Question

Refer to Exercise \(11.46 .\) The means of two of the factor level combinations- say, \(\mathrm{A}_{1} \mathrm{~B}_{1}\) and \(\mathrm{A}_{2} \mathrm{~B}_{1}-\) are \(\bar{x}_{1}=8.3\) and \(\bar{x}_{2}=6.3,\) respectively. Find a \(95 \%\) confidence interval for the difference between the two corresponding population means.

Short answer

Answer: The sample sizes (n₁ and n₂) and sample variances (s²₁ and s²₂) are missing in the problem. We need these values to calculate the standard error, which is required to determine the 95% confidence interval for the difference between the population means.

Step-by-step solution

Identify the given values

We are given the sample means: \(\bar{x}_{1} = 8.3\) \(\bar{x}_{2} = 6.3\) We are asked to find a 95% confidence interval for the difference, i.e., \((\bar{x}_{1}-\bar{x}_{2})\).

Calculate the standard error of the difference

To find the confidence interval, we need to estimate the standard error for the difference between the two population means. We are not given information about the sample sizes, sample variances, or any other information needed to calculate the standard error directly. Therefore, we will assume equal variances and use the following formula: \(SE(\bar{x}_{1}-\bar{x}_{2})=\sqrt{\frac{(n_1-1) s^{2}_{1} + (n_2 - 1)s^{2}_{2}}{n_1+n_2-2}} \cdot \sqrt{\frac{1}{n_1}+\frac{1}{n_2}}\) We do not have enough information provided in the problem to calculate the standard error with this formula. The solution cannot be completed without this information.

Find the appropriate t-score

Assuming we had the standard error value, we would then need to find the appropriate t-score for a 95% confidence interval. To do this, we would use the degrees of freedom calculated as follows: \(df = n_1 + n_2 - 2\) The t-score can then be found using a t-distribution table or online calculator with the 95% confidence level and degrees of freedom.

Calculate the confidence interval

After finding the standard error and the t-score, we can calculate the confidence interval for the difference between the population means using the following formula: \((\bar{x}_{1}-\bar{x}_{2}) \pm t_{\alpha/2} \cdot SE(\bar{x}_{1}-\bar{x}_{2})\) Since we were not provided with the necessary information in the problem, such as sample sizes and variances, we are unable to calculate the 95% confidence interval for the difference between the population means.

Key concepts

Population Mean Difference

The concept of population mean difference is critical when comparing two different groups. It represents the difference between the average values (means) of two populations. For example, if we have two groups A and B, and their means are represented as \( \bar{x}_1 \) and \( \bar{x}_2 \), then their population mean difference is calculated as \( \bar{x}_1 - \bar{x}_2 \). This difference is an estimate of how much one group outperforms or underperforms the other. In practical terms, understanding this difference helps in determining if a significant change or effect exists between the two groups. A small difference could imply that both groups have similar performances, whereas a large difference might indicate a noticeable effect. This concept is especially useful in fields like clinical trials, marketing studies, and educational assessments where comparing two groups' performances is necessary.

Standard Error Calculation

The standard error (SE) is a statistical metric that describes the accuracy with which a sample represents a population. When comparing two populations, it tells us how reliable the estimate of the population mean difference is. To calculate the standard error for the mean difference \( \bar{x}_1 - \bar{x}_2 \), there are specific formulas and assumptions involved.One common formula involves assuming equal variances between the two groups. This assumption allows us to pool the variances and use the formula: \[SE(\bar{x}_1-\bar{x}_2) = \sqrt{\frac{(n_1-1) s^{2}_{1} + (n_2 - 1)s^{2}_{2}}{n_1+n_2-2}} \cdot \sqrt{\frac{1}{n_1}+\frac{1}{n_2}}\]Here, \( n_1 \) and \( n_2 \) are the sample sizes, and \( s^{2}_{1} \) and \( s^{2}_{2} \) are the sample variances for the two groups. The smaller the standard error, the more precise our estimate of the population mean difference becomes. Unfortunately, without specific sample sizes or variances provided, we cannot compute the standard error directly from hypothetical data.

T-Distribution

The t-distribution is a critical component in statistics, especially when you're working with smaller sample sizes or when the population variance is unknown. It resembles the normal distribution but has heavier tails, which means it accounts for the extra variability expected with smaller samples. When constructing confidence intervals for population mean differences, the t-distribution helps in determining how far the sample mean difference \( \bar{x}_1 - \bar{x}_2 \) is from the true population mean difference.Using the t-distribution requires a specific t-score, which depends on the confidence level specified (like 95% or 99%) and the degrees of freedom. Degrees of freedom are calculated as \( df = n_1 + n_2 - 2 \) when comparing two means. This score can be found in statistical tables or via online calculators tailored to your specific confidence level and sample parameters. Once determined, the t-score is used in the confidence interval formula to provide the range within which we expect the true population mean difference to lie.

Assuming Equal Variances

The assumption of equal variances, often referred to as homogeneity of variance, is an important element in comparative studies. This assumption states that the variance within each of the two groups being compared is roughly the same. When it's valid, it simplifies the computation of the standard error and t-tests by allowing the combination of variances into a single pooled variance estimate. Without this assumption, more complex methods like the Welch's t-test might be required, which don’t assume equal variances. However, assuming equal variances simplifies calculations and is often considered valid when preliminary analysis (like Levene's test) indicates equality. In practical applications, this assumption can speed up the process of analysis and doesn't distort results significantly as long as variance differences are minimal. By assuming equal variances, statisticians can apply simpler, more straightforward calculations, facilitating the comparison between two groups' means confidently and efficiently.