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A random sample of nine students was selected to test for the effectiveness of a special course designed to improve memory. The following table gives the scores in a memory test given to these students before and after this course. $$ \begin{array}{l|lllllllll} \hline \text { Before } & 43 & 57 & 48 & 65 & 81 & 49 & 38 & 69 & 58 \\ \hline \text { After } & 49 & 56 & 55 & 77 & 89 & 57 & 36 & 64 & 69 \\ \hline \end{array} $$ a. Construct a \(95 \%\) confidence interval for the mean \(\mu_{d}\) of the population paired differences, where a paired difference is defined as the difference between the memory test scores of a student before and after attending this course. b. Test at the \(1 \%\) significance level whether this course makes any statistically significant improvement in the memory of all students.

Step by step solution

01

Compute the differences

For each student, find the difference between the memory score after and before the course. These differences can be computed as follows: \n49-43 = 6, 56-57 = -1, 55-48 = 7, 77-65 = 12, 89-81 = 8, 57-49 = 8, 36-38 = -2, 64-69 = -5, 69-58 = 11

02

Calculate the mean (d_bar) and standard deviation (s_d) of differences

Compute the mean and standard deviation of these differences. Mean can be calculated by adding all the differences and dividing by the number of differences. The standard deviation is computed by finding the square root of the variance. Variance is the average squared deviation from the mean.

03

Formulate and calculate 95% confidence interval

The 95% confidence interval can be computed using the formula: \((d-bar - t* s_d/\sqrt{N}, d-bar + t* s_d/\sqrt{N})\), where 't' is the t-value corresponding to 95% confidence level with N-1 degrees of freedom, 'd-bar' is the mean difference, 's_d' is the standard deviation of differences and 'N' is the number of pairs.

04

Hypothesis testing using paired t-test

Null Hypothesis (H_0) is that the course makes no statistically significant improvement, i.e., the difference in the means is zero. Alternate Hypothesis (H_A) is that the course makes a statistically significant improvement, i.e., the difference in the means is not zero. The p-value corresponding to the computed t-statistics will be compared with the significance level (0.01). If the p-value is less than the significance level, then null hypothesis is rejected indicating that there is statistically significant improvement.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Confidence Interval

A confidence interval is a statistical tool that gives a range of plausible values for a population parameter. In the context of our exercise, we are looking at the mean of the population paired differences. These differences come from the memory test scores before and after the course.
For our exercise, a 95% confidence interval means we are 95% confident that the true mean difference falls within this interval. Creating this interval involves:

• Finding the mean of the differences between scores.
• Calculating the standard deviation of these differences.
• Using a t-distribution (because of small sample size) to find the appropriate t-value.

The formula for the interval is: \(d-bar \pm t \times \frac{s_d}{\sqrt{N}}\), where 'd-bar' is the mean difference, 't' is the t-value, and 's_d' is the standard deviation of differences. This interval helps in understanding the magnitude and direction of memory improvement.

Mean Difference

The mean difference is a simple yet essential concept in our paired t-test. It represents the average change in memory scores after the course compared to before. In this scenario, each student's memory improvement or decline is calculated, and these differences are averaged.
To find the mean difference, follow these steps:

• Subtract the "before" score from the "after" score for each student.
• Add up all these differences to get the total sum.
• Divide the total by the number of students to obtain the mean difference.

If the mean difference is positive, it indicates an overall improvement. A negative mean difference would suggest a decline in memory performance after the course.

Hypothesis Testing

Hypothesis testing is the process of making decisions based on data analysis. In this scenario, we consider whether the memory improvement course has a statistically significant effect on students' scores. Here's how it works:

• We begin with two hypotheses: the null hypothesis (H_0) suggests no improvement, while the alternative hypothesis (H_A) suggests there is improvement.
• Using the paired differences, calculate the test statistic with the formula: \(t = \frac{d-bar}{(s_d/\sqrt{N})}\), where 'd-bar' is the mean difference, and 's_d' is the standard deviation of differences.
• Determine the p-value, which tells us the probability of observing our data if the null hypothesis is true.

If the p-value is less than the chosen significance level (0.01), we reject the null hypothesis. This would indicate that the course likely has a real impact on improving memory.

Memory Improvement Course

Memory improvement courses are designed to enhance cognitive abilities through structured exercises and techniques. In this exercise, we assess the efficacy of such a course by comparing memory test scores before and after participation.
Features of a good memory improvement course include:

• Evidence-based techniques like mnemonics, visualization, and spaced repetition.
• Regular assessments to track progress, as seen in the "before" and "after" tests.
• Interactive and engaging content to maintain participant interest.

Through statistical analysis, like the paired t-test used here, educators and program designers can confirm whether their course effectively improves memory performance, ensuring that participants truly benefit from their efforts.