/*! 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 106 A \(95 \%\) confidence interval ... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

A \(95 \%\) confidence interval for \(\mu\) using the sample results \(\bar{x}=84.6, s=7.8,\) and \(n=42\)

Short Answer

Expert verified
The 95% confidence interval for the given sample data is calculated to be approximately \[84.6 \pm 1.96(\frac{7.8}{\sqrt{42}})\].

Step by step solution

01

Identify given values

Identify and write down the given values. The sample mean (\(\bar{x}\)) is 84.6, the sample standard deviation (s) is 7.8, and the sample size (n) is 42. The confidence level is 95%.
02

Calculate the standard error

Calculate the standard error (SE). The standard error is calculated as the standard deviation divided by the square root of the sample size, i.e., \( SE = \frac{s}{\sqrt{n}} = \frac{7.8}{\sqrt{42}}\).
03

Determine Z score

At a 95% confidence level in a normal distribution, the Z score (Z) is ±1.96. This is the Z value that cuts off the top 2.5% and bottom 2.5% of the data, leaving 95% in the middle.
04

Calculate the Confidence Interval

Calculate the confidence interval for μ using the formula \(\bar{x} \pm Z(SE)\). Substituting the values we get, Confidence Interval is \[84.6 \pm 1.96(\frac{7.8}{\sqrt{42}})\]

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.

Standard Error
To understand the concept of Standard Error (SE), think about how well your sample represents the larger population. When you have a collection of data points, like scores or measurements in a study, they can describe a larger group. The sample standard deviation (s) gives an idea of how spread out the data points are from the average within the sample. The Standard Error then tells us how much these sample means would differ from the true population mean if we took numerous samples.The formula for calculating the Standard Error (SE) is:\[ SE = \frac{s}{\sqrt{n}} \]where:
  • s: Sample standard deviation.
  • n: Sample size, the number of observations in the sample.
By dividing the standard deviation by the square root of the sample size, the Standard Error scales the sample's variability, showing how "noisy" our sample is. A lower SE suggests that the sample mean is a more accurate reflection of the population mean. For larger samples, the SE generally decreases, making our estimate more precise.
Z Score
The Z Score is a crucial part of determining a confidence interval. Think of it as a conversion of your data into standard deviation units. The Z Score helps to understand where a data point fits into a normal distribution by saying how far away it is, in standard deviations, from the mean. For a confidence interval calculation, a Z Score shows how many standard deviations above or below the population mean our sample mean might lie, considering the desired confidence level. For a standard normal distribution:
  • Z Score: At a 95% confidence interval, the associated Z Score is ±1.96. This tells us that the range within 1.96 standard deviations on either side of the mean will capture 95% of the sample means.
In practical terms, when calculating confidence intervals: 1. Multiply the Standard Error by 1.96 for a 95% confidence level. 2. Then, add and subtract this product from the sample mean to find the range of the confidence interval. The Z Score's value changes depending on the confidence level chosen. For example, a 99% confidence level would require a larger Z Score since you want more certainty that the true population mean is captured.
Sample Mean
The Sample Mean, denoted as \( \bar{x} \), is the average value of all the collected data points in a sample. Calculating the sample mean is straightforward and involves adding all the data points together and dividing them by the number of points, or observations, in the sample.In the context of a confidence interval, the Sample Mean serves as the midpoint of the interval. It acts as the best estimate of the population mean we are trying to understand.Here's how you calculate the sample mean:\[ \bar{x} = \frac{\sum x_i}{n} \]where:
  • \( \sum x_i \): Sum of all sample observations.
  • n: Total number of observations.
This concept is significant as it uses a sample to make an inference about the larger population from which the sample is drawn.The accuracy and relevance of this estimate heavily rely on the representativeness and size of the sample.The larger and more representative your sample, the more reliable your inference about the population mean will be.

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

In Exercises 6.1 to \(6.6,\) if random samples of the given size are drawn from a population with the given proportion: (a) Find the mean and standard error of the distribution of sample proportions. (b) If the sample size is large enough for the Central Limit Theorem to apply, draw a curve showing the shape of the sampling distribution. Include at least three values on the horizontal axis. Samples of size 30 from a population with proportion 0.27

Difference in mean commuting distance (in miles) between commuters in Atlanta and commuters in St. Louis, using \(n_{1}=500, \bar{x}_{1}=18.16,\) and \(s_{1}=13.80\) for Atlanta and \(n_{2}=500, \bar{x}_{2}=14.16,\) and \(s_{2}=10.75\) for St. Louis.

A sample with \(n=10, \bar{x}=508.5,\) and \(s=21.5\)

Percent of Free Throws Made Usually, in sports, we expect top athletes to get better over time. We expect future athletes to run faster, jump higher, throw farther. One thing has remained remarkably constant, however. The percent of free throws made by basketball players has stayed almost exactly the same for 50 years. \(^{5}\) For college basketball players, the percent is about \(69 \%,\) while for players in the NBA (National Basketball Association) it is about \(75 \%\). (The percent in each group is also very similar between male and female basketball players.) In each case below, find the mean and standard deviation of the distribution of sample proportions of free throws made if we take random samples of the given size. (a) Samples of 100 free throw shots in college basketball (b) Samples of 1000 free throw shots in college basketball (c) Samples of 100 free throw shots in the \(\mathrm{NBA}\) (d) Samples of 1000 free throw shots in the \(\mathrm{NBA}\)

Random samples of the given sizes are drawn from populations with the given means and standard deviations. For each scenario: (a) Find the mean and standard error of the distribution of differences in sample means, \(\bar{x}_{1}-\bar{x}_{2}\) (b) If the sample sizes are large enough for the Central Limit Theorem to apply, draw a curve showing the shape of the sampling distribution. Include at least three values on the horizontal axis. Samples of size 100 from Population 1 with mean 87 and standard deviation 12 and samples of size 80 from Population 2 with mean 81 and standard deviation 15

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.