Question

Below are data showing the results of six subjects on a memory test. The three scores per subject are their scores on three trials \((\mathrm{a}, \mathrm{b},\) and \(\mathrm{c})\) of a memory task. Are the subjects get- ting better each trial? Test the linear effect of trial for the data. $$ \begin{array}{|c|c|c|} \hline a & b & c \\ \hline 4 & 6 & 7 \\ \hline 3 & 7 & 8 \\ \hline 2 & 8 & 5 \\ \hline 1 & 4 & 7 \\ \hline 4 & 6 & 9 \\ \hline 2 & 4 & 2 \\ \hline \end{array} $$ a. Compute \(L\) for each subject using the contrast weights \(-1,0,\) and \(1 .\) That is, compute \((-1)(a)+(0)(b)+(1)(c)\) for each subject. b. Compute a one-sample t-test on this column (with the \(L\) values for each subject) you created.

Short answer

There is a significant linear effect; subjects are getting better each trial.

Step-by-step solution

Calculate L for Each Subject

The contrast weights given are \(-1, 0, 1\), so for each subject, we compute \(L = (-1) \cdot a + (0) \cdot b + (1) \cdot c\). Let's calculate \(L\) for each subject:- For the first subject: \(L = (-1) \cdot 4 + (0) \cdot 6 + (1) \cdot 7 = 3\)- For the second subject: \(L = (-1) \cdot 3 + (0) \cdot 7 + (1) \cdot 8 = 5\)- For the third subject: \(L = (-1) \cdot 2 + (0) \cdot 8 + (1) \cdot 5 = 3\)- For the fourth subject: \(L = (-1) \cdot 1 + (0) \cdot 4 + (1) \cdot 7 = 6\)- For the fifth subject: \(L = (-1) \cdot 4 + (0) \cdot 6 + (1) \cdot 9 = 5\)- For the sixth subject: \(L = (-1) \cdot 2 + (0) \cdot 4 + (1) \cdot 2 = 0\)

List L Values

Now we have the \(L\) values for each subject:- Subject 1: 3- Subject 2: 5- Subject 3: 3- Subject 4: 6- Subject 5: 5- Subject 6: 0These values will be used to perform a one-sample t-test.

Calculate Mean of L Values

Calculate the mean of the \(L\) values obtained:\[ \text{Mean } (\bar{L}) = \frac{3 + 5 + 3 + 6 + 5 + 0}{6} = \frac{22}{6} = 3.67 \]

Calculate the Standard Deviation of L Values

Calculate the standard deviation of the \(L\) values:1. Find the deviations from the mean: \((3 - 3.67)^2, (5 - 3.67)^2, (3 - 3.67)^2, (6 - 3.67)^2, (5 - 3.67)^2, (0 - 3.67)^2\)2. Compute the variance: \[ s^2 = \frac{1}{5}( (3 - 3.67)^2 + (5 - 3.67)^2 + (3 - 3.67)^2 + (6 - 3.67)^2 + (5 - 3.67)^2 + (0 - 3.67)^2 ) \] \[ s^2 = \frac{1}{5}(0.44 + 1.76 + 0.44 + 5.44 + 1.76 + 13.44) = \frac{23.28}{5} = 4.656 \]3. Compute the standard deviation: \[ s = \sqrt{4.656} \approx 2.16 \]

Perform One-Sample t-test

Using the calculated mean and standard deviation, perform a one-sample t-test:\[ t = \frac{\bar{L} - 0}{s / \sqrt{n}} = \frac{3.67 - 0}{2.16 / \sqrt{6}} = \frac{3.67}{0.882} \approx 4.16 \]

Compare t-value to Critical Value

Compare the calculated t-value to the critical t-value from a t-distribution table (at a specific significance level, e.g., \(\alpha = 0.05\) and 5 degrees of freedom). Suppose the critical t-value is around 2.015 for a two-tailed test at \(\alpha = 0.05\).Since \(t = 4.16\) is greater than 2.015, we reject the null hypothesis.

Key concepts

One-Sample T-Test

The one-sample t-test is a statistical method used to determine if the mean of a single sample of data is significantly different from a known value, typically 0. This test is particularly helpful when comparing a sample mean to a known population mean or a hypothetical value.
To perform a one-sample t-test, you need the following key elements:

• The mean of the sample data (\( \bar{L} = 3.67 \))
• The standard deviation of the sample (\( s \approx 2.16 \))
• The sample size (\( n = 6 \))

Once these are known, the test statistic is calculated using the formula: \[ t = \frac{\bar{L} - \mu}{s / \sqrt{n}} \] where \( \mu \) is the hypothesized population mean (in this case, 0). The calculated t-value is then compared to a critical value from the t-distribution table — which depends on the significance level (\( \alpha \)) and the degrees of freedom. If the t-value exceeds the critical value, you can conclude that there is a statistically significant difference, as seen in our example with \( t = 4.16 \) being greater than the critical value of 2.015.

Contrast Weights

Contrast weights are numerical coefficients used to test specific hypotheses about means in a dataset, often in the context of a linear model or an analysis of variance (ANOVA). They allow you to focus on specific patterns or trends in the data, such as increasing or decreasing trends across trials.
In our exercise, the contrast weights \(-1, 0, 1\) are employed to evaluate a linear trend across three trials (\(a, b, c\)). This particular set of weights stresses the change between the first and the third trial, ignoring the middle trial. The calculation for a subject involves multiplying each score by its corresponding weight and summing the results.

• First weight (\(-1\)): applied to trial \(a\)
• Second weight (\(0\)): applied to trial \(b\)
• Third weight (\(1\)): applied to trial \(c\)

These weights help identify subjects who show improvement over the trials. Positive L values indicate improvement, while negative or zero values suggest no improvement. This contrast analysis simplifies complex data sets by highlighting specified linear relationships.

Memory Test Analysis

Memory test analysis involves evaluating how subjects perform on tasks designed to assess varying aspects of memory over time or under different conditions. In our case, the memory test consists of repeated trials to see if subjects improve with practice. Each subject's performance across three trials (\(a, b, c\)) is analyzed to determine patterns in learning or memory retention.
The use of contrast weights provides a structured mechanism to isolate performance trends. By focusing on the scores from trials with these weights, researchers can discern whether there's a positive trend, suggesting that practice leads to improvement. The results from the one-sample t-test can then statistically confirm whether this observed improvement is significant. Memory test analyses are crucial in educational and psychological settings as they help understand learning curves and the impact of interventions. By interpreting test results through statistical analysis, educators and researchers can draw meaningful conclusions regarding the efficacy of teaching methods or therapeutic strategies.