners and tension in children is statistically significant at significance level $\alpha=0.05$ with a two-tailed test. 2. **Given data:** - Sample size $n=15$ - Correlation coefficient $r=0.582$ - Significance level $\alpha=0.05$ (two-tailed) 3. **Formula for testing significance of correlation:** We use the $t$-test for correlation: $$ t = \frac{r \sqrt{n-2}}{\sqrt{1-r^2}} $$ where $t$ follows a $t$-distribution with $df = n-2$ degrees of freedom. 4. **Calculate degrees of freedom:** $$ df = 15 - 2 = 13 $$ 5. **Calculate $t$ statistic:** First calculate $1-r^2$: $$ 1 - r^2 = 1 - (0.582)^2 = 1 - 0.338724 = 0.661276 $$ Calculate numerator: $$ r \sqrt{n-2} = 0.582 \times \sqrt{13} = 0.582 \times 3.605551 = 2.098 $$ Calculate denominator: $$ \sqrt{1-r^2} = \sqrt{0.661276} = 0.8131 $$ Calculate $t$: $$ t = \frac{2.098}{0.8131} = 2.58 $$ 6. **Determine critical $t$ value:** For $df=13$ and two-tailed $\alpha=0.05$, critical $t$ is approximately $\pm 2.160$ (from $t$-distribution tables). 7. **Compare calculated $t$ with critical $t$:** $$ 2.58 > 2.160 $$ Since calculated $t$ is greater than critical $t$, we reject the null hypothesis. 8. **Conclusion:** There is a statistically significant relationship between tension in partners and tension in children at the 0.05 significance level. **Final answer:** The correlation $r=0.582$ is significant with $t=2.58$, $df=13$, $p < 0.05$.