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A student project by Heather Kral studied students on "lifestyle floors" of a dormitory in comparison to students on other floors. On a lifestyle floor the students share a common major, and there are a faculty coordinator and resident assistant from that department. Here are the grade point averages of 30 students on lifestyle floors (L) and 30 students on other floors \((\mathrm{N})\) : L: \(2.00,2.25,2.60,2.90,3.00,3.00,3.00,3.00\), \(3.00,3.20,3.20,3.25,3.30,3.30,3.32,3.50\), \(3.50,3.60,3.60,3.70,3.75,3.75,3.79,3.80\), \(3.80,3.90,4.00,4.00,4.00,4.00\). \(\mathrm{N}: 1.20,2.00,2.29,2.45,2.50,2.50,2.50,2.50\), \(2.65,2.70,2.75,2.75,2.79,2.80,2.80,2.80\), \(2.86,2.90,3.00,3.07,3.10,3.25,3.50,3.54\), \(3.56,3.60,3.70,3.75,3.80,4.00\). Notice that the lifestyle grade point averages have a large number of repeats and the distribution is skewed, so there is some question about normality. a. Obtain a \(95 \%\) confidence interval for the difference of population means using the method based on the theorem of Section 10.2. b. Obtain a bootstrap sample of 999 differences of means. Check the bootstrap distribution for normality using a normal probability plot. c. Use the standard deviation of the bootstrap distribution along with the mean and \(t\) critical value from (a) to get a \(95 \%\) confidence interval for the difference of means. d. Use the bootstrap sample and the percentile method to obtain a \(95 \%\) confidence interval for the difference of means. e. Compare your three confidence intervals. If they are very similar, why do you think this is the case? f. Interpret your results. Is there a substantial difference between lifestyle and other floors? Why do you think the difference is as big as it is?

Step by step solution

01

Calculate Mean and Standard Deviation

First, we calculate the mean and standard deviation of the GPA for both the lifestyle (L) and non-lifestyle (N) groups. For Lifestyle: Mean \( \bar{x}_L = 3.39 \) and Standard Deviation \( s_L = 0.543 \). For Non-Lifestyle: Mean \( \bar{x}_N = 2.89 \) and Standard Deviation \( s_N = 0.582 \).

02

Compute the Difference in Means

Calculate the difference in the sample means between the two groups: \( \bar{x}_L - \bar{x}_N = 3.39 - 2.89 = 0.50 \).

03

Confidence Interval Based on Theory

Using the information from Step 1, the standard error (SE) for the difference in means is \( SE = \sqrt{\frac{s_L^2}{n_L} + \frac{s_N^2}{n_N}} = \sqrt{\frac{0.543^2}{30} + \frac{0.582^2}{30}} = 0.135 \). The degrees of freedom is approximately 57. T-score for 95% confidence is about 2.005 from t-distribution table. Thus, the confidence interval (CI) is \( 0.50 \pm 2.005 \times 0.135 = (0.227, 0.773) \).

04

Bootstrap Resampling

Implement a bootstrap method by generating 999 samples of differences from the provided datasets. For each sample, calculate the mean difference. This creates a distribution of mean differences.

05

Check Normality of Bootstrap Distribution

Use a normal probability plot on the bootstrap sample differences to evaluate normality. If the data points lie approximately along a straight line, the data can be considered normally distributed.

06

Confidence Interval Using Bootstrap SD

Calculate the standard deviation of the bootstrap distribution. Use this bootstrap SD with the mean and t critical value from Step 3 to compute a new 95% CI. This will often give similar results to the theoretical CI if the boostrapped distribution is approximately normal.

07

Confidence Interval Using Percentile Method

Obtain the 2.5th and 97.5th percentiles of the bootstrap distribution. This forms the 95% confidence interval by determining the range within which 95% of the bootstrap mean differences fall.

08

Compare Confidence Intervals

Compare each of the CIs found in Steps 3, 6, and 7. If they are similar, it indicates that the assumptions about the normality of data or the central limit theorem hold true in this context for the given sample size.

09

Interpretation of Results

If all the intervals consistently support a positive difference above zero, it suggests a substantial difference between the lifestyle and non-lifestyle floors. This difference could be due to the environment and shared academic and social support among students on lifestyle floors, which potentially enhances their academic performance.

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Key Concepts

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

Confidence Interval

In statistics, a confidence interval (CI) offers an estimated range of values that is likely to include a population parameter with a certain level of confidence, typically 95%. It helps us understand the precision of our sample statistics and is used to express the reliability of an estimate.
When calculating the confidence interval for the difference of means, like in this exercise, we first determine the mean and standard deviation of each group. Then, we calculate the standard error, which measures the variability of the sample mean difference.

• The standard error (SE) formula for the difference between two independent means is given by \[SE = \sqrt{\frac{s_L^2}{n_L} + \frac{s_N^2}{n_N}}\]where \(s_L\) and \(s_N\) are the standard deviations of the lifestyle and non-lifestyle groups, and \(n_L\), \(n_N\) are their respective sample sizes.
• With the calculated SE, we then multiply it by the critical value from the t-distribution (for 95% confidence, it is typically around 2.005) to find the margin of error.
• The CI is finally computed as \[\text{Mean Difference} \pm (\text{T-score} \times \text{SE}).\]

For this exercise, the resulting confidence interval signifies that with 95% confidence, the true difference in GPA means between the lifestyle and non-lifestyle groups falls within the calculated range.

Bootstrap Resampling

Bootstrap resampling is a powerful statistical tool used to estimate the distribution of a sample statistic by repeatedly sampling with replacement from the original data. This method helps overcome problems like non-normality or small sample sizes.
When you deal with distributions that might not fit into the normal curve snugly, bootstrap methods come to the rescue. By taking many samples from your data and computing the statistic of interest repeatedly, you build a distribution from which you can conclude potential values for your parameter.

• For instance, in this exercise, generating 999 bootstrap samples involves randomly selecting 30 grade point averages with replacement for each lifestyle and non-lifestyle group to calculate the difference in their means.
• This method structures a distribution of 999 mean differences, offering a robust foundation to examine the stability and variability of the estimates.
• The specially constructed bootstrap distribution can then be assessed for normality using tools like normal probability plots, which aids in ensuring statistical inferences remain valid.

Bootstrap approaches are versatile, which is why they are widely used in modern statistical analysis.

Difference of Means

The difference of means is a straightforward yet crucial concept in statistical analysis. It represents the gap between the average values of two groups, in this case, GPA scores between students living on lifestyle floors and those residing on other floors.
To find this difference, you first calculate the average (mean) of each group separately.

• The formula for the difference of means is simply the mean of the first group minus the mean of the second group: \[\bar{x}_L - \bar{x}_N\]where \(\bar{x}_L\) is the mean of the lifestyle group, and \(\bar{x}_N\) is the mean of the non-lifestyle group.
• This statistic can hint at whether a meaningful or significant disparity exists between two populations.

In the exercise example, the difference was 0.50, suggesting a potential distinction in academic performance. Understanding whether this gap is statistically significant or just a chance artifact involves further inferential procedures, such as calculating confidence intervals or conducting hypothesis tests based on the data distribution.

Sample Distribution

A sample distribution refers to the probability distribution of a given sample statistic, such as the mean, which helps in making inferences about the population from which the sample is drawn. Understanding sample distribution is key to applying inferential statistics like confidence intervals.
Ideally, we want our sample distribution to mirror the population distribution. However, that's not always possible, especially if the sample size is small or the data contains outliers or is skewed. This is where Central Limit Theorem (CLT) steps in to save the day, stating that the distribution of the sample mean will tend toward a normal distribution as the sample size increases, regardless of the original distribution shape.

• In this exercise, Heather Kral's data showed skewness, raising questions about whether the normality assumption holds.
• Bootstrap resampling was used here to generate a sample distribution of the mean differences even when normality was questionable.

When the sample size is adequate and distribution roughly normal, classical statistical methods work well. But if conditions challenge normality assumptions, alternative methods like bootstrapping can ensure accurate statistical inference.