/*! 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 32 Find \(z^{\circ}\) for a \(93 \%... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Find \(z^{\circ}\) for a \(93 \%\) confidence interval using Table A or your calculator. Show your method

Short Answer

Expert verified
The critical value \(z^{\circ}\) for a 93% confidence interval is approximately 1.81.

Step by step solution

01

Understanding Confidence Intervals

The confidence interval level is given as 93%. This means we are interested in finding the critical value, typically denoted as \(z^{\circ}\), that corresponds to this confidence level for a standard normal distribution.
02

Identify the Significance Level

The confidence level is 93%, so the total area in the tails is \(1 - 0.93 = 0.07\). This area is split between the two tails of the normal distribution, giving us \(0.035\) in each tail. Therefore, we need to find the \(z^{\circ}\) value such that the area to the left is \(0.965 = 1 - 0.035\).
03

Locate the Critical Value Using a Z-table or Calculator

Using a standard normal distribution table (Z-table) or a calculator that provides cumulative normal distribution functions, find the \(z^{\circ}\) value corresponding to a cumulative probability of 0.965.
04

Find the Z-score

Consult the Z-table or use a calculator. If the cumulative probability is 0.965, the corresponding \(z^{\circ}\) value that fulfills this is approximately \(z^{\circ} = 1.81\).

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.

Critical Value
The critical value is a key component in constructing confidence intervals. It represents the point on a standard normal distribution beyond which a given percentage of data falls.
For instance, if you're working with a 93% confidence interval, you're essentially saying that 93% of your data lies within these bounds, while the remaining 7% is in the tails (3.5% in each tail in this symmetrical distribution case). Thus, the critical value helps in determining how far from the mean our confidence interval should extend.
In this exercise, the critical value, denoted as \( z^{\circ} \) for a 93% confidence level, is approximately 1.81. This means that 93% of data under the standard normal distribution spreads out from \(-1.81\) to \(1.81\) around the mean, covering the central portion of the dataset.
Significance Level
The significance level is directly related to the concept of confidence intervals. While a 93% confidence level tells us the proportion of data captured by the interval, the significance level quantifies the proportion that falls outside.
Mathematically, it is calculated as \(1 - \text{confidence level}\). For a 93% confidence level, this gives us a significance level of 7% or \(0.07\).
This 7% is split between the two tails of the distribution: 3.5% on the left and 3.5% on the right. These tails represent the area where data points are considered to be significantly different from what is expected under the normal curve.
Standard Normal Distribution
The standard normal distribution is a special case of the normal distribution, with a mean of 0 and a standard deviation of 1. It's a foundational concept in statistics, often used as a basis for standardizing other distributions.
This distribution allows comparison of different datasets on a common scale, helping us find probabilities and data points such as our critical values using standardized tools like the Z-table.
When working with confidence intervals, the standard normal distribution provides a baseline where 68% of data falls within one standard deviation, 95% within two, and so forth. In the context of our problem, it helps locate the 93% confidence level across its span, marking the extremes with our critical value of \( \pm 1.81 \).
Z-table
The Z-table is an essential tool in statistics that aids in finding the critical values in a standard normal distribution. It lists the cumulative probabilities associated with standard Z-scores.
When constructing confidence intervals like in our exercise, you determine which Z-score corresponds to a particular cumulative probability (e.g., 0.965 for 93% confidence) by consulting the Z-table.
For instance, a Z-score of 1.81 aligns with 0.965 probability, indicating that approximately 96.5% of data falls below this point, leaving 3.5% in the tail on one side. Using this tool ensures accuracy in statistical analysis, providing a straightforward way to interpret complex data distributions.

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

When two lights close together blink alternately, we "see" one light moving back and forth if the time between blinks is short. What is the longest interval of time between blinks that preserves the illusion of motion? Ask subjects to turn a knob that slows the blinking until they "see" two lights rather than one light moving. A report gives the results in the form "mean plus or minus the standard error of the mean." \({ }^{22}\) Data for 12 subjects are summarized as \(251 \pm 45\) (in milliseconds). (a) Find the sample standard deviation \(s_{x}\) for these measurements. Show your work. (b) A hasty reader believes that the interval given in the report is a \(95 \%\) confidence interval for the population mean. Find the actual confidence level for the given interval.

High school students who take the SAT Math exam a second time generally score higher than on their first try. Past data suggest that the score increase has a standard deviation of about 50 points. How large a sample of high school students would be needed to estimate the mean change in SAT score to within 2 points with \(95 \%\) confidence? Show your work.

A medical study finds that \(\bar{x}=114.9\) and \(s_{x}=9.3\) for the seated systolic blood pressure of the 27 members of one treatment group. What is the standard error of the mean? Interpret this value in context.

Check whether each of the conditions is met for calculating a confidence interval for the population proportion \(\bar{p}\). The small round holes you often see in sea shells were drilled by other sea creatures, who ate the former dwellers of the shells. Whelks often drill into mussels, but this behavior appears to be more or less common in different locations. Researchers collected whelk eggs from the coast of Oregon, raised the whelks in the laboratory, then put each whelk in a container with some delicious mussels. Only 9 of 98 whelks drilled into a mussel. \({ }^{11}\) The researchers want to estimate the proportion \(p\) of Oregon whelks that will spontaneously drill into mussels.

You have an SRS of 23 observations from a large population. The distribution of sample values is roughly symmetric with no outliers. What critical value would you use to obtain a \(98 \%\) confidence interval for the mean of the population? (a) 2.177 (b) 2.183 (c) 2.326 (d) 2.500 (e) 2.508

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.