/*! 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 24 Determine the \(t\) -percentile ... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Determine the \(t\) -percentile that is required to construct each of the following one-sided confidence intervals: (a) Confidence level \(=95 \%,\) degrees of freedom \(=14\) (b) Confidence level \(=99 \%,\) degrees of freedom \(=19\) (c) Confidence level \(=99.9 \%\), degrees of freedom \(=24\)

Short Answer

Expert verified
(a) 1.761, (b) 2.539, (c) 3.745

Step by step solution

01

Understand the Problem

We need to find the critical t-value (or t-percentile) for each of the given confidence levels and degrees of freedom (df). This value is needed to construct a one-sided confidence interval.
02

Determine t-Percentile Requirements

For a one-sided confidence interval with a confidence level of x%, the t-percentile corresponds to 1 - x% in the t-distribution table. For example, a 95% confidence level means looking up the 0.05 (1 - 0.95) percentile.
03

Find t-Value for (a)

With a 95% confidence level and 14 degrees of freedom, consult a t-distribution table to find the t-value at the 0.05 significance level. This value is known as \( t_{0.05,14} \).
04

Look Up t-Value in t-Table for (a)

For 14 degrees of freedom and a 0.05 significance level, the t-value is approximately 1.761.
05

Find t-Value for (b)

With a 99% confidence level and 19 degrees of freedom, use a t-distribution table to find the t-value at the 0.01 significance level. This is \( t_{0.01,19} \).
06

Look Up t-Value in t-Table for (b)

For 19 degrees of freedom and a 0.01 significance level, the t-value is approximately 2.539.
07

Find t-Value for (c)

With a 99.9% confidence level and 24 degrees of freedom, check a t-distribution table for the t-value at the 0.001 significance level. This is \( t_{0.001,24} \).
08

Look Up t-Value in t-Table for (c)

For 24 degrees of freedom and a 0.001 significance level, the t-value is approximately 3.745.

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 Interval
A confidence interval gives us a range in which we expect the true value of a parameter to fall, with a certain level of confidence. When dealing with statistics, this parameter is often the mean of a population. A one-sided confidence interval, as used in the problem, is interested in finding either an upper or lower limit rather than both. Thus, a 95% confidence interval means we are 95% sure the population parameter lies below a certain point.

To construct a confidence interval, we need several components: the mean of the sample data, the standard deviation, and crucially, the t-value corresponding to the desired confidence level and degrees of freedom.

The lower or upper bound of the confidence interval is calculated as the sample mean plus or minus this t-value multiplied by the standard deviation divided by the square root of the sample size.
T-Value
The t-value is a critical number from the t-distribution that helps construct confidence intervals. It allows us to account for variability and sample size when estimating a population parameter. It serves as a multiplier to ensure that our confidence interval has the correct width to be statistically valid.

When searching for a t-value, we look for it in a t-distribution table. Each row of this table corresponds to degrees of freedom, and each column corresponds to a significance level, which is the complement of the confidence level (1 - confidence level).

For instance, when finding the t-value for a 95% confidence level with 14 degrees of freedom, we search for the value corresponding to a 0.05 significance level. This value tells us how many sample standard errors away from the mean you must go to achieve your desired confidence level.
Degrees of Freedom
Degrees of freedom (df) is a fundamental concept in statistics, often related to the number of values in a calculation that are free to vary. It is critical in determining the appropriate t-value from the t-distribution table. The degrees of freedom typically equal the total number of observations minus one for a single sample.

For example, if you have a sample size of 15 data points, the degrees of freedom would be 14. Higher degrees of freedom generally mean a closer approximation to the normal distribution.

The concept is essential because the shape of the t-distribution changes with the degrees of freedom. More degrees of freedom result in a narrower and taller t-distribution. Thus, the degrees of freedom influence the t-value we use in constructing confidence intervals.
T-Percentile
T-percentile is another term for the critical t-value found using a t-distribution table. It indicates the probability that a value from the t-distribution is less than or equal to a specific t-value. This concept helps us determine the boundaries of confidence intervals.

In a one-sided confidence interval, finding the t-percentile involves identifying the t-value that corresponds to the complement of the confidence level (1 - confidence level). This value determines whether a sample mean significantly differs from the population mean or whether the sample data is consistent with the assumed mean.

For example, in a 99% one-sided confidence interval with 19 degrees of freedom, you will find the t-percentile at a 0.01 level, balancing out the 1% chance of exceeding the limits of the interval. Thus, t-percentiles are an indispensable part of hypothesis testing and confidence interval construction.

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

A postmix beverage machine is adjusted to release a certain amount of syrup into a chamber where it is mixed with carbonated water. A random sample of 25 beverages was found to have a mean syrup content of \(\bar{x}=1.10\) fluid ounce and a standard deviation of \(s=0.015\) fluid ounce. Find a \(95 \%\) CI on the mean volume of syrup dispensed.

A random sample has been taken from a normal distribution and the following confidence intervals constructed using the same data: (38.02,61.98) and (39.95,60.05) (a) What is the value of the sample mean? (b) One of these intervals is a \(95 \% \mathrm{CI}\) and the other is a \(90 \%\)

The brightness of a television picture tube can be evaluated by measuring the amount of current required to achieve a particular brightness level. A sample of 10 tubes results in \(\bar{x}=317.2\) and \(s=15.7\). Find (in microamps) a \(99 \%\) confidence interval on mean current required. State any necessary assumptions about the underlying distribution of the data.

The life in hours of a 75 -watt light bulb is known to be normally distributed with \(\sigma=25\) hours. A random sample of 20 bulbs has a mean life of \(\bar{x}=1014\) hours. (a) Construct a \(95 \%\) two-sided confidence interval on the mean life (b) Construct a \(95 \%\) lower-confidence bound on the mean life. Compare the lower bound of this confidence interval with the one in part (a).

The yield of a chemical process is being studied. From previous experience, yield is known to be normally distributed and \(\sigma=3\). The past five days of plant operation have resulted in the following percent yields: \(91.6,88.75,90.8,89.95,\) and \(91.3 .\) Find a \(95 \%\) two-sided confidence interval on the true mean yield.

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.