/*! 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 6 If the original \(x\) distributi... [FREE SOLUTION] | 91Ó°ÊÓ

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If the original \(x\) distribution has a relatively small standard deviation, the confidence interval for \(\mu\) will be relatively short.

Short Answer

Expert verified
A small standard deviation results in a more precise, shorter confidence interval for \\(\\mu\\).

Step by step solution

01

Understanding the Relationship

A confidence interval for the mean \(\mu\) is affected by the variability in the data, represented by the standard deviation \(s\). If \(s\) is small, the data points are closely clustered around the mean, leading to a shorter confidence interval.
02

Examine the Formula

The formula for a confidence interval for the mean is given by \[ \bar{x} \pm z \left(\frac{s}{\sqrt{n}}\right) \], where \z\ is the z-score corresponding to the desired confidence level, \ is the sample size, and \s\ is the standard deviation. A smaller \s\ results in a smaller margin of error \(z\frac{s}{\sqrt{n}}\).
03

Impact of a Small Standard Deviation

When the standard deviation \(s\) is small, the term \(\frac{s}{\sqrt{n}}\) becomes smaller, thereby reducing the margin of error. This leads to a shorter confidence interval around the sample mean \(\bar{x}\).
04

Concluding the Impact on Confidence Interval

Since the confidence interval is shorter, the estimate of the population mean \(\mu\) is more precise. Hence, with a small standard deviation, we can say we are more confident that the computed interval contains the true mean.

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Key Concepts

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

Standard Deviation
The standard deviation is a crucial measure in statistics that indicates how spread out the data points are in a data set. Imagine you have a bunch of numbers, like test scores. The standard deviation tells you how much those scores deviate, or differ, from the average score (the mean). If the standard deviation is small, it means that most of the scores are clustered close to the average. Proportionally, if it's large, the scores vary greatly from the mean.

The significance of the standard deviation is highlighted when calculating confidence intervals. A small standard deviation results in a more precise estimate of the average, as the data points are not widely scattered. This precision is desirable in statistical analyses, as it allows conclusions to be drawn with greater confidence.

In practical terms: - **Small Standard Deviation:** Data are tightly packed around the mean. - **Large Standard Deviation:** Data are widely spread over a range.

This measurement plays a key role in determining the width of confidence intervals, ultimately affecting the certainty of our findings.
Mean
The mean is often referred to as the "average" and is a fundamental concept in statistics and everyday life. It is calculated by adding up all the numbers in a data set and then dividing by the number of values. Knowing the mean gives you a general idea of the "center" of your data.

For example, if you're looking at the test scores of a class, the mean score tells you what you can expect an average student to receive. It's a simple yet powerful measure that is commonly used to summarize data with a single number.

In relation to confidence intervals, the mean (\( \bar{x} \)) is at the center, with intervals extending on both sides to capture the population mean. This interval tells us that we estimate the true average, taking into account some level of uncertainty. The precision of this estimate depends not only on the mean itself but also on how spread out our data are.
Margin of Error
The margin of error is an essential component of any confidence interval, providing a range within which the true population parameter (like a mean) is expected to fall. It quantifies uncertainty about this estimate.

The formula for determining the margin of error in a confidence interval for the mean is \( z \left( \frac{s}{\sqrt{n}} \right) \). Each component plays a role:- **\( z \):** The z-score, which reflects the confidence level (like 95%).- **\( s \):** The standard deviation, indicating data variability.- **\( n \):** Sample size, where a larger sample results in a smaller fraction and, thus, a smaller margin of error.

When the standard deviation (\( s \)) is smaller or the sample size (\( n \)) is larger, the margin of error decreases, which leads to a shorter, more exact confidence interval.

This means that the smaller the margin of error, the more precise is our estimate of the population mean, providing a sharper focus on data analysis conclusions.

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Most popular questions from this chapter

Over the past several months, an adult patient has been treated for tetany (severe muscle spasms). This condition is associated with an average total calcium level below \(6 \mathrm{mg} / \mathrm{dl}(\) Reference: Manual of Laboratory and Diagnostic Tests, F. Fischbach). Recently, the patient's total calcium tests gave the following readings (in \(\mathrm{mg} / \mathrm{dl}\) ). \(9.3\) \(\begin{array}{llllll}8.8 & 10.1 & 8.9 & 9.4 & 9.8 & 10.0\end{array}\) \(\begin{array}{lll}9.9 & 11.2 & 12.1\end{array}\) (a) Use a calculator to verify that \(\bar{x}=9.95\) and \(s \approx 1.02\). (b) Find a \(99.9 \%\) confidence interval for the population mean of total calcium in this patient's blood. (c) Based on your results in part (b), do you think this patient still has a calcium deficiency? Explain.

(a) Suppose a \(95 \%\) confidence interval for the difference of means contains both positive and negative numbers. Will a \(99 \%\) confidence interval based on the same data necessarily contain both positive and negative numbers? Explain. What about a \(90 \%\) confidence interval? Explain. (b) Suppose a \(95 \%\) confidence interval for the difference of proportions contains all positive numbers. Will a \(99 \%\) confidence interval based on the same data necessarily contain all positive numbers as well? Explain. What about a \(90 \%\) confidence interval? Explain.

A random sample of medical files is used to estimate the proportion \(p\) of all people who have blood type \(B\). (a) If you have no preliminary estimate for \(p\), how many medical files should you include in a random sample in order to be \(85 \%\) sure that the point estimate \(\hat{p}\) will be within a distance of \(0.05\) from \(p ?\) (b) Answer part (a) if you use the preliminary estimate that about 8 out of 90 people have blood type B. (Reference: Manual of Laboratory and Diagnostic Tests, F. Fischbach.)

Do you want to own your own candy store? Wow! With some interest in running your own business and a decent credit rating, you can probably get a bank loan on startup costs for franchises such as Candy Express, The Fudge Company, Karmel Corn, and Rocky Mountain Chocolate Factory. Startup costs (in thousands of dollars) for a random sample of candy stores are given below (Source: Entrepreneur Magazine, Vol. 23, No. 10\()\). \(\begin{array}{lllllllll}95 & 173 & 129 & 95 & 75 & 94 & 116 & 100 & 85\end{array}\) Use a calculator with mean and sample standard deviation keys to verify that \(\bar{x} \approx 106.9\) thousand dollars and \(s \approx 29.4\) thousand dollars. Find a \(90 \%\) confidence interval for the population average startup costs \(\mu\) for candy store franchises.

Jobs and productivity! How do banks rate? One way to answer this question is to examine annual profits per employee. Forbes Top Companies, edited by J. T. Davis (John Wiley \& Sons), gave the following data about annual profits per employee (in units of one thousand dollars per employee) for representative companies in financial services. Companies such as Wells Fargo, First Bank System, and Key Banks were included. Assume \(\sigma \approx 10.2\) thousand dollars. $$ \begin{array}{llllllllll} 42.9 & 43.8 & 48.2 & 60.6 & 54.9 & 55.1 & 52.9 & 54.9 & 42.5 & 33.0 & 33.6 \\ 36.9 & 27.0 & 47.1 & 33.8 & 28.1 & 28.5 & 29.1 & 36.5 & 36.1 & 26.9 & 27.8 \\ 28.8 & 29.3 & 31.5 & 31.7 & 31.1 & 38.0 & 32.0 & 31.7 & 32.9 & 23.1 & 54.9 \\ 43.8 & 36.9 & 31.9 & 25.5 & 23.2 & 29.8 & 22.3 & 26.5 & 26.7 & & \end{array} $$ (a) Use a calculator or appropriate computer software to verify that, for the preceding data, \(\bar{x} \approx 36.0\). (b) Let us say that the preceding data are representative of the entire sector of (successful) financial services corporations. Find a \(75 \%\) confidence interval for \(\mu\), the average annual profit per employee for all successful banks. (c)Let us say that you are the manager of a local bank with a large number of employees. Suppose the annual profits per employee are less than 30 thousand dollars per employee. Do you think this might be somewhat low compared with other successful financial institutions? Explain by referring to the confidence interval you computed in part (b). (d) Suppose the annual profits are more than 40 thousand dollars per employee. As manager of the bank, would you feel somewhat better? Explain by referring to the confidence interval you computed in part (b). (e) Repeat parts (b), (c), and (d) for a \(90 \%\) confidence level.

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