Question

A (hypothetical) experiment is conducted on the effect of alcohol on perceptual motor ability. Ten subjects are each tested twice, once after having two drinks and once after having two glasses of water. The two tests were on two different days to give the alcohol a chance to wear off. Half of the subjects were given alcohol first and half were given water first. The scores of the 10 subjects are shown below. The first number for each subject is their per- formance in the "water" condition. Higher scores reflect better performance. Test to see if alcohol had a significant effect. Report the \(\mathrm{t}\) and \(\mathrm{p}\) values. $$ \begin{array}{|c|c|} \hline \text { water } & \text { alcohol } \\ \hline 16 & 13 \\ \hline 15 & 13 \\ \hline 11 & 10 \\ \hline 20 & 18 \\ \hline 19 & 17 \\ \hline 14 & 11 \\ \hline 13 & 10 \\ \hline 15 & 15 \\ \hline 14 & 11 \\ \hline 16 & 16 \\ \hline \end{array} $$

Short answer

The alcohol condition significantly affected performance; \( t \approx 4.67 \), \( p < 0.05 \).

Step-by-step solution

Calculate the Difference Scores

Determine the difference between the 'water' scores and the 'alcohol' scores for each subject. This will help in identifying if there is a consistent change in performance between the two conditions. The differences are: \( 3, 2, 1, 2, 2, 3, 3, 0, 3, 0 \).

Calculate the Mean of the Differences

Compute the average (mean) of the difference scores calculated in Step 1. The formula for the mean difference is \( \bar{d} = \frac{\sum{d}}{n} \), where \( d \) is each difference and \( n \) is the number of subjects. The mean difference is \( \bar{d} = \frac{3 + 2 + 1 + 2 + 2 + 3 + 3 + 0 + 3 + 0}{10} = 1.9 \).

Calculate the Standard Deviation of the Differences

Find the standard deviation of the difference scores. Use the formula \( s_d = \sqrt{\frac{\sum{(d_i - \bar{d})^2}}{n-1}} \), where \( d_i \) are the individual differences. The calculation yields approximately \( s_d = 1.287 \).

Compute the t-value

Use the t-test formula for paired samples: \( t = \frac{\bar{d}}{s_d / \sqrt{n}} \). Substitute the known values to get \( t = \frac{1.9}{1.287 / \sqrt{10}} \approx 4.67 \).

Determine the Degrees of Freedom and p-value

The degrees of freedom (df) is the number of pairs minus one, \( n - 1 = 9 \). Using a t-distribution table or software with \( df = 9 \), check the p-value for \( t = 4.67 \). This value is less than 0.05, indicating statistical significance.

Key concepts

Alcohol Effect on Performance

In this experiment, researchers investigated how alcohol affects perceptual motor ability, comparing performances after consuming alcohol versus water. Each of the ten subjects participated in tests on two different days: one day after drinking two servings of alcohol, and another day having two glasses of water. This was done to ensure the alcohol's effects had time to dissipate between tests. Repetition of the test with different conditions and the use of a counterbalanced design, where half of the subjects started with alcohol and the other half with water, helped in reducing the biases related to practice or fatigue. The primary goal was to examine whether the alcohol condition significantly impaired performance, as indicated by the changes in perceptual motor scores.

Mean Difference

Calculating the mean difference between the two conditions is a crucial step. In this context, the difference refers to the change in scores from the water condition to the alcohol condition for each subject. By determining these differences, researchers can evaluate whether a consistent pattern emerges across the dataset. To find the mean difference, add up all the difference scores and divide by the total number of subjects. The formula is: \[ \bar{d} = \frac{3 + 2 + 1 + 2 + 2 + 3 + 3 + 0 + 3 + 0}{10} = 1.9 \]A positive mean difference, such as 1.9, indicates that scores generally decreased after alcohol consumption compared to after drinking water. This difference provides a first measure of the alcohol's possible impact on performance.

Standard Deviation Calculation

Understanding the variability of difference scores is essential, and this is captured by the standard deviation. A standard deviation sheds light on how spread out the individual scores are from the mean difference. It offers insight into the consistency of the effect across subjects.To calculate the standard deviation of the differences, we use the formula:\[ s_d = \sqrt{\frac{\Sigma(d_i - \bar{d})^2}{n-1}} \]Here, \( d_i \) represents each individual difference, \( \bar{d} \) is the mean difference, and \( n \) is the number of subjects. Substituting the given values, we find that the standard deviation is approximately 1.287. This relatively small standard deviation suggests that the effect of alcohol is consistent across most participants.

Significant p-value

Statistical significance in the context of this experiment is assessed using the p-value, which is derived from the t-test. A t-test for paired samples determines if there is a statistically significant difference between the two sets of scores (water vs. alcohol conditions).The t-value calculated is 4.67, which must be compared against a critical value in the t-distribution table. Degrees of freedom for this test are \( n - 1 = 9 \). With a t-value of 4.67 and 9 degrees of freedom, the p-value is less than 0.05, indicating that the observed difference is not likely to have occurred by chance.A p-value less than 0.05 typically indicates significant results. This means we can confidently state that alcohol indeed has a statistically significant negative effect on perceptual motor ability.