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In statistical terms, a point estimate is a single value estimate of a parameter of a population. In this context, it refers to estimating the difference between two population means \( \mu_1 \) and \( \mu_2 \). Instead of considering a complete dataset, you use sample data to make this estimate.
You calculate the point estimate by finding the difference between two sample means. From the problem, the two sample means are given as \( \bar{x}_1 = 13.97 \) and \( \bar{x}_2 = 15.55 \).
To find this point estimate, simply perform the subtraction:

• \( \bar{x}_1 - \bar{x}_2 = 13.97 - 15.55 = -1.58 \)

This result, -1.58, is the point estimate of the difference between the population means \( \mu_1 - \mu_2 \). It's a straightforward calculation providing a single best guess value.

The standard error is a quantification of the variability or spread of a sample statistic around the true population parameter. Specifically, for the difference between two sample means, it tells us how precise our point estimate is.
Here's the formula for calculating the standard error for the difference in means:

• \[ SE = \sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}} \]

- \( s_1 = 3.78 \) and \( s_2 = 3.26 \) are the sample standard deviations.
- \( n_1 = 21 \) and \( n_2 = 20 \) are the sample sizes.
Substituting these values, calculate the standard error:

• \[ SE = \sqrt{\frac{3.78^2}{21} + \frac{3.26^2}{20}} = 1.157 \]

This value, 1.157, measures the expected dispersion of the point estimate from the actual difference in population means. A smaller standard error indicates a more reliable estimate.

The t-distribution is crucial when dealing with smaller sample sizes or unknown population variances. It is similar to the normal distribution but has heavier tails, which means it is more accommodating for variation due to smaller samples.
When building confidence intervals, especially for small sample sizes, the t-distribution helps account for the additional uncertainty. In this exercise, we're constructing a 95% confidence interval for the difference in means using the t-distribution.
The degree of freedom (df) for our t-distribution is calculated as the smallest of \( n_1 - 1 \) or \( n_2 - 1 \). Here, it would be:

• \( \text{df} = \text{min}(21-1, 20-1) = 19 \)

Using the t-distribution table, the t-value for 19 degrees of freedom at a 95% confidence level is approximately 2.093. This value helps us expand the point estimate into a confidence interval.

The sample mean is a statistical measure summarizing the central tendency of a sample. It is simply the arithmetic average of the values in the sample.
In our context, we have two samples summarized by their sample means:

• \( \bar{x}_1 = 13.97 \)
• \( \bar{x}_2 = 15.55 \)

These sample means are vital because they serve as the best estimators of their respective population means \( \mu_1 \) and \( \mu_2 \).
From these, we derive the point estimate for the difference in population means. Moreover, these sample means form the foundation for calculating the standard error, constructing confidence intervals, and determining the reliability of our estimations. By calculating the point estimate, standard error, and using the t-distribution, we can make statistical inferences about the population from which the samples were taken.