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The following results are for independent random samples taken from two populations. \(\begin{array}{ll}\text { Sample } 1 & \text { Sample } 2 \\\ n_{1}=20 & n_{2}=30 \\ \bar{x}_{1}=22.5 & \bar{x}_{2}=20.1 \\ s_{1}=2.5 & s_{2}=4.8\end{array}\) a. What is the point estimate of the difference between the two population means? b. What is the degrees of freedom for the \(t\) distribution? c. \(\quad\) At \(95 \%\) confidence, what is the margin of error? d. What is the \(95 \%\) confidence interval for the difference between the two population means?

Short Answer

Expert verified
a) The point estimate is 2.4. b) Degrees of freedom ≈ 46. c) Margin of error is 2.093. d) 95% confidence interval: (0.307, 4.493).

Step by step solution

01

Calculate Point Estimate

The point estimate for the difference between the two population means is obtained by subtracting the sample means. Thus, \( \bar{x}_1 - \bar{x}_2 = 22.5 - 20.1 = 2.4 \).
02

Determine Degrees of Freedom

The degrees of freedom (df) for the two-sample t-test can be computed using the formula for the Welch-Satterthwaite equation: \[ df = \frac{\left( \frac{s_1^2}{n_1} + \frac{s_2^2}{n_2} \right)^2}{\frac{\left( \frac{s_1^2}{n_1} \right)^2}{n_1 - 1} + \frac{\left( \frac{s_2^2}{n_2} \right)^2}{n_2 - 1}} \]Substituting the values, we have \[ df = \frac{\left( \frac{2.5^2}{20} + \frac{4.8^2}{30} \right)^2}{\frac{\left( \frac{2.5^2}{20} \right)^2}{19} + \frac{\left( \frac{4.8^2}{30} \right)^2}{29}} \]Calculating each part let's solver step by step:\( s_1 = 2.5 \Rightarrow s_1^2 = 6.25 \Rightarrow \frac{s_1^2}{n_1} = \frac{6.25}{20} = 0.3125\)\( s_2 = 4.8 \Rightarrow s_2^2 = 23.04 \Rightarrow \frac{s_2^2}{n_2} = \frac{23.04}{30} = 0.768 \) Thus,\[ \text{numerator fraction: } \left( 0.3125 + 0.768 \right)^2 = 1.0805^2 = 1.1674 \] \( \text{{denominator}} = \frac{(0.3125)^2}{19} + \frac{(0.768)^2}{29} = \frac{0.09765625}{19} + \frac{0.589824}{29} = 0.0051445 + 0.0203387 = 0.0254782 \)Divide\( df \approx \frac{1.1674}{0.0254782} \approx 45.83 \)The rounded df value is approximately 46.
03

Calculate the Margin of Error

To find the margin of error, we need the critical value of the t-distribution at 95% confidence level and df = 46. Use a t-table or calculator to find \( t^* \approx 2.013 \).The standard error (SE) of the difference between means is calculated by: \[ \text{SE} = \sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}} \]\( \Rightarrow \text{SE} = \sqrt{\frac{2.5^2}{20} + \frac{4.8^2}{30}} = \sqrt{0.3125 + 0.768} = \sqrt{1.0805} = 1.0399 \).Thus, the margin of error = \( t^* \times \text{SE} = 2.013 \times 1.0399 \approx 2.093 \).The margin of error is 2.093.
04

Calculate 95% Confidence Interval

Finally, calculate the 95% confidence interval for the difference between the means using:\[ (\bar{x}_1 - \bar{x}_2) \pm \text{Margin of Error} \]The point estimate is 2.4, and the margin of error is 2.093.Therefore, confidence interval is:\[ 2.4 \pm 2.093 \rightarrow (2.4 - 2.093, 2.4 + 2.093) \rightarrow (0.307, 4.493) \]

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Key Concepts

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

Point Estimate
When comparing two population means, the point estimate helps to identify the estimated difference between them based on sample data. It is the single best guess of the actual value. In this exercise, we are given two independent random samples with the following details:
  • Sample 1 has a mean of 22.5
  • Sample 2 has a mean of 20.1
The point estimate for the difference between these two population means is found by subtracting the mean of Sample 2 from the mean of Sample 1. To put it simply, it is calculated as:\[\bar{x}_1 - \bar{x}_2 = 22.5 - 20.1 = 2.4\]This means that, based on the sample data, it is estimated that Sample 1 has a mean that is 2.4 units greater than Sample 2. Point estimates, like this difference, provide a starting point for further statistical analysis, such as constructing a confidence interval.
Degrees of Freedom
Degrees of freedom (df) are a crucial aspect of many statistical calculations, including the two-sample t-test used here. They represent the number of independent values or quantities which can vary freely. For a two-sample t-test, the degrees of freedom help in determining the critical value from the t-distribution, which is used to construct confidence intervals.In this exercise, we utilize the Welch-Satterthwaite equation to calculate the degrees of freedom for two independent samples. It takes into account the sample variances and sizes:\[df = \frac{\left( \frac{s_1^2}{n_1} + \frac{s_2^2}{n_2} \right)^2}{\frac{\left( \frac{s_1^2}{n_1} \right)^2}{n_1 - 1} + \frac{\left( \frac{s_2^2}{n_2} \right)^2}{n_2 - 1}}\]After performing the detailed calculations, we find the degrees of freedom to be approximately 46. This is a rounded value and is essential for accurately determining the t critical value needed for the margin of error and confidence interval calculations.
Margin of Error
The margin of error provides an indication of the potential error in the point estimate. It defines the range in which we expect our population parameter to lie, with a certain level of confidence, typically 95% or 99%.To calculate the margin of error for the difference between two population means, we use the critical value from the t-distribution and the standard error of the difference. The critical value, denoted as \(t^*\), is determined based on the confidence level and degrees of freedom. Here, with df = 46 and a 95% confidence level, \(t^* \approx 2.013\).The standard error (SE) is calculated using:\[\text{SE} = \sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}}\]For our samples, this comes out to approximately 1.0399.The margin of error is then obtained by multiplying \(t^*\) with the SE:\[\text{Margin of Error} = t^* \times \text{SE} = 2.013 \times 1.0399 \approx 2.093\]This means we can be 95% confident that the true difference between the population means lies within 2.093 units of our point estimate, providing a clearer picture of the variability present in the estimate. This step aids in making informed decisions based on the data.

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Most popular questions from this chapter

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