/*! 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 9 \(t\) -models, part I Using the ... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

\(t\) -models, part I Using the \(t\) -tables, software, or a calculator, estimate a. the critical value of \(t\) for a \(90 \%\) confidence interval with \(d f=17\) b. the critical value of \(t\) for a \(98 \%\) confidence interval with \(d f=88\)

Short Answer

Expert verified
The critical value of t for a 90% confidence interval with df of 17 is approximately 1.740 and for a 98% confidence interval with df of 88 is approximately 2.626.

Step by step solution

01

Identify the degrees of freedom and confidence interval

First, identify the given degrees of freedom (df) and the confidence intervals for which the critical values need to be found. In part (a), df is 17 and the confidence interval is 90%. In part (b), df is 88 and the confidence interval is 98%.
02

Use a T-table, calculator, or software to find the critical value

Next, you should use the identified degree of freedom and the corresponding confidence interval to look up the critical value in a t-table, or by using a calculator or statistical software. The critical value is the t-value such that the area to its right is equal to (1 - confidence level)/2.
03

Find the critical value for 90% confidence interval with df of 17

For the 90% confidence level and df of 17, the critical value is approximately 1.740.
04

Find the critical value for 98% confidence interval with df of 88

For the 98% confidence level and df of 88, the critical value is approximately 2.626.

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.

T-distribution
When working with sample data to estimate population parameters, statisticians use various distributions. One particularly important distribution is the t-distribution, which resembles the normal distribution but accounts for more variability. This is especially relevant when dealing with small sample sizes.

The t-distribution is a family of curves, each determined by its degrees of freedom (df), which relates to the sample size. For a given sample size, the degrees of freedom are typically the sample size minus one. As the df increases, the t-distribution curve gets closer to the normal distribution curve.

In practical terms, if we're estimating population means from small samples, the t-distribution gives more accurate probabilities than the normal distribution. This is crucial for tasks like constructing confidence intervals and hypothesis testing where precise probability estimations are needed.
Confidence Intervals
The concept of confidence intervals is fundamental in statistics as it offers a range of values that is likely to contain a population parameter, such as the mean, with a certain level of confidence.

For instance, a 90% confidence interval would mean that if you were to draw multiple samples and compute the confidence interval for each, about 90% of the intervals would contain the true population mean. It's like saying, 'We are 90% confident the mean falls within this range.' However, it doesn't mean there is a 90% chance the specific interval calculated from our sample contains the mean. It's about the confidence in the process, not in the interval itself.

To construct a confidence interval for the mean, using the t-distribution, one needs the mean of the sample, the standard error of the mean (which includes the sample standard deviation and size), and the critical value of t for the desired confidence level and degrees of freedom.
Degrees of Freedom
The term degrees of freedom might seem abstract, but it's a pivotal part of estimating population parameters. In statistics, degrees of freedom (df) refer to the number of independent values in a calculation that are free to vary.

Understanding degrees of freedom helps in determining the appropriate distribution to use, like the t-distribution. For example, when calculating a sample standard deviation, one value is lost to the sample mean, resulting in n - 1 degrees of freedom, where n is the sample size.

Why does this matter? Because the degrees of freedom affect the shape of the t-distribution, and thus the critical value of t. For fewer degrees of freedom, the t-distribution will be more spread out with heavier tails, indicating more uncertainty. As the degrees of freedom increase, the curve becomes more bell-shaped, akin to the normal distribution.
Statistical Tables
Statistical tables are handy tools that contain pre-calculated values for various statistical distributions, including the t-distribution. Before computers, these tables were crucial for performing statistical calculations. Even today, they serve as a quick reference guide for finding values like critical points for given degrees of freedom and confidence levels.

For example, to find the critical value of t for a 90% confidence interval with 17 degrees of freedom, we refer to a t-table. We go down the column for the desired confidence level and across the row for the given df. The table gives us the corresponding t-value, which is the point beyond which lies 5% of the distribution's area on each tail for a two-tailed test.

While software and calculators now perform these calculations instantly, understanding how to read these tables helps grasp the principles behind statistical testing and confidence interval estimation. It is still a valuable skill for interpreting and checking results against the potential errors of automated tools.

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

GPAs A college's data about the incoming freshmen indicate that the mean of their high school GPAs was \(3.4,\) with a standard deviation of 0.35 ; the distribution was roughly mound-shaped and only slightly skewed. The students are randomly assigned to freshman writing seminars in groups of 25\. What might the mean GPA of one of these seminar groups be? Describe the appropriate sampling distribution model-shape, center, and spread-with attention to assumptions and conditions. Make a sketch using the \(68-95\) \(99.7 \mathrm{P}\)

Teachers Software analysis of the salaries of a random sample of 288 Nevada teachers produced the confidence interval shown below. Which conclusion is correct? What's wrong with the others? with \(90.00 \%\) Confidence, \(t\) -interval for \(\mu: 43454<\mu(\) TchPay \()<45398\) a. If we took many random samples of 288 Nevada teachers, about 9 out of 10 of them would produce this confidence interval. b. If we took many random samples of Nevada teachers, about 9 out of 10 of them would produce a confidence interval that contained the mean salary of all Nevada teachers. c. About 9 out of 10 Nevada teachers earn between \(\$ 43,454\) and \(\$ 45,398 .\) d. About 9 out of 10 of the teachers surveyed earn between \(\$ 43,454\) and \(\$ 45,398\). e. We are \(90 \%\) confident that the average teacher salary in the United States is between \(\$ 43,454\) and \(\$ 45,398\).

\(t\) -models, part IV Describe how the critical value of \(t\) for a \(95 \%\) confidence interval changes as the number of degrees of freedom increases.

Cattle Livestock are given a special feed supplement to see if it will promote weight gain. Researchers report that the 77 cows studied gained an average of 56 pounds, and that a \(95 \%\) confidence interval for the mean weight gain this supplement produces has a margin of error of ±11 pounds. Some students wrote the following conclusions. Did anyone interpret the interval correctly? Explain any misinterpretations. a. \(95 \%\) of the cows studied gained between 45 and 67 pounds. b. We're \(95 \%\) sure that a cow fed this supplement will gain between 45 and 67 pounds. c. We're \(95 \%\) sure that the average weight gain among the cows in this study was between 45 and 67 pounds. d. The average weight gain of cows fed this supplement will be between 45 and 67 pounds \(95 \%\) of the time. e. If this supplement is tested on another sample of cows, there is a \(95 \%\) chance that their average weight gain will be between 45 and 67 pounds.

Tips A waiter believes the distribution of his tips has a model that is slightly skewed to the right, with a mean of \(\$ 9.60\) and a standard deviation of \(\$ 5.40 .\) a. Explain why you cannot determine the probability that a given party will tip him at least \(\$ 20\). b. Can you estimate the probability that the next 4 parties will tip an average of at least \(\$ 15 ?\) Explain. c. Is it likely that his 10 parties today will tip an average of at least \(\$ 15 ?\) Explain.

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.