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Consider a normal population distribution with the value of \(\sigma\) known. a. What is the confidence level for the interval \(\bar{x} \pm\) \(2.81 \sigma / \sqrt{n}\) ? b. What is the confidence level for the interval \(\bar{x} \pm\) \(1.44 \sigma / \sqrt{n} ?\) c. What value of \(z_{\mathrm{a} / 2}\) in the CI formula (7.5) results in a confidence level of \(99.7 \%\) ? d. Answer the question posed in part (c) for a confidence level of \(75 \%\).

Short Answer

Expert verified
a) 99.5%, b) 85%, c) 3.00, d) 1.15.

Step by step solution

01

Understanding the Confidence Interval Formula

The formula for constructing a confidence interval when the population standard deviation is known is \(\bar{x} \pm z_{\alpha/2} \frac{\sigma}{\sqrt{n}}\). Here, \(\bar{x}\) is the sample mean, \(\sigma\) is the population standard deviation, \(n\) is the sample size, and \(z_{\alpha/2}\) is the z-score that corresponds to the desired level of confidence.
02

Solving Part (a) for Confidence Interval \(\bar{x} \pm 2.81 \frac{\sigma}{\sqrt{n}}\)

For this interval, the z-score \(z_{\alpha/2}\) is 2.81. This value corresponds to a critical point in the z-distribution. Using standard normal distribution tables or calculator, find the confidence level that matches \(z=2.81\), which is approximately 99.5%.
03

Solving Part (b) for Confidence Interval \(\bar{x} \pm 1.44 \frac{\sigma}{\sqrt{n}}\)

For this interval, the z-score \(z_{\alpha/2}\) is 1.44. Find the confidence level associated with \(z=1.44\). Using the z-distribution, a z-score of 1.44 corresponds to a confidence level of approximately 85%.
04

Solving Part (c) for \(z_{\mathrm{a} / 2}\) with 99.7% Confidence Level

For a confidence level of 99.7%, use the standard normal distribution to find the z-score \(z_{\alpha/2}\) that corresponds to the middle 99.7% of the data. The z-score for a 99.7% confidence interval is approximately 3.00.
05

Solving Part (d) for \(z_{\mathrm{a} / 2}\) with 75% Confidence Level

For a confidence level of 75%, determine the z-score \(z_{\alpha/2}\) such that 75% of the distribution lies within \(\pm z_{\alpha/2}\). This z-score is approximately 1.15.

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Key Concepts

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

Normal Distribution
When dealing with confidence intervals, it is essential to understand normal distribution as it is the basis for calculating these intervals when the population standard deviation is known. A normal distribution is a continuous probability distribution that is symmetric around its mean, creating a bell-shaped curve.
This distribution has some key characteristics:
  • The mean, median, and mode are all equal.
  • It is perfectly symmetrical about the mean.
  • Empirical rule applies: 68% of data falls within one standard deviation (\(\sigma\)) of the mean, 95% falls within two, and 99.7% within three.
Normal distributions are described by their mean (\(\mu\)) and standard deviation (\(\sigma\)). These parameters determine the location and spread of the data.Confidence intervals often assume a normal distribution, especially when sample sizes are large. This underlying assumption simplifies calculations using known statistical methods.
Z-score
The z-score plays a pivotal role in the world of statistics and confidence intervals. It measures how many standard deviations an element is from the mean. By converting values into z-scores, we standardize different data sets for comparison.The formula for a z-score is: \[ z = \frac{x - \mu}{\sigma} \]where \(x\) is the value in question, \(\mu\) is the mean, and \(\sigma\) is the standard deviation.
In the context of confidence intervals, the z-score \(z_{\alpha/2}\) determines the number of standard deviations required to construct the interval. For example, a z-score of 2.81 is associated with a confidence level of 99.5% from the step-by-step solution. This value tells us that the sample mean is within 2.81 standard deviations (divided by the square root of the sample size) from the population mean, offering a high degree of certainty in our interval estimate.
Population Standard Deviation
The population standard deviation, denoted as \(\sigma\), is a measure of the dispersion of a set of data from its mean. Unlike sample standard deviation, the population standard deviation considers all data points in a population.In context, knowing the population standard deviation is crucial because it directly influences the width of the confidence interval. The smaller the standard deviation, the narrower the interval, indicating that the data points tend to be close to the mean. Conversely, a larger standard deviation results in a wider interval, reflecting more variability.
Confidence Intervals involving known \(\sigma\) utilize the formula: \[\bar{x} \pm z_{\alpha/2} \frac{\sigma}{\sqrt{n}}\]where \(\bar{x}\) is the sample mean, \(z_{\alpha/2}\) is the z-score, and \(n\) is the sample size. Understanding \(\sigma\) ensures that interpretations and predictions derived from the data are accurate.
Confidence Level
A confidence level is a percentage that signifies how sure you can be that a parameter lies within a confidence interval. Higher confidence levels mean a higher degree of certainty, but also result in wider confidence intervals. The concept of confidence level is deeply intertwined with the z-score. Different z-score values correspond to different confidence levels. For instance, a 99.7% confidence level has a z-score of 3, reflecting more certainty and capturing more variability of the data. While a 75% confidence level has a z-score of about 1.15, indicating lesser certainty, it also results in a narrower interval.
Choosing a confidence level depends on the context and how much risk one is willing to take. In critical situations, a higher confidence level may be desired, even if it means a less precise estimate, given by a wider interval.

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Most popular questions from this chapter

The financial manager of a large department store chain selected a random sample of 200 of its credit card customers and found that 136 had incurred an interest charge during the previous year because of an unpaid balance. a. Compute a \(90 \%\) CI for the true proportion of credit card customers who incurred an interest charge during the previous year. b. If the desired width of the \(90 \%\) interval is .05, what sample size is necessary to ensure this? c. Does the upper limit of the interval in part (a) specify a \(90 \%\) upper confidence bound for the proportion being estimated? Explain.

A random sample of 487 nonsmoking women of normal weight (body mass index between \(19.8\) and 26.0) who had given birth at a large metropolitan medical center was selected ("The Effects of Cigarette Smoking and Gestational Weight Change on Birth Outcomes in Obese and NormalWeight Women," Amer. J. of Public Health, 1997: 591-596). It was determined that \(7.2 \%\) of these births resulted in children of low birth weight (less than \(2500 \mathrm{~g}\) ). Calculate an upper confidence bound using a confidence level of \(99 \%\) for the proportion of all such births that result in children of low birth weight.

Suppose that a random sample of 50 bottles of a particular brand of cough syrup is selected and the alcohol content of each bottle is determined. Let \(\mu\) denote the average alcohol content for the population of all bottles of the brand under study. Suppose that the resulting \(95 \%\) confidence interval is \((7.8,9.4)\). a. Would a \(90 \%\) confidence interval calculated from this same sample have been narrower or wider than the given interval? Explain your reasoning. b. Consider the following statement: There is a \(95 \%\) chance that \(\mu\) is between \(7.8\) and \(9.4\). Is this statement correct? Why or why not? c. Consider the following statement: We can be highly confident that \(95 \%\) of all bottles of this type of cough syrup have an alcohol content that is between \(7.8\) and \(9.4\). Is this statement correct? Why or why not? d. Consider the following statement: If the process of selecting a sample of size 50 and then computing the corresponding \(95 \%\) interval is repeated 100 times, 95 of the resulting intervals will include \(\mu\). Is this statement correct? Why or why not?

It is important that face masks used by firefighters be able to withstand high temperatures because firefighters commonly work in temperatures of \(200-500^{\circ} \mathrm{F}\). In a test of one type of mask, 11 of 55 masks had lenses pop out at \(250^{\circ}\). Construct a \(90 \%\) CI for the true proportion of masks of this type whose lenses would pop out at \(250^{\circ}\).

Determine the following: a. The 95 th percentile of the chi-squared distribution with \(v=10\) b. The 5 th percentile of the chi-squared distribution with \(v=10\) c. \(P\left(10.98 \leq \chi^{2} \leq 36.78\right)\), where \(\chi^{2}\) is a chi- squared rv with \(v=22\) d. \(P\left(\chi^{2}<14.611\right.\) or \(\left.\chi^{2}>37.652\right)\), where \(\chi^{2}\) is a chisquared rv with \(v=25\)

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