/*! 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 3 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 normally distributed populations. $$ \begin{array}{lll} n_{1}=18 & \bar{x}_{1}=7.82 & \sigma_{1}=2.35 \\ n_{2}=15 & \bar{x}_{2}=5.99 & \sigma_{2}=3.17 \end{array} $$ a. What is the point estimate of \(\mu_{1}-\mu_{2}\) ? b. Construct a \(99 \%\) 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 1.83. The 99% confidence interval for \(\mu_1 - \mu_2\) is 1.83 ± 2.618 or equivalently (-0.788, 4.448). The margin of error for this estimate is 2.62.

Step by step solution

01

Point Estimate Of \(\mu_1 - \mu_2\)

A point estimate of a population parameter is a single value of a statistic. For instance, the sample mean, \(\bar{x}\), is a point estimate of the population mean, \(\mu\). In this scenario, a point estimate of \(\mu_1 - \mu_2\) is given by \(\bar{x}_1 - \bar{x}_2\). Therefore, \(\bar{x}_1 - \bar{x}_2 = 7.82 - 5.99 = 1.83.\)
02

Determining the Standard Error

The standard error of the difference between the means is given by the formula \(SE = \sqrt{(\sigma_1^2/n1) + (\sigma_2^2/n2)}\), where \(\sigma_1^2\) and \(\sigma_2^2\) are the variances of the first and second populations, respectively, and n1, n2 are the sizes of the first and the second samples. Substitute the given values into this formula to get \(SE = \sqrt{(2.35^2/18) + (3.17^2/15)} = 1.017.\)
03

Constructing a 99% Confidence Interval for \(\mu_1 - \mu_2\)

A 99% confidence interval is computed as \(\bar{x}_1 - \bar{x}_2 ± Z(α/2) * SE\), where Z(α/2) is the Z-score that corresponds to the desired confidence level (for a 99% confidence interval, Z(α/2) is approximately 2.576). Thus, the 99% confidence interval for \(\mu_1 - \mu_2\) is \(1.83 ± 2.576 * 1.017 = 1.83 ± 2.618.\)
04

Calculating the Margin of Error

The margin of error, E for this estimate is given by \(E = Z * SE\), where Z is the Z-score discussed in the previous step. So, \(E = 2.576 * 1.017 = 2.62.\)

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Key Concepts

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

Point Estimate
A point estimate gives us a simple number to represent a population parameter. When we talk about the difference between two population means, \(\mu_1 - \mu_2\), the point estimate is determined using sample means. In this exercise, the point estimate for the difference is calculated as \(\bar{x}_1 - \bar{x}_2\). This means we subtract the mean of the second sample from the first, providing \(1.83\) as the point estimate.

Think of the point estimate as the best guess or approximation based on the sample data we have. It's a handy way to summarize the data even though we know it doesn't give the complete picture of the population difference. To get deeper insights and account for variability, we need to delve into confidence intervals and other statistical tools.
Confidence Interval
The confidence interval provides a range of values within which we expect the true population parameter \(\mu_1 - \mu_2\) to fall, with a certain level of confidence. For this exercise, we created a 99% confidence interval. This means we are 99% sure that the true population difference lies within this interval.

Constructing a confidence interval involves the point estimate and something called the margin of error. We use the formula \(\bar{x}_1 - \bar{x}_2 \pm Z(\alpha/2) \times SE\), where \(Z(\alpha/2)\) corresponds to the critical value from the Z-distribution (2.576 for 99%) and \(SE\) is the standard error. Here, \(1.83 \pm 2.618\) gives us the confidence interval.
  • Lower limit = 1.83 - 2.618 = -0.788
  • Upper limit = 1.83 + 2.618 = 4.448
Thus, the interval \[ -0.788, 4.448 \] shows where we estimate the true difference might be.
Margin of Error
The margin of error tells us how much the point estimate might vary by providing a buffer around it. It's crucial to understanding how precise our point estimate is. For the exercise, the margin of error is determined using the formula \(E = Z \times SE\), where \(Z\) is the Z-score (2.576 for 99%) and \(SE\) is the standard error.

The margin of error calculated here is \(2.62\). This figure shows the range above or below the point estimate that defines the confidence interval. It tells us, with high confidence, how much the estimate could potentially vary, thus offering assurance about the accuracy of our estimate.
Standard Error
The standard error measures the spread of the sample means and gives us an idea of how much sampling variability exists. When comparing two means, the standard error helps us understand the reliability of the point estimate. In this exercise, the calculated standard error was \(1.017\), using the formula \(SE = \sqrt{(\sigma_1^2/n1) + (\sigma_2^2/n2)}\).

The standard error draws from knowledge about the sample sizes and their variances. Smaller standard errors signify more reliable estimates, which is crucial when building confidence intervals and setting margins of error. It acts as a guide for the overall stability and accuracy of our statistical inference, bridging the gap between abstract numbers and real-world interpretations.

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Most popular questions from this chapter

A company claims that its medicine, Brand A, provides faster relief from pain than another company's medicine, Brand B. A researcher tested both brands of medicine on two groups of randomly selected patients. The results of the test are given in the following table. The mean and standard deviation of relief times are in minutes. \begin{tabular}{cccc} \hline Brand & Sample Size & Mean of Relief Times & Standard Deviation of Relief Times \\ \hline A & 25 & 44 & 11 \\ B & 22 & 49 & 9 \\ \hline \end{tabular} Assume that the two populations are normally distributed with unknown but equal standard deviations. a. Construct a \(99 \%\) confidence interval for the difference between the mean relief times for the two brands of medicine. b. Test at a \(1 \%\) significance level whether the mean relief time for Brand \(\mathrm{A}\) is less than that for Brand B.

A town that recently started a single-stream recycling program provided 60 -gallon recycling bins to 25 randomly selected houscholds and 75 -gallon recycling bins to 22 randomly selected households. The total volume of recycling over a 10 -week period was measured for each of the households. The average total volumes were 382 and 415 gallons for the households with the \(60-\) and 75 -gallon bins, respectively. The sample standard deviations were \(52.5\) and \(43.8\) gallons, respectively. Assume that the 10 -week total volumes of recycling are approximately normally distributed for both groups and that the population standard deviations are equal. a. Construct a \(98 \%\) confidence interval for the difference in the mean volumes of 10 -week recycling for the households with the \(60-\) and 75 -gallon bins. b. Using a \(2 \%\) significance level, can you conclude that the average 10 -week recycling volume of all households having 60 -gallon containers is different from the average volume of all households that have 75 -gallon containers?

As mentioned in Exercise \(10.26\), a town that recently started a single-stream recycling program provided 60 -gallon recycling bins to 25 randomly selected households and 75 -gallon recycling bins to 22 randomly selected households. The average total volumes of recycling over a 10 -week period were 382 and 415 gallons for the two groups, respectively, with standard deviations of \(52.5\) and \(43.8\) gallons, respectively. Suppose that the standard deviations for the two populations are not equal. a. Construct a \(98 \%\) confidence interval for the difference in the mean volumes of 10 -week recyclying for the households with the 60 - and 75 -gallon bins. b. Using a \(2 \%\) significance level, can you conclude that the average 10 -week recycling volume of all households having 60 -gallon containers is different from the average 10 -week recycling volume of all households that have 75 -gallon containers? c. Suppose that the sample standard deviations were \(59.3\) and \(33.8\) gallons, respectively. Redo parts a and b. Discuss any changes in the results.

Two local post offices are interested in knowing the average number of Christmas cards that are mailed out from the towns that they serve. A random sample of 80 households from Town A showed that they mailed an average of \(28.55\) Christmas cards with a standard deviation of \(10.30 .\) The corresponding values of the mean and standard deviation produced by a random sample of 58 households from Town B were \(33.67\) and \(8.97\) Christmas cards. Assume that the distributions of the numbers of Christmas cards mailed by all households from both these towns have the same population standard deviation. a. Construct a \(95 \%\) confidence interval for the difference in the average numbers of Christmas cards mailed by all households in these two towns. b. Using a \(10 \%\) significance level, can you conclude that the average number of Christmas cards mailed out by all households in Town \(\mathrm{A}\) is different from the corresponding average for Town B?

Briefly explain the meaning of independent and dependent samples. Give one example of each.

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