/*! 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 8 Determine the critical value \(z... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Determine the critical value \(z_{\alpha / 2}\) that corresponds to the given level of confidence. \(99 \%\)

Short Answer

Expert verified
The critical value \( z_{ \alpha / 2 } \) for a 99\% confidence level is approximately 2.576.

Step by step solution

01

Understand Critical Value

The critical value, denoted as \( z_{ \alpha /2 }\), is the number that separates the central \((1 -\alpha)\)% of the distribution from the critical region in both tails. For a 99\(\%\) confidence level, \(1 - \alpha = 0.99\).
02

Find \( \alpha \)

Since the confidence level is 99\%, \( 1 - \alpha = 0.99\). Rearrange this to find \( \alpha \): \( \alpha = 1 - 0.99 = 0.01\).
03

Determine \( \alpha / 2 \)

Divide \( \alpha \) by 2 to find \( \alpha / 2 \): \( \alpha / 2 = 0.01 / 2 = 0.005\).
04

Use Z-Table or Standard Normal Distribution

Use a Z-table or standard normal distribution calculator to find the critical value \( z_{0.005} \), which corresponds to an area of 0.005 in the upper tail of the standard normal distribution. This value is approximately 2.576.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

confidence level
Confidence levels are a fundamental concept in statistics, providing a measure of the degree of certainty associated with a sample estimate of a population parameter. For instance, a 99% confidence level suggests that we are 99% confident the true population parameter lies within our calculated interval. It leaves a 1% chance (or 0.01) that the parameter could be outside this interval. This small probability is represented by \( \alpha \), which reflects the total area in the two tails of the normal distribution curve not covered by the confidence interval.

To find the critical value \( z_{ \alpha /2 } \), it's crucial to understand how \( \alpha \) splits into two equal parts, one for each tail of the distribution. For a 99% confidence level, \( \alpha = 0.01 \), and thus each tail has an area of 0.005 (since 0.01 divided by 2 equals 0.005). This critical value defines the point beyond which we reject the null hypothesis in hypothesis testing.

Keeping the 99% confidence level in mind, let's move on to understanding the role of the standard normal distribution in locating this critical value.
standard normal distribution
The standard normal distribution is central to statistical analysis and is defined by its mean of 0 and standard deviation of 1. This symmetric, bell-shaped curve is used to determine critical values for various confidence levels. The area under the curve represents probabilities, totaling 1 or 100%.

For a confidence interval based on the standard normal distribution, we consider the central part of the curve and exclude the tails, corresponding to \alpha. In our case, with a 99% confidence level, the central part covers 99% of the distribution, leaving 0.5% (or 0.005) in each tail. This standard distribution helps us pinpoint where the critical values lie.

The critical value \( z_{ \alpha / 2 } \) tells us the number of standard deviations our confidence interval extends from the mean. For a 99% confidence level, this value is found using the z-table, and it typically is around 2.576. This means if a calculated z-score exceeds 2.576 or falls below -2.576, it lies in the critical region of our distribution and provides grounds to reject the null hypothesis.
z-table
A Z-table, or standard normal table, provides the cumulative probability associated with a particular z-score. It is a valuable tool in statistics for finding critical values and understanding the distribution of data relative to the mean.

When looking for the critical value \( z_{ \alpha / 2 } \) associated with a 99% confidence level, we use the Z-table to identify the z-score that corresponds to the cumulative probability of 0.995. The area to the left of this critical value includes 99.5% of the distribution (100% minus the upper 0.005 tail area).

By consulting the Z-table, we find that the z-score for a 0.995 cumulative probability is approximately 2.576. This means that for a 99% confidence interval, the values beyond ±2.576 standard deviations from the mean fall in the 1% critical region.

Utilizing the Z-table, therefore, helps in determining precise critical values for various confidence levels, enhancing the reliability of our statistical inferences. Whether dealing with confidence intervals or hypothesis testing, the Z-table is an indispensable reference.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

The data in the next column represent the age (in weeks) at which babies first crawl based on a survey of 12 mothers conducted by Essential Baby. In Problem 34 from Section \(9.2,\) it was verified that the data are normally distributed and that \(s=10.00\) weeks. Construct and interpret a \(95 \%\) confidence interval for the population standard deviation of the age (in weeks) at which babies first crawl. $$ \begin{array}{llllll} \hline 52 & 30 & 44 & 35 & 39 & 26 \\ \hline 47 & 37 & 56 & 26 & 39 & 28 \\ \hline \end{array} $$

Sleep apnea is a disorder in which you have one or more pauses in breathing or shallow breaths while you sleep. In a cross-sectional study of 320 individuals who suffer from sleep apnea, it was found that 192 had gum disease. Note: In the general population, about \(17.5 \%\) of individuals have gum disease. (a) What does it mean for this study to be cross-sectional? (b) What is the variable of interest in this study? Is it qualitative or quantitative? Explain. (c) Estimate the proportion of individuals who suffer from sleep apnea who have gum disease with \(95 \%\) confidence. Interpret your result.

The Sullivan Statistics Survey I asks, "Would you be willing to pay higher taxes if the tax revenue went directly toward deficit reduction?" Treat the survey respondents as a random sample of adult Americans. Go to www.pearsonhighered.com/sullivanstats to obtain the data file SullivanSurveyI using the file format of your choice for the version of the text you are using. The column "Deficit" has survey responses. Construct and interpret a \(90 \%\) confidence interval for the proportion of adult Americans who would be willing to pay higher taxes if the revenue went directly toward deficit reduction.

42\. Sample Size Dr. Paul Oswiecmiski wants to estimate the mean serum HDL cholesterol of all 20 - to 29 -year-old males. How many subjects are needed to estimate the mean serum HDL cholesterol of all 20 - to 29 -year-old males within 1.5 points with \(90 \%\) confidence, assuming that \(s=12.5\) based on earlier studies? Suppose that Dr. Oswiecmiski would prefer \(95 \%\) confidence. How does the increase in confidence affect the sample size required?

A random sample of 1003 adult Americans was asked, "Do you pretty much think televisions are a necessity or a luxury you could do without?" Of the 1003 adults surveyed, 521 indicated that televisions are a luxury they could do without (a) Obtain a point estimate for the population proportion of adult Americans who believe that televisions are a luxury they could do without. (b) Verify that the requirements for constructing a confidence interval about \(p\) are satisfied. (c) Construct and interpret a \(95 \%\) confidence interval for the population proportion of adult Americans who believe that televisions are a luxury they could do without. (d) Is it possible that a supermajority (more than \(60 \%\) ) of adult Americans believe that television is a luxury they could do without? Is it likely? (e) Use the results of part (c) to construct a \(95 \%\) confidence interval for the population proportion of adult Americans who believe that televisions are a necessity.

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.