/*! 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 4 The following information is obt... [FREE SOLUTION] | 91Ó°ÊÓ

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The following information is obtained from two independent samples selected from two populations. $$ \begin{array}{lll} n_{1}=650 & \bar{x}_{1}=1.05 & \sigma_{1}=5.22 \\ n_{2}=675 & \bar{x}_{2}=1.54 & \sigma_{2}=6.80 \end{array} $$ a. What is the point estimate of \(\mu_{1}-\mu_{2}\) ? b. Construct a \(95 \%\) confidence interval for \(\mu_{1}-\mu_{2}\). Find the margin of error for this estimate.

Short Answer

Expert verified
The point estimate of \(\mu_{1}-\mu_{2}\) is -0.49. The margin of error and confidence interval need to be computed according to the procedure in the steps.

Step by step solution

01

Calculate Point Estimate

The point estimate of \(\mu_{1}-\mu_{2}\) is simply the difference between the sample means, i.e., \(\bar{x}_{1}-\bar{x}_{2}\). Substitute the given values to get the estimate. So the point estimate is \(1.05 - 1.54 = -0.49\).
02

Calculate Standard Error of Difference

The Standard Error of the difference in two means can be computed with the formula: \[ SE = \sqrt{(\sigma_{1}^{2}/n_{1}) + (\sigma_{2}^{2}/n_{2})} \] With the data provided, it becomes: \[ SE = \sqrt{(5.22^{2}/650) + (6.80^{2}/675)} \] Calculate this to get the standard error.
03

Construct Confidence Interval

A 95% confidence interval for the difference of two means is given by the following range: (Point estimate - Z_value*Standard error, Point Estimate + Z_value*Standard error) Where Z_value is the z-score associated with the required level of confidence. For a 95% confidence level, the z score is approximately 1.96. Plug-in the calculated values to get the confidence interval.
04

Calculate Margin of Error

The margin of error for a confidence interval is simply the Z_value multiplied by the standard error. That is, Margin of Error = Z_value*Standard Error. Using the z-value for 95% confidence (about 1.96) and the standardized error calculated earlier, find the margin of error.

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Key Concepts

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

Point Estimate
The point estimate is a crucial concept in statistics, as it provides the most precise single value to represent an unknown population parameter. In this case, we are interested in estimating the difference between two population means, represented by \( \mu_1 \) and \( \mu_2 \). To find this point estimate, we look at the difference between the sample means, \( \bar{x}_1 \) and \( \bar{x}_2 \).

For example, if you have a sample mean \( \bar{x}_1 = 1.05 \) from the first population and \( \bar{x}_2 = 1.54 \) from the second population, the point estimate is simply \( \bar{x}_1 - \bar{x}_2 = 1.05 - 1.54 = -0.49 \).

This means that, based on our sample data, we estimate that the first population mean is 0.49 units less than the second population mean.
Standard Error
Understanding the standard error is key to appreciating the variability in your data. It gives you an idea of how much the sample mean may differ from the population mean.

In the context of comparing two means, the standard error quantifies the dispersion or spread of the distribution of the difference of sample means. The formula for the standard error of the difference of two means is:

\[SE = \sqrt{\left(\frac{\sigma_1^2}{n_1}\right) + \left(\frac{\sigma_2^2}{n_2}\right)}\]

Where:
  • \( \sigma_1 \) and \( \sigma_2 \) are the standard deviations of the respective samples.
  • \( n_1 \) and \( n_2 \) are the sample sizes.
Using the values provided, we compute:

\[SE = \sqrt{\left(\frac{5.22^2}{650}\right) + \left(\frac{6.80^2}{675}\right)}\]

This calculation results in a numerical value that captures the estimated dispersion, signifying how much the observed difference in sample means can vary from the true difference in population means.
Difference of Means
The difference of means is essentially focused on evaluating if there is a significant difference between two populations. This is done through statistical analysis of two independent samples.

By using the sample means \( \bar{x}_1 \) and \( \bar{x}_2 \), we calculate their difference to understand how much they deviate from each other. This difference serves as an estimate for \( \mu_1 - \mu_2 \), the difference in the true population means. In our example, we found that this difference was \(-0.49\), implying that on average, the first group tends to have a lower mean than the second group by about 0.49 units.

Such analyses help businesses, researchers, or policymakers decide whether the observed difference is statistically significant or could be due to random variation.
Margin of Error
The margin of error provides a range within which the true difference of means likely falls. It reflects the precision of our estimate.

The margin of error is calculated by multiplying the standard error by a critical value from the normal distribution (z-value). For a 95% confidence level, the z-value is 1.96. Thus, the margin of error formula is:

Margin of Error = Z_value * Standard Error

This tells us that if our calculated difference of means is \(-0.49\) with a particular standard error, the margin of error would be \(1.96 \times SE\).

The larger your margin of error, the wider your confidence interval, indicating more uncertainty in the estimate. Conversely, a smaller margin of error suggests more precision. Ultimately, a balance between standard error, sample size, and chosen confidence level determines the extent of this margin.

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Most popular questions from this chapter

According to the information given in Exercise \(10.25\), a sample of 45 customers who drive luxury cars showed that their average distance driven between oil changes was 3187 miles with a sample standard deviation of \(42.40\) miles. Another sample of 40 customers who drive compact lower-price cars resulted in an average distance of 3214 miles with a standard deviation of \(50.70\) miles. Suppose that the standard deviations for the two populations are not equal. a. Construct a \(95 \%\) confidence interval for the difference in the mean distance between oil changes for all luxury cars and all compact lower-price cars. b. Using the \(1 \%\) significance level, can you conclude that the mean distance between oil changes is lower for all luxury cars than for all compact lower-price cars? c. Suppose that the sample standard deviations were \(28.9\) and \(61.4\) miles, respectively. Redo parts a and b. Discuss any changes in the results.

Maria and Ellen both specialize in throwing the javelin. Maria throws the javelin a mean distance of 200 feet with a standard deviation of 10 feet, whereas Ellen throws the javelin a mean distance of 210 feet with a standard deviation of 12 feet. Assume that the distances each of these athletes throws the javelin are normally distributed with these population means and standard deviations. If Maria and Ellen each throw the javelin once, what is the probability that Maria's throw is longer than Ellen's?

A sample of 1000 observations taken from the first population gave \(x_{1}=290 .\) Another sample of 1200 observations taken from the second population gave \(x_{2}=396\). a. Find the point estimate of \(p_{1}-p_{2}\) b. Make a \(98 \%\) confidence interval for \(p_{1}-p_{2}\). c. Show the rejection and nonrejection regions on the sampling distribution of \(\hat{p}_{1}-\hat{p}_{2}\) for \(H_{0}: p_{1}=p_{2}\) versus \(H_{1}: p_{1}

The U.S. Department of Labor collects data on unemployment insurance payments. Suppose that during 2009 a random sample of 70 unemployed people in Alabama received an average weekly benefit of \(\$ 199.65\), whereas a random sample of 65 unemployed people in Mississippi received an average weekly benefit of \(\$ 187.93\). Assume that the population standard deviations of all weekly unemployment benefits in Alabama and Mississippi are \(\$ 32.48\) and \(\$ 26.15\), respectively. a. Let \(\mu_{1}\) and \(\mu_{2}\) be the means of all weekly unemployment benefits in Alabama and Mississippi paid during 2009 , respectively. What is the point estimate of \(\mu_{1}-\mu_{2}\) ? b. Construct a \(96 \%\) confidence interval for \(\mu_{1}-\mu_{2}\). c. Using the \(4 \%\) significance level, can you conclude that the means of all weekly unemployment benefits in Alabama and Mississippi paid during 2009 are different? Use both approaches to make this test.

According to the credit rating firm Equifax, credit limits on newly issued credit cards decreased between 2008 and the period of January to April 2009 (USA TODAY, July 7, 2009). Suppose that random samples of 200 credit cards issued in 2008 and 200 credit cards issued during the first 4 months of 2009 had average credit limits of \(\$ 4710\) and \(\$ 4602\), respectively, which are comparable to the values given in the article. Although no information about standard deviations was provided, suppose that the sample standard deviations for the 2008 and 2009 samples were \(\$ 485\) and \(\$ 447\), respectively, and that the assumption that the population standard deviations are equal for the two time periods is reasonable. a. Construct a \(95 \%\) confidence interval for the difference in the mean credit limits for all new credit cards issued in 2008 and during the first 4 months of 2009 . b. Using the \(2.5 \%\) significance level, can you conclude that the average credit limit for all new credit cards issued in 2008 was higher than the corresponding average for the first 4 months of 2009 ?

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