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Find the critical value \(z_{\alpha / 2}\) that corresponds to the given confidence level. \(99.5 \%\)

Short Answer

Expert verified
The critical value \(\text{z}_{\alpha / 2}\) for a 99.5% confidence level is approximately 2.807.

Step by step solution

01

Understand the Confidence Level

The given confidence level is 99.5%. This means that the total area under the standard normal distribution curve represents the confidence level, which is 0.995.
02

Determine the Significance Level

The significance level \(\beta\) is equal to \(1 - \text{confidence level}\). For a 99.5% confidence level, \(\beta = 1 - 0.995 = 0.005\).
03

Calculate the Significance Level for One Tail

Since the critical value \(\text{z}_{\alpha / 2}\) splits the total significance level into two tails, each tail will have an area of \(\frac{\beta}{2} = \frac{0.005}{2} = 0.0025\).
04

Find the Critical Value

Using the standard normal distribution table or a calculator, find the z-value that corresponds to an area of 0.0025 in the tail. The critical value \(\text{z}_{\alpha / 2}\) that corresponds to this area is approximately 2.807.

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Key Concepts

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

Confidence Level
The confidence level is a measure of how certain we are about our estimate. It is usually expressed as a percentage. For example, a 99.5% confidence level means we are 99.5% certain that our estimate includes the true population parameter. This percentage translates to a decimal for calculations, so 99.5% becomes 0.995. The remaining area under the curve (0.5%) is the uncertainty we are willing to accept. It helps us determine the range within which we expect our population parameter to fall. To better understand this, imagine you have a box of chocolates and you want to ensure 99.5% of them are perfect. If you draw a sample and 99.5% of the time your sample includes only perfect chocolates, you are using a 99.5% confidence level.
Significance Level
The significance level, denoted as \(\beta\), quantifies the probability of making an error when our sample does not include the true population parameter. It is calculated as \(\beta = 1 - \text{Confidence Level}\). For example, a 99.5% confidence level has a significance level of \(0.005\). This value is split into two tails in the critical value calculation, making each tail have an area of 0.0025. Understanding the significance level helps to determine how much risk we are willing to take in our estimates. If you think of it as taking a gamble, the significance level is your chance of losing. So, a lower significance level means you are taking less risk.
Standard Normal Distribution
The standard normal distribution is a special form of the normal distribution with a mean of 0 and a standard deviation of 1. This curve is symmetrical, and most of its values lie within three standard deviations from the mean. It is crucial in statistics because it helps us calculate probabilities and make inferences about populations. For example, in our problem, we use it to get the z-value corresponding to a specified area under the curve. The total area under the standard normal distribution is 1, and each side of the mean covers 50% of the data. Think of it like a bell where most values cluster around the center, and fewer values are at the extremes.
Z-Value
The z-value, also known as the z-score, is the number of standard deviations a data point is from the mean. In other words, it shows how far away, in standard deviation units, a value is from the center of the standard normal distribution. For instance, if we want to find a critical value that corresponds to a certain area in the tail, we look up the z-value in standard normal distribution tables or use a calculator. In our case, for a 99.5% confidence level and 0.0025 in each tail, the critical z-value is about 2.807. This means we are looking for a point on the curve that is 2.807 standard deviations away from the mean. If you think of the z-value like marks on a measuring tape, it tells us precisely where to draw our line to include the desired area.

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Most popular questions from this chapter

In a test of weight loss programs, 40 adults used the Atkins weight loss program. After 12 months, their mean weight loss was found to be \(2.1 \mathrm{lb}\), with a standard deviation of \(4.8\) lb. Construct a \(90 \%\) confidence interval estimate of the standard deviation of the weight loss for all such subjects. Does the confidence interval give us information about the effectiveness of the diet?

In a test of weight loss programs, 40 adults used the Atkins weight loss program. After 12 months, their mean weight loss was found to be \(2.1 \mathrm{lb}\), with a standard deviation of \(4.8\) lb. Construct a \(90 \%\) confidence interval estimate of the mean weight loss for all such subjects. Does the Atkins program appear to be effective? Does it appear to be practical?

Instead of using Table \(7-2\) for determining the sample size required to estimate a population standard deviation \(\sigma\), the following formula can be used $$ n=\frac{1}{2}\left(\frac{z_{\alpha / 2}}{d}\right)^{2} $$ where \(z_{\alpha / 2}\) corresponds to the confidence level and \(d\) is the decimal form of the percentage error. For example, to be \(95 \%\) confident that \(s\) is within \(15 \%\) of the value of \(\sigma\), use \(z_{\alpha / 2}=1.96\) and \(d=0.15\) to get a sample size of \(n=86\). Find the sample size required to estimate \(\sigma\), assuming that we want \(98 \%\) confidence that \(s\) is within \(15 \%\) of \(\sigma\).

In a program designed to help patients stop smoking, 198 patients were given sustained care, and \(82.8 \%\) of them were no longer smoking after one month. Among 199 patients given standard care, \(62.8 \%\) were no longer smoking after one month (based on data from "Sustained Care Intervention and Postdischarge Smoking Cessation Among Hospitalized Adults"" by Rigotti et al., Journal of the American Medical Association, Vol. 312, No. 7). Construct the two \(95 \%\) confidence interval estimates of the percentages of success. Compare the results. What do you conclude?

You plan to conduct a survey to estimate the percentage of adults who have had chickenpox. Find the number of people who must be surveyed if you want to be \(90 \%\) confident that the sample percentage is within two percentage points of the true percentage for the population of all adults. a. Assume that nothing is known about the prevalence of chickenpox. b. Assume that about \(95 \%\) of adults have had chickenpox. c. Does the added knowledge in part (b) have much of an effect on the sample size?

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