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Question

A researcher investigating fetal alcohol spectrum disorder (FASD) enrolled 7 children into a choline supplement trial. The children completed the O’Connor Finger Dexterity Test before and after a 90-day period in which their diet was supplemented with choline. Lower scores indicate better dexterity. Analyze the scores by calculating an effect size measure and test whether dexterity changed using α = .05. Include a 95% confidence interval and a conclusion.

Before After

11.2 9.6

12.3 10.9

9.9 8.4

10.9 9.8

9.9 9.4

11.5 10.6

10.7 9.3

Answer 1

High-level steps:

1. Calculate the difference scores: Subtract the "After" scores from the "Before" scores for each child.
2. Calculate the mean and standard deviation of the difference scores.
3. Perform a paired t-test: This test will determine if the mean difference is significantly different from zero.
4. Calculate the effect size (Cohen's d): This will measure the size of the effect of the choline supplement.
5. Calculate the 95% confidence interval for the mean difference.
6. Draw a conclusion based on the p-value and the confidence interval.

Step 1: Calculate the difference scores

Difference = Before − After (third column is the Difference column)

11.2	9.6	1.6
12.3	10.9	1.4
9.9	8.4	1.5
10.9	9.8	1.1
9.9	9.4	0.5
11.5	10.6	0.9
10.7	9.3	1.4

Step 2: Calculate the mean and standard deviation of the difference scores

1. Take the sum of the Difference column and divide by 7, which gets you a value of 1.2.
2. The standard deviation of the differences is approximately 0.39.

Step 3: Perform a paired t-test

1. t-statistic = mean difference / (standard deviation of the differences / sqrt(number of pairs))
2. t-statistic = 1.2 / (0.39 / sqrt(7)) = 8.14 (approximately)

Step 4: Calculate effect size (Cohen's d)

1. We use Cohen's d (over Pearson's r) because it is designed for comparing two groups
2. Cohen's d = mean difference / standard deviation of the differences = 1.2 / 0.39 = 3.08 (approximately)

Step 5: Calculate the 95% confidence interval for the mean difference

1. CI = mean difference +- critical value of t-distribution * (standard deviation of the differences / sqrt(number of pairs)
2. CI = 1.2 +- 2.4469 * (0.39 / sqrt(7)) = 1.2 +- 0.361 (approximately)
3. 2.4469 is obtained via a critical t-value calculator using 6 degrees of freedom (number of pairs minus 1) and a significance level of 95%, two-tailed test (we use two-tailed because we're determining if dexterity simply changed)
4. 95% Confidence Interval: (0.839, 1.561)

Step 6: Draw the conclusion

1. Since our t-statistic of 8.14 is greater than the critical t-value of 2.4469 (p-value less than alpha of 0.05), we reject the null hypothesis that there is no significant difference in dexterity before and after the choline supplementation.
2. The p-value associated with a t statistic of 8.14 is approximately 0.0002, which can be found using a statistics calculator or t-distribution table.
3. The effect size (Cohen's d) of 3.08 indicates a very large effect of the choline supplementation on dexterity.
4. The 95% confidence interval for the mean difference (0.839, 1.561) suggests that we can be 95% confident that the true mean improvement in dexterity scores lies within this range.
5. In summary, the choline supplementation significantly improved the dexterity of the children in the trial, as evidenced by the lower scores on the O’Connor Finger Dexterity Test after the supplementation period.

Hope this was helpful.

Answer 2

Hey Jolene! Here is a step by step solution below. Do let me know if you have any questions or need further clarifications.

Summary of the method: To analyze the scores from the O’Connor Finger Dexterity Test before and after the choline supplement trial, we'll calculate the effect size measure and perform a paired t-test to determine if there is a significant change in dexterity using an alpha level of 0.05. We will also calculate a 95% confidence interval for the mean difference and draw a conclusion based on the results.

Step-by-Step Calculation

To analyze the scores from the O’Connor Finger Dexterity Test before and after the choline supplement trial, we will calculate the effect size measure and perform a paired t-test to determine if there is a significant change in dexterity using an alpha level of 0.05. We will also calculate a 95% confidence interval for the mean difference and draw a conclusion based on the results.

1. Calculate the differences:

Before - After

11.2 - 9.6 = 1.6

12.3 - 10.9 = 1.4

9.9 - 8.4 = 1.5

10.9 - 9.8 = 1.1

9.9 - 9.4 = 0.5

11.5 - 10.6 = 0.9

10.7 - 9.3 = 1.4

2. Differences (d):

1.6, 1.4, 1.5, 1.1, 0.5, 0.9, 1.4

3. Calculate the mean of the differences (d-bar):

Mean difference = (1.6 + 1.4 + 1.5 + 1.1 + 0.5 + 0.9 + 1.4) / 7

Mean difference = 8.4 / 7 = 1.2

4. Calculate the standard deviation of the differences:

First, find each squared difference from the mean:

(1.6 - 1.2)^2 = 0.16

(1.4 - 1.2)^2 = 0.04

(1.5 - 1.2)^2 = 0.09

(1.1 - 1.2)^2 = 0.01

(0.5 - 1.2)^2 = 0.49

(0.9 - 1.2)^2 = 0.09

(1.4 - 1.2)^2 = 0.04

Sum of squared differences = 0.16 + 0.04 + 0.09 + 0.01 + 0.49 + 0.09 + 0.04 = 0.92

Variance = 0.92 / (7 - 1) = 0.92 / 6 = 0.1533

Standard deviation = sqrt(0.1533) = 0.3915

5. Calculate the standard error of the mean difference:

Standard error = standard deviation / sqrt(n)

Standard error = 0.3915 / sqrt(7) = 0.3915 / 2.6458 = 0.148

6. Calculate the t-statistic:

t = mean difference / standard error

t = 1.2 / 0.148 = 8.108

7. Degrees of freedom:

df = n - 1 = 7 - 1 = 6

8. Find the critical t-value for alpha = 0.05 (two-tailed) and df = 6:

Using a t-table or calculator, the critical t-value is approximately 2.447.

9. Compare the t-statistic with the critical t-value:

8.108 > 2.447, so the result is significant.

10. Calculate the 95% confidence interval for the mean difference:

CI = mean difference ± (critical t-value * standard error)

CI = 1.2 ± (2.447 * 0.148)

CI = 1.2 ± 0.362

So, the 95% confidence interval is (0.838, 1.562).

Conclusion

Since the t-statistic (8.108) is greater than the critical t-value (2.447) and the 95% confidence interval (0.838, 1.562) does not include zero, we reject the null hypothesis. This indicates that there is a significant improvement in dexterity after the 90-day choline supplement trial. The mean difference of 1.2 with a confidence interval that does not span zero supports this conclusion. The children showed significantly better dexterity (lower scores) after the trial.