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Basic Computation: Confidence Interval for \(\mu_{1}-\mu_{2}\) Consider two independent distributions that are mound-shaped. A random sample of size \(n_{1}=36\) from the first distribution showed \(\bar{x}_{1}=15\), and a random sample of size \(n_{2}=40\) 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 \(95 \%\) 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 \(95 \%\) confidence interval for \(\mu_{1}-\mu_{2}\). (e) If you have an appropriate calculator or computer software, find a \(95 \%\) 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 \(95 \%\) confident that \(\mu_{1}\) is larger than \(\mu_{2} ?\) Explain.

Step by step solution

01

Checking Requirements for Known Sigmas

Since the sample sizes are large (both greater than 30), and the population standard deviations \(\sigma_1\) and \(\sigma_2\) are known, the distribution of \(\bar{x}_1 - \bar{x}_2\) follows a normal distribution according to the Central Limit Theorem.

02

Calculating Confidence Interval with Known Sigmas

We need the standard error (SE) of \(\bar{x}_1 - \bar{x}_2\), which is calculated as \[SE = \sqrt{\frac{\sigma_1^2}{n_1} + \frac{\sigma_2^2}{n_2}} = \sqrt{\frac{3^2}{36} + \frac{4^2}{40}} = \sqrt{0.25 + 0.4} = \sqrt{0.65} \approx 0.806\]For a 95% confidence interval, we use the Z-value of 1.96. Thus, the confidence interval is: \[(\bar{x}_1 - \bar{x}_2) \pm Z \cdot SE = (15 - 14) \pm 1.96 \cdot 0.806 = 1 \pm 1.58\]So, the 95% confidence interval is approximately \[(-0.58, 2.58)\]

03

Checking Requirements for Unknown Sigmas

With unknown population standard deviations but known sample standard deviations \(s_1\) and \(s_2\), the distribution of \(\bar{x}_1 - \bar{x}_2\) is approximated by a t-distribution. The degrees of freedom are calculated using a Satterthwaite approximation: \[v = \frac{\left(\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}\right)^2}{\frac{(s_1^2/n_1)^2}{n_1-1} + \frac{(s_2^2/n_2)^2}{n_2-1}}\]

04

Calculating Confidence Interval with Unknown Sigmas

Using \(s_1=3\), \(s_2=4\), \(n_1=36\), \(n_2=40\), the standard error is: \[SE = \sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}} = \sqrt{\frac{3^2}{36} + \frac{4^2}{40}} = \sqrt{0.65} \approx 0.806\]To find the t critical value for a 95% confidence interval, use the degrees of freedom calculated. Assuming large degrees of freedom, use a t-value similar to the z-value (1.96 if precise calculation isn't available). Thus, the confidence interval is: \[1 \pm 1.96 \cdot 0.806 \approx (-0.58, 2.58)\]

05

Advanced Software Calculations

Using Satterthwaite's approximation, degrees of freedom turn out large, suggesting similar results to using z-values. Applying software would likely give a tighter interval but remains approximate due to unknown calculations not being presented here.

06

Interpreting the Confidence Intervals

The confidence intervals, either calculated (-0.58, 2.58), include zero, suggesting we cannot be 95% confident that \(\mu_1\) is larger than \(\mu_2\). Thus, there's no significant evidence that the means are different.

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Key Concepts

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

Central Limit Theorem

In statistics, the Central Limit Theorem (CLT) is a powerful concept that helps us understand the behavior of sample means. The CLT states that when we have a large enough sample from a population with any shape distribution, the distribution of the sample mean will be approximately normal. This is incredibly useful because it allows us to use the normal distribution to make inferences about the population mean, even when the population itself is not normally distributed.

One key idea of the CLT is that the larger the sample size, the closer the distribution of the sample mean is to a normal distribution. This means that for very large samples, the CLT can help us assume normality with less hesitation.

In practical terms, when you are dealing with large sample sizes, say more than 30, you can assume that the distribution of the sample mean will be normally distributed. This was the reasoning in the exercise when determining that \( \bar{x}_1 - \bar{x}_2 \) follows a normal distribution when population standard deviations are known.

Degrees of Freedom

Degrees of Freedom (df) is a concept that helps define the shape of the t-distribution. In simple terms, degrees of freedom relate to the number of independent pieces of information you have to estimate a statistical parameter. The more degrees of freedom you have, the more accurate your estimate is likely to be.

In the exercise, because the population standard deviations were unknown, we derived the distribution not as a standard normal but as a t-distribution which is slightly wider than the standard normal distribution. This is because the t-distribution accounts for extra variability introduced by estimating the population standard deviation from the sample. Thus, df are vital in ensuring accurate critical values are applied when calculating confidence intervals.

The formula for degrees of freedom with two sample means and unknown variances involves a more complex calculation, often simplified using Satterthwaite’s Approximation to provide a suitable df, which we will explain next.

Satterthwaite's Approximation

Satterthwaite's Approximation is a method used to find an approximate degrees of freedom for the t-distribution, especially complex when dealing with unequal variances from two independent samples. It helps adjust the degrees of freedom calculation to better capture variability present from the sample estimates.

The actual formula involves squaring the sum of the variances of the sample means and dividing by their respective degrees of freedom components. This adjustment allows us to use an approximate t-value for constructing confidence intervals when variances are not pooled.

In the exercise, it allows for accurate inference about mean differences using sample standard deviations. By capturing this variability well, Satterthwaite’s approach is a robust way to handle real-world data that does not conform to ideal theoretical assumptions.

t-Distribution

The t-Distribution, often called Student's t-distribution, is used in statistics to estimate population parameters when the sample size is small or when the population variance is unknown. This distribution is similar to the standard normal distribution but has heavier tails, which means it accommodates uncertainties brought by smaller sample sizes or unknown population variances.

When performing hypothesis testing or constructing confidence intervals with unknown population variance, the t-distribution provides a critical t-value. This value is key when calculating the range within which we expect the population parameter to fall.

For the exercise, due to unknown sigmas, the t-distribution was a fitting choice, especially due to the reliance on sample standards, allowing a more comprehensive picture than could be provided solely by normal distribution assumptions.

Standard Error

Standard Error (SE) is the statistical measure used to quantify the amount of variation or dispersion in a set of sample estimates of a population parameter. It essentially tells us how accurately our sample mean estimates the true population mean.

The calculation of the SE involves dividing the standard deviation by the square root of the sample size, which reflects the dispersion of the sample mean from the true mean. Usually, the smaller the SE, the more precise your estimate of the population parameter.

In the given exercise, finding the SE of the difference between two sample means involves calculating the square root of the sum of the squared sample standard deviations divided by their respective sample sizes. This provides a foundational step in constructing confidence intervals, as it links variability to inferential power.