/*! 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 Find the critical value \(z_{\al... [FREE SOLUTION] | 91Ó°ÊÓ

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

Short Answer

Expert verified
The critical value \(z_{0.005}\) for a 99% confidence level is \(2.5758\).

Step by step solution

01

Understand the Confidence Level

The problem asks for the critical value corresponding to a 99% confidence level. This means that 99% of the data lies within the confidence interval, leaving 0.5% on each tail of the normal distribution.
02

Determine Alpha (\( \boldsymbol{\boldsymbol{\theta}} \boldsymbol{\boldsymbol{\theta}} \boldsymbol{\boldsymbol{\theta}}alpha\theta\boldsymbol{})).

The alpha (α) can be calculated by subtracting the confidence level from 1. This gives us: \(\theta = 1 - 0.99 = 0.01\). Therefore, for the two tails, \( \frac{\theta}{2} = 0.005\).
03

Use Standard Normal Distribution Table

Locate the critical value \(z_{\theta / 2}\) for the tail probability \(0.005\) in the standard normal distribution table (z-table). For \(z_{0.005}\), the corresponding z-value is \(2.5758\).
04

Interpret the Result

The critical value found from the z-table, \(2.5758\), represents the z-score where 0.5% of the distribution is beyond it in each tail, capturing 99% of the data within the interval.

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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, often expressed as a percentage, indicates how certain we can be that a parameter lies within the confidence interval derived from our sample data.
For example, a confidence level of 99% means that if we repeated an experiment or study 100 times, we would expect the true parameter to fall within the calculated interval 99 times out of 100.
Higher confidence levels provide more certainty but result in wider confidence intervals, while lower confidence levels generate narrower intervals.
This is essential when making decisions based on statistical data as it provides a measure of reliability for our estimates.
Alpha Level
The alpha level (α) is the probability of rejecting the null hypothesis when it is actually true. It's also known as the significance level and complements the confidence level.
Mathematically, the alpha level can be found by subtracting the confidence level from 1. For a 99% confidence level, the alpha level would be \(\theta = 1 - 0.99 = 0.01\).ewline The alpha level is then divided by 2, because it is split equally between the two tails of the normal distribution. So in this case, \(\frac{\theta}{2} = 0.005\).
It's used to determine critical values in hypothesis tests; small alpha levels mean stricter criteria for rejecting the null hypothesis, reducing the risk of Type I errors (false positives).
Normal Distribution
The normal distribution, often called the bell curve, is a continuous probability distribution that is symmetrical around the mean.
Most data points are concentrated around the mean, and the probabilities for values taper off as you move further from the mean.
In this distribution, approximately 68% of the data lies within one standard deviation from the mean, 95% within two, and 99.7% within three, following the Empirical Rule.
In our exercise, we use the normal distribution to determine critical values by locating specific z-scores that correspond to defined tail probabilities, like 0.005 for each tail in a 99% confidence level scenario.
These critical values help us capture the middle percentage of the data, giving us the z-scores needed for constructing confidence intervals.

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Most popular questions from this chapter

Comparing Waiting Lines a. The values listed below are waiting times (in minutes) of customers at the Jefferson Valley Bank, where customers enter a single waiting line that feeds three teller windows. Construct a \(95 \%\) confidence interval for the population standard deviation \(\sigma\). $$ \begin{array}{llllllllll} 6.5 & 6.6 & 6.7 & 6.8 & 7.1 & 7.3 & 7.4 & 7.7 & 7.7 & 7.7 \end{array} $$ b. The values listed below are waiting times (in minutes) of customers at the Bank of Providence, where customers may enter any one of three different lines that have formed at three teller windows. Construct a \(95 \%\) confidence interval for the population standard deviation \(\sigma\). $$ \begin{array}{llllllllll} 4.2 & 5.4 & 5.8 & 6.2 & 6.7 & 7.7 & 7.7 & 8.5 & 9.3 & 10.0 \end{array} $$ c. Interpret the results found in parts (a) and (b). Do the confidence intervals suggest a difference in the variation among waiting times? Which arrangement seems better: the single-line system or the multiple-line system?

The drug Eliquis (apixaban) is used to help prevent blood clots in certain patients. In clinical trials, among 5924 patients treated with Eliquis, 153 developed the adverse reaction of nausea (based on data from Bristol-Myers Squibb Co.). Construct a \(99 \%\) confidence interval for the proportion of adverse reactions.

The Wechsler IQ test is designed so that the mean is 100 and the standard deviation is 15 for the population of normal adults. Find the sample size necessary to estimate the mean IQ score of college professors. We want to be \(99 \%\) confident that our sample mean is within 4 IQ points of the true mean. The mean for this population is clearly greater than 100 . The standard deviation for this population is less than 15 because it is a group with less variation than a group randomly selected from the general population; therefore, if we use \(\sigma=15\) we are being conservative by using a value that will make the sample size at least as large as necessary. Assume then that \(\sigma=15\) and determine the required sample size. Does the sample size appear to be practical?

A sociologist plans to conduct a survey to estimate the percentage of adults who believe in astrology. How many people must be surveyed if we want a confidence level of \(99 \%\) and a margin of error of four percentage points? a. Assume that nothing is known about the percentage to be estimated. b. Use the information from a previous Harris survey in which \(26 \%\) of respondents said that they believed in astrology.

In a survey of 1002 people, \(70 \%\) said that they voted in a recent presidential election (based on data from ICR Research Group). Voting records show that \(61 \%\) of eligible voters actually did vote. a. Find a \(98 \%\) confidence interval estimate of the proportion of people who say that they voted. b. Are the survey results consistent with the actual voter turnout of \(61 \%\) ? Why or why not?

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