A popular myth is that for identical twins, one is the "nice one" and one is the "mean one." Fourteen pairs of identical twins were randomly selected and given a personality test for measuring "niceness" or "meanness" where the higher score implies "niceness" and a lower score implies "meanness." At the $1 \%$ significance level, is there a difference in the personality scores between the identical twins.
First Born Score & 43 & 54 & 49 & 53 & 49 & 52 & 58 & 51 & 59 & 48 & 45 & 49 & 47 & 46
Second Born Score & 52 & 47 & 50 & 46 & 50 & 51 & 50 & 45 & 46 & 47 & 49 & 54 & 50 & 54
Answer (2)
Using a paired t-test, we calculated the mean difference and the standard deviation of the personality scores for 14 pairs of identical twins. The t-statistic was found to be 0.4905, which is less than the critical t-value at the 1% significance level. Therefore, we fail to reject the null hypothesis, concluding no significant difference in personality scores.
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Calculate the differences in personality scores between the twins.
Find the mean and standard deviation of the differences: d ˉ = 0.8571 , s d = 6.5381 .
Compute the t-statistic: t = 0.4905 .
Compare the t-statistic to the critical t-value and conclude: Fail to reject the null hypothesis. No significant difference
Explanation
Problem Analysis We are given the personality scores of 14 pairs of identical twins and want to determine if there is a statistically significant difference in their scores at a 1% significance level. We will perform a paired t-test to compare the means of the two groups.
Calculate Differences First, calculate the difference in scores for each twin pair (First Born - Second Born): Differences: -9, 7, -1, 7, -1, 1, 8, 6, 13, 1, -4, -5, -3, -8
Calculate Mean Difference Next, calculate the mean of these differences: d ˉ = 14 − 9 + 7 − 1 + 7 − 1 + 1 + 8 + 6 + 13 + 1 − 4 − 5 − 3 − 8 = 14 12 = 0.8571
Calculate Standard Deviation Then, calculate the standard deviation of the differences: s d = n − 1 ∑ i = 1 n ( d i − d ˉ ) 2
Standard Deviation Value Using the python calculation tool, the standard deviation of the differences is found to be: s d = 6.5381
Calculate T-Statistic Now, calculate the t-statistic: t = s d / n d ˉ = 6.5381/ 14 0.8571 = 1.7475 0.8571 = 0.4905
Degrees of Freedom Determine the degrees of freedom: df = n − 1 = 14 − 1 = 13
Find Critical T-Value Find the critical t-value for a two-tailed test with a significance level of α = 0.01 and df = 13 . Using the python calculation tool, the critical t-value is found to be: t cr i t i c a l = 3.0123
Compare T-Statistic and Critical T-Value Compare the calculated t-statistic to the critical t-value: ∣ t ∣ = 0.4905 < t cr i t i c a l = 3.0123 Since the absolute value of the t-statistic is less than the critical t-value, we fail to reject the null hypothesis.
Conclusion Conclusion: There is not enough evidence to conclude that there is a significant difference in the personality scores between the identical twins at a 1% significance level.
Examples
In psychological studies, paired t-tests are often used to compare the effects of an intervention on related subjects, such as twins or matched pairs. For example, researchers might want to investigate whether a specific training program has a different impact on the cognitive abilities of identical twins. By comparing the differences in test scores before and after the intervention, they can determine if the program has a statistically significant effect.
