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Chapter 7: Problem 7

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

Short Answer

Expert verified
The critical value \(z_{a / 2}\) for a 99.5% confidence level is approximately 2.81.

Step by step solution

01

Understand the Confidence Level

The given confidence level is 99.5%. This means that we want to find the critical value corresponding to the middle 99.5% of the standard normal distribution.
02

Determine the Value of \(\alpha\)

The confidence level of 99.5% implies that the level of significance \(\alpha\) is 1 - 0.995 = 0.005.
03

Find the Value of \(\alpha/2\)

Since the confidence level involves two tails in the standard normal distribution, we divide \(\alpha\) by 2. Thus, \(\alpha/2 = 0.005/2 = 0.0025\).
04

Locate the Critical Value on the Z-Table

Using a Z-table or statistical software, find the z-score that corresponds to the area to the left of the critical value. Specifically, find the z-score where the cumulative area to the left is 1 - 0.0025 = 0.9975.
05

Identify \(z_{a/2}\)

From the Z-table, the z-score corresponding to 0.9975 is approximately 2.81. Therefore, \(z_{a/2} \approx 2.81\).

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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 an essential concept in statistics. It represents the percentage of all possible samples that can be expected to include the true population parameter.
For example, a 99.5% confidence level means that if we took 1000 different samples and computed the confidence interval for each sample, we would expect 995 of those intervals to contain the population parameter we're estimating.
  • A higher confidence level indicates a higher degree of certainty but results in a wider confidence interval.
  • A lower confidence level indicates more risk of error but results in a narrower interval.

In most cases, popular confidence levels are 90%, 95%, and 99%. In our example, with a 99.5% confidence level, we want to locate the middle 99.5% of the standard normal distribution.
Standard Normal Distribution
The standard normal distribution is a normal distribution with a mean of 0 and a standard deviation of 1. It's also referred to as the Z-distribution.
  • It is symmetrical around the mean.
  • It serves as a reference to calculate probabilities and critical values.
In the context of finding critical values, the standard normal distribution helps determine where a specific proportion of data lies.
When we talk about a confidence interval, we are often interested in finding the z-score (or critical value) that corresponds to a certain area under the curve. In our example with a 99.5% confidence level, we need to find the critical z-value that leaves 0.5% (or 0.0025) in the tails of the distribution combined.
That's why the standard normal distribution is so useful: it allows us to convert between raw scores and probabilities.
Z-table
The Z-table, or standard normal table, helps to find the area under the curve of the standard normal distribution.
  • This area represents probabilities and is essential to finding critical values or z-scores.
  • The Z-table shows the cumulative probability up to a specific z-score from the mean (0).

To find the critical value for a 99.5% confidence level, you need to:
1. Determine the cumulative probability equivalent to the tail area. For a 99.5% confidence level, this involves finding the z-score with a cumulative probability of 0.9975 (since 1 - 0.0025 = 0.9975).
2. Look up this cumulative probability in the Z-table.
In our example, the Z-table tells us that the z-score corresponding to a cumulative area of 0.9975 is approximately 2.81.
By understanding how to use the Z-table, you can easily find critical values for a wide range of confidence levels.

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Most popular questions from this chapter

Assume that we want to construct a confidence interval. Do one of the following, as appropriate: (a) Find the critical value \(t_{\alpha / 2}\) (b) find the critical value \(z_{\alpha / 2},\) or \((c)\) state that neither the normal distribution nor the \(t\) distribution applies. Here are summary statistics for randomly selected weights of newborn girls: \(n=205, \bar{x}=30.4 \mathrm{hg}, s=7.1 \mathrm{hg}\) (based on Data Set 4 "Births" in Appendix B). The confidence level is \(95 \%\)

Why does the bootstrap method require sampling with replacement? What would happen if we used the methods of this section but sampled without replacement?

Finding Confidence Intervals. In Exercises \(9-16,\) assume that each sample is a simple random sample obtained from a population with a normal distribution. Body Temperature Data Set 3 "Body Temperatures" in Appendix B includes a sample of 106 body temperatures having a mean of \(98.20^{\circ} \mathrm{F}\) and a standard deviation of \(0.62^{\circ} \mathrm{F}\) (for day 2 at 12 AM). Construct a \(95 \%\) confidence interval estimate of the standard deviation of the body temperatures for the entire population.

Find the critical value \(z_{a / 2}\) that corresponds to the given confidence level. $$99 \%$$

Use the given data to find the minimum sample size required to estimate a population proportion or percentage. In a study of government financial aid for college students, it becomes necessary to estimate the percentage of full-time college students who earn a bachelor's degree in four years or less. Find the sample size needed to estimate that percentage. Use a 0.05 margin of error, and use a confidence level of \(95 \% .\) a. Assume that nothing is known about the percentage to be estimated. b. Assume that prior studies have shown that about \(40 \%\) of full-time students earn bachelor's degrees in four years or less. c. Does the added knowledge in part (b) have much of an effect on the sample size?
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