/*! 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 35 The formula used to calculate a ... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 12: Problem 35

The formula used to calculate a confidence interval for the mean of a normal population when \(n\) is small is $$ \bar{x} \pm(t \text { critical value }) \frac{s}{\sqrt{n}} $$ What is the appropriate \(t\) critical value for each of the following confidence levels and sample sizes? a. \(95 \%\) confidence, \(n=15\) b. \(99 \%\) confidence, \(n=20\) c. \(90 \%\) confidence, \(n=26\)

Short Answer

Expert verified
For the given confidence levels and sample sizes, the appropriate t critical values are: a. \(95\%\) confidence, \(n=15\): \(t=2.145\) b. \(99\%\) confidence, \(n=20\): \(t=2.861\) c. \(90\%\) confidence, \(n=26\): \(t=1.708\)

Step by step solution

01

Understanding the formula

The formula given calculates the confidence interval for the mean of a normal population when the sample size \(n\) is small: \[ \bar{x} \pm (t \text { critical value }) \frac{s}{\sqrt{n}} \] Here, \(\bar{x}\) is the sample mean, \(s\) is the sample standard deviation, and \(n\) is the sample size. The t critical value depends on the desired confidence level and the degrees of freedom, which is given by \(n-1\).
02

Using a t-distribution table

To find the appropriate t critical values for each of the given confidence levels and sample sizes, we will use a t-distribution table. We need to look up the t critical value based on the desired confidence level (in the column headings) and the degrees of freedom (in the row headings). a. \(95\%\) confidence, \(n=15\) Degrees of freedom: \(n - 1 = 15 - 1 = 14\) Using a t-distribution table, the corresponding t critical value is: \(t_{14, 0.975} = 2.145\) b. \(99\%\) confidence, \(n=20\) Degrees of freedom: \(n - 1 = 20 - 1 = 19\) Using a t-distribution table, the corresponding t critical value is: \(t_{19, 0.995} = 2.861\) c. \(90\%\) confidence, \(n=26\) Degrees of freedom: \(n - 1 = 26 - 1 = 25\) Using a t-distribution table, the corresponding t critical value is: \(t_{25, 0.95} = 1.708\)
03

Writing the final answers

a. For a \(95\%\) confidence level and \(n=15\), the appropriate t critical value is \(t=2.145\). b. For a \(99\%\) confidence level and \(n=20\), the appropriate t critical value is \(t=2.861\). c. For a \(90\%\) confidence level and \(n=26\), the appropriate t critical value is \(t=1.708\).

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 91影视!

Key Concepts

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

Understanding the T-Distribution
The t-distribution is an important concept when dealing with small sample sizes in statistics. Unlike the normal distribution, the t-distribution adjusts for the added uncertainty associated with estimating the population standard deviation from a small sample.
This distribution is symmetric and bell-shaped, like the normal distribution, but has heavier tails. Those heavier tails are what allow for better estimates when dealing with smaller samples.
  • The t-distribution is primarily used in hypothesis testing and in calculating confidence intervals.
  • It is especially useful in cases where the sample size is less than 30, or when the population standard deviation is unknown.
  • As the sample size increases, the t-distribution begins to resemble the normal distribution more closely.
To choose the correct t-distribution, we rely on degrees of freedom, which makes it flexible for different sample sizes and confidence levels.
Determining Appropriate Sample Sizes
Sample size, noted as \(n\), plays a critical role in statistical analysis since it affects the reliability of our estimates. Smaller samples have less representation of the population, therefore increasing variability.
When calculating confidence intervals, sample size affects the width of the interval; smaller samples typically result in wider confidence intervals, reflecting the higher uncertainty in our estimations.
  • In the t-distribution formula \(\bar{x} \pm (t \text{ critical value}) \frac{s}{\sqrt{n}}\), \(n\) is crucial in determining the margin of error \((ME)\), which influences the interval width.
  • The larger the sample size, the more the t-distribution approaches a normal distribution and the less impact the sample size has on the margin of error.
  • In practical terms, selecting an appropriate sample size is often a balance between resource limitations and desired confidence in results.
Understanding the impact of sample size helps in interpreting statistical results and making informed decisions based on statistical analysis.
Role of Degrees of Freedom in a Sample
Degrees of freedom (df) are a concept used in statistics that refer to the number of independent values or quantities that can vary in an analysis, while still adhering to certain constraints. When creating confidence intervals with the t-distribution, degrees of freedom are calculated as \(n-1\), where \(n\) is the sample size.
Degrees of freedom impact the shape of the t-distribution:
  • As degrees of freedom increase, the t-distribution curve becomes closer to a standard normal distribution curve.
  • A lower number of degrees of freedom results in a wider and flatter curve, indicating more variability and uncertainty.
  • The degrees of freedom adjust the t critical value, thereby influencing the resulting confidence interval's precision.
This adjustment ensures that the confidence intervals used are suitable for the specific data involved, maintaining their accuracy across different sample sizes.

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 authors of the paper "Driving Performance While Using a Mobile Phone: A Simulation Study of Greek Professional Drivers" (Transportation Research Part \(F\) [2016]: 164-170) describe a study to evaluate the effect of mobile phone use by taxi drivers in Greece. Fifty taxi drivers drove in a driving simulator where they were following a lead car. The drivers were asked to carry on a conversation on a mobile phone while driving, and the following distance (the distance between the taxi and the lead car) was recorded. The sample mean following distance was 3.20 meters and the sample standard deviation was 1.11 meters. a. Construct and interpret a \(95 \%\) confidence interval for \(\mu,\) the population mean following distance while talking on a mobile phone for the population of taxi drivers. b. What assumption must be made in order to generalize this confidence interval to the population of all taxi drivers in Greece?

The paper referenced in the previous exercise also reported that for a representative sample of 68 second-year students at the university, the sample mean procrastination score was 41.00 and the sample standard deviation was 6.82 . a. Construct a \(95 \%\) confidence interval estimate of \(\mu,\) the population mean procrastination scale for second-year students at this college. b. How does the confidence interval for the population mean score for second- year students compare to the confidence interval for first-year students calculated in the previous exercise? What does this tell you about the difference between first-year and second-year students in terms of mean procrastination score?

What percentage of the time will a variable that has a \(t\) distribution with the specified degrees of freedom fall in the indicated region? a. 10 df, between -2.23 and 2.23 b. 24 df, between -2.80 and 2.80 c. 24 df, to the right of 2.80

Medical research has shown that repeated wrist extension beyond 20 degrees increases the risk of wrist and hand injuries. Each of 24 students at Cornell University used a proposed new computer mouse design, and while using the mouse, each student's wrist extension was recorded. Data consistent with summary values given in the paper "Comparative Study of Two Computer Mouse Designs" (Cornell Human Factors Laboratory Technical Report RP7992) are given. Use these data to test the hypothesis that the mean wrist extension for people using this new mouse design is greater than 20 degrees. Are any assumptions required in order for it to be appropriate to generalize the results of your test to the population of all Cornell students? To the population of all university students? $$ \begin{array}{lllllll} 27 & 28 & 24 & 26 & 27 & 25 & 25 \\ 24 & 24 & 24 & 25 & 28 & 22 & 25 \\ 24 & 28 & 27 & 26 & 31 & 25 & 28 \\ 27 & 27 & 25 & & & & \end{array} $$

Acrylic bone cement is sometimes used in hip and knee replacements to secure an artificial joint in place. The force required to break an acrylic bone cement bond was measured for six specimens, and the resulting mean and standard deviation were 306.09 Newtons and 41.97 Newtons, respectively. Assuming that it is reasonable to believe that breaking force has a distribution that is approximately normal, use a confidence interval to estimate the mean breaking force for acrylic bone cement.
See all solutions

Recommended explanations on Math Textbooks

Geometry

Read Explanation

Calculus

Read Explanation

Statistics

Read Explanation

Decision Maths

Read Explanation

Probability and Statistics

Read Explanation

Discrete Mathematics

Read Explanation
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.