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Confidence Interval for \(\mu_{1}-\mu_{2}\) Consider two independent normal distributions. A random sample of size \(n_{1}=20\) from the first distribution showed \(\bar{x}_{1}=12\) and a random sample of size \(n_{2}=25\) from the second distribution showed \(\bar{x}_{2}=14\). (a) Check Requirements If \(\sigma_{1}\) and \(\sigma_{2}\) are known, what distribution does \(\bar{x}_{1}-\bar{x}_{2}\) follow? Explain. (b) Given \(\sigma_{1}=3\) and \(\sigma_{2}=4\), find a \(90 \%\) confidence interval for \(\mu_{1}-\mu_{2}\) (c) Check Requirements Suppose \(\sigma_{1}\) and \(\sigma_{2}\) are both unknown, but from the random samples, you know \(s_{1}=3\) and \(s_{2}=4 .\) What distribution approximates the \(\bar{x}_{1}-\bar{x}_{2}\) distribution? What are the degrees of freedom? Explain. (d) With \(s_{1}=3\) and \(s_{2}=4\), find a \(90 \%\) confidence interval for \(\mu_{1}-\mu_{2} .\) (e) If you have an appropriate calculator or computer software, find a \(90 \%\) confidence interval for \(\mu_{1}-\mu_{2}\) using degrees of freedom based on Satterthwaite's approximation. (f) Interpretation Based on the confidence intervals you computed, can you be \(90 \%\) confident that \(\mu_{1}\) is smaller than \(\mu_{2}\) ? Explain.

Step by step solution

01

Check Requirements When  and  are Known

When the standard deviations  and  are known, the sampling distribution of - follows a normal distribution. This is because the means of each sample are distributed normally due to the Central Limit Theorem, and the samples are assumed to come from normal distributions.

02

Calculate Confidence Interval using Normal Distribution

Given that  and  are known, we will use the standard normal distribution to find the confidence interval. The formula for the confidence interval is:\[(\bar{x}_{1} - \bar{x}_{2}) \pm Z_{\alpha/2} \sqrt{\frac{\sigma_{1}^{2}}{n_{1}} + \frac{\sigma_{2}^{2}}{n_{2}}}\]Substituting the values: - \(\bar{x}_{1} = 12\), \(\bar{x}_{2} = 14\)- \(\sigma_{1} = 3\), \(\sigma_{2} = 4\)- \(Z_{\alpha/2} = 1.645\) for 90% confidence levelThe confidence interval is:\[(12 - 14) \pm 1.645 \sqrt{\frac{3^{2}}{20} + \frac{4^{2}}{25}} = -2 \pm 1.645 \times 1.3686 = -2 \pm 2.252\]So, the interval is approximately \([-4.252, 0.252]\).

03

Requirements When  and  are Unknown

When  and  are unknown but estimates are given with sample standard deviations \(s_{1}\) and \(s_{2}\), the sampling distribution of \(\bar{x}_{1}-\bar{x}_{2}\) is approximated by a t-distribution. This is because the exact standard deviations of the population are unknown and are estimated from the sample itself.

04

Degrees of Freedom Calculation

The degrees of freedom for the t-distribution when \(s_{1}\) and \(s_{2}\) are unknown can be estimated using: \[ df = \frac{\left(\frac{s_{1}^{2}}{n_{1}} + \frac{s_{2}^{2}}{n_{2}}\right)^{2}}{\frac{\left(\frac{s_{1}^{2}}{n_{1}}\right)^{2}}{n_{1}-1} + \frac{\left(\frac{s_{2}^{2}}{n_{2}}\right)^{2}}{n_{2}-1}} \]Plugging the values yields:\[ df = \frac{\left(\frac{3^{2}}{20} + \frac{4^{2}}{25}\right)^{2}}{\frac{\left(\frac{3^{2}}{20}\right)^{2}}{19} + \frac{\left(\frac{4^{2}}{25}\right)^{2}}{24}} = 32.21\]Approximated to 32. Thus, using 32 degrees of freedom.

05

Confidence Interval using t-Distribution

With \(s_{1}=3\), \(s_{2}=4\), and calculated degrees of freedom being 32, we use the t-distribution to find the confidence interval. The t-value for 90% confidence and 32 df is approximately 1.694.The formula for the confidence interval is:\[(\bar{x}_{1} - \bar{x}_{2}) \pm t_{\alpha/2} \sqrt{\frac{s_{1}^{2}}{n_{1}} + \frac{s_{2}^{2}}{n_{2}}}\]Substituting the values:\[(-2) \pm 1.694 \sqrt{\frac{3^{2}}{20} + \frac{4^{2}}{25}} = -2 \pm 1.694 \times 1.3686 = -2 \pm 2.317\]So, the interval is approximately \([-4.317, 0.317]\).

06

Confidence Interval using Satterthwaite's Approximation

Again using an approximate 32 degrees of freedom and the same t-value 1.694, the result is consistent:\[-2 \pm 1.694 \times 1.369 \approx -2 \pm 2.317\]The confidence interval is again approximately from \([-4.317, 0.317]\).

07

Interpretation of Confidence Interval

Both confidence intervals calculated (using the normal distribution and Satterthwaite's approximation) include values less than 0 and greater than 0. This means that zero falls within both confidence intervals, suggesting that there is not enough evidence to be 90% confident that the true mean \(\mu_{1}\) is smaller than \(\mu_{2}\).

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Key Concepts

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

t-distribution

The t-distribution is an essential concept in statistics, especially when dealing with small sample sizes and unknown population standard deviations. Unlike the normal distribution which is symmetrical and bell-shaped, the t-distribution is slightly more spread out. It has heavier tails, which means there is a higher probability of values further from the mean. This accounts for the increased variability observed in smaller samples. The t-distribution becomes crucial when estimating parameters like the mean difference between two groups when standard deviations are unknown. In such cases, sample standard deviations (s) replace population standard deviations (σ). When we use the t-distribution to create confidence intervals, we effectively account for the uncertainty introduced through estimating the population standard deviation with a sample statistic. As the sample size increases, the t-distribution gradually approaches the normal distribution.

degrees of freedom

Degrees of freedom (df) is a concept used to describe the number of independent values or quantities that can vary in an analysis without breaking any constraints. In statistical terms, degrees of freedom often represent the number of values in a calculation that are free to vary.In the context of the exercise, degrees of freedom are important when using the t-distribution. The degree of freedom affects the shape of the t-distribution curve, and a correct calculation ensures the confidence interval accurately reflects the variability of the samples. When calculating degrees of freedom for two independent samples with sample standard deviations, we use the formula:\[ df = \frac{\left(\frac{s_{1}^{2}}{n_{1}} + \frac{s_{2}^{2}}{n_{2}}\right)^{2}}{\frac{\left(\frac{s_{1}^{2}}{n_{1}}\right)^{2}}{n_{1}-1} + \frac{\left(\frac{s_{2}^{2}}{n_{2}}\right)^{2}}{n_{2}-1}} \]This formula comes into play when the standard deviations are unknown, allowing an estimation of the correct t-distribution.

Satterthwaite's approximation

Satterthwaite's approximation is a method to approximate the degrees of freedom for the t-distribution when dealing with samples that have unequal variances. When standard deviations of populations are unknown, and we estimate them using samples, Satterthwaite's formula provides a reliable way to determine an equivalent t-distribution by calculating a weighted average of the sample variances. This method is especially useful in the classic case of two-sample tests for means, as it adjusts for differences in sample sizes and variances between the two groups. The approximation effectively transforms the degrees of freedom into a non-integer value, typically rounded to the nearest whole number for practical use. This adjustment allows researchers and statisticians to more accurately calculate the confidence interval and make inferences about the population means based on the sample data, making it an indispensable tool when managing unequal variances.

normal distribution

The normal distribution, commonly known as the Gaussian distribution, is a fundamental concept in statistics. It is characterized by its symmetrical bell-shaped curve and is described by two parameters: the mean and the standard deviation. When the standard deviations of populations are known, and the sample sizes are sufficiently large, the mean differences between samples typically follow a normal distribution. This assumption allows us to use the Z-distribution for calculating confidence intervals, utilizing the known population standard deviations. The standard normal distribution, a special case of the normal distribution, has a mean of 0 and a standard deviation of 1. This implies that data points are measured as z-scores, which reflect how many standard deviations a point is from the mean. For hypothesis testing and confidence intervals, the normal distribution provides a framework for understanding data spread and variability, under the Central Limit Theorem. The theorem suggests that for a large enough sample size, the means of samples from any distribution will approach a normal distribution.