/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 1 The following results come from ... [FREE SOLUTION] | 91Ó°ÊÓ

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The following results come from two independent random samples taken of two populations. \(\begin{array}{ll}\text { Sample } 1 & \text { Sample } 2 \\ n_{1}=50 & n_{2}=35 \\ \bar{x}_{1}=13.6 & \bar{x}_{2}=11.6 \\ \sigma_{1}=2.2 &\sigma_{2}=3.0\end{array}\) a. What is the point estimate of the difference between the two population means? b. Provide a \(90 \%\) confidence interval for the difference between the two population means. c. Provide a \(95 \%\) confidence interval for the difference between the two population means.

Short Answer

Expert verified
Point estimate: 2.0; 90% CI: (0.816, 3.184); 95% CI: (0.589, 3.411).

Step by step solution

01

Identify Point Estimate

The point estimate for the difference between the two population means is calculated by subtracting the sample mean of sample 2 from the sample mean of sample 1. So, the point estimate is given by \( \bar{x}_1 - \bar{x}_2 \). Substituting the given values, we have \( 13.6 - 11.6 = 2.0 \). The point estimate is 2.0.
02

Calculate Standard Error

The standard error (SE) of the difference between two means from independent samples is calculated using the formula: \( SE = \sqrt{\frac{\sigma_1^2}{n_1} + \frac{\sigma_2^2}{n_2}} \). Plugging in the values: \( SE = \sqrt{\frac{2.2^2}{50} + \frac{3.0^2}{35}} \approx 0.720 \).
03

Find Z-Score for 90% Confidence Interval

For a 90% confidence interval, the Z-score is approximately 1.645. This value can be found using a standard normal distribution table or Z-score calculator.
04

Calculate 90% Confidence Interval

The confidence interval is given by \( (\bar{x}_1 - \bar{x}_2) \pm Z \times SE \). Substituting the appropriate values, the 90% confidence interval is \( 2.0 \pm 1.645 \times 0.720 \approx 2.0 \pm 1.184 \). Therefore, it ranges from approximately 0.816 to 3.184.
05

Find Z-Score for 95% Confidence Interval

For a 95% confidence interval, the Z-score is approximately 1.96. This value is also obtained using a standard normal distribution table or Z-score calculator.
06

Calculate 95% Confidence Interval

Using the same formula \( (\bar{x}_1 - \bar{x}_2) \pm Z \times SE \), for the 95% confidence interval we have \( 2.0 \pm 1.96 \times 0.720 \approx 2.0 \pm 1.411 \). So this interval ranges from approximately 0.589 to 3.411.

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Key Concepts

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

Point Estimate
In statistics, the point estimate is all about getting an educated guess on the parameter of a population. It's like trying to pinpoint the target when we don't have our eyes directly on it.
A point estimate helps us to determine the exact number representing a population characteristic, such as a mean or proportion, based on our sample data. In the context of the given exercise, our goal is to find a point estimate for the difference between the means of two populations. We find it by subtracting one sample mean from another:
  • Sample mean of Population 1 (\( \bar{x}_1 \)) is 13.6.
  • Sample mean of Population 2 (\( \bar{x}_2 \)) is 11.6.
The calculation is straightforward: \[ Point \ Estimate = \bar{x}_1 - \bar{x}_2 = 13.6 - 11.6 = 2.0 \]So, our point estimate for the difference in population means is 2.0.
Standard Error
Standard error is a crucial concept that provides insight into how much our sample mean deviates from the actual population mean. It tells us about the reliability and stability of our point estimate.Let's break down the calculation. The standard error for the difference between two independent sample means is given by:\[ SE = \sqrt{\frac{\sigma_1^2}{n_1} + \frac{\sigma_2^2}{n_2}} \]Where:
  • \(\sigma_1\) and \(\sigma_2\) are the standard deviations of the samples.
  • n\(n_1\) and \(n_2\) are the sample sizes.
Using the provided data:\[ SE = \sqrt{\frac{2.2^2}{50} + \frac{3.0^2}{35}} \approx 0.720 \]This value, approximately 0.720, indicates the variability of our point estimate. The smaller the standard error, the more reliable our point estimate.
Population Means
Population means are fundamental in determining the average behavior of a large group. Let's consider this in two parts:
  • Sample mean is a crucial reflection of the population mean.
  • It helps infer what the population is like based on a relatively small subset of data.
In the exercise, the sample means of the two populations were given as 13.6 and 11.6. Since it's not feasible to measure everything in a gigantic population, we rely on these sample means to make educated guesses about the actual population means. The ultimate aim of the exercise was to estimate the difference in these population means. It illustrates how sample data can be effectively used to understand bigger groups without checking every single element.
Z-Score
Z-scores make it possible to compute the confidence interval, providing a range within which the true population parameter lies. A Z-score translates a data point's position into the context of a standard normal distribution. Here are the key aspects of Z-score usage in calculating confidence intervals:
  • The Z-score determines the distance between the sample mean and the population mean.
  • For a 90% confidence interval, a Z-score of approximately 1.645 is used.
  • For a 95% confidence interval, a Z-score of approximately 1.96 is essential.
Calculating these intervals involves the formula:\[ CI = (\bar{x}_1 - \bar{x}_2) \pm Z \times SE \]Using the aforementioned Z-scores and the standard error calculated, we achieve two different confidence intervals. By adjusting the Z-score, we can change how sure we want to be about our interval: the higher the Z-score, the higher the confidence and wider the interval.This concept is central to making predictions based on sample data, helping us understand how confident we can be in our estimates.

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Most popular questions from this chapter

How much is the cost of a hospital stay increasing? The mean cost of one day in a semiprivate room was reported to be \(\$ 4848\) in 2005 and \(\$ 5260\) in 2006 (The Wall Street Journal, January 2,2007 . Assume the estimate for 2005 is a sample mean based on a sample size of 80 and the estimate for 2006 is a sample mean based on a sample size of 60 a. Develop a point estimate of the increase in the cost of a semiprivate hospital room from 2005 to 2006 b. Historical data indicate that a population standard deviation of \(\$ 800\) is a reasonable assumption for both years. Compute the margin of error for your estimate in part (a). Use \(95 \%\) confidence c. Develop a \(95 \%\) confidence interval estimate of the increase in cost for a semiprivate room.

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