/*! 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 10 Assume that all variables are ap... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 7: Problem 10

Assume that all variables are approximately normally distributed. The number of students who belong to the dance company at each of several randomly selected small universities is shown here. Estimate the true population mean size of a university dance company with \(99 \%\) confidence. $$ \begin{array}{llllllll} 21 & 25 & 32 & 22 & 28 & 30 & 29 & 30 \\ 47 & 26 & 35 & 26 & 35 & 26 & 28 & 28 \\ 32 & 27 & 40 & & & & & \end{array} $$

Short Answer

Expert verified
The 99% confidence interval for the true mean size of a university dance company is (20.52, 29.06).

Step by step solution

01

Understanding the Problem

We need to find a 99% confidence interval for the true population mean of the university dance company sizes. We assume that the number of students is normally distributed.
02

Calculate the Sample Mean

First, sum up all the provided student numbers and divide by the count of numbers to find the average. The numbers given are: 21, 25, 32, 22, 28, 30, 29, 30, 47, 26, 35, 26, 35, 26, 28, 28, 32, 27, and 40. Calculate the sum: \( 471 \), then divide by 19 (the number of data points): \[ \bar{x} = \frac{471}{19} = 24.79 \]
03

Calculate the Sample Standard Deviation

To find the sample standard deviation, first find the variance. For each number, subtract the sample mean and square the result, then average these squares (except the division here is by \(n-1\) since it's a sample). Calculate the squares and sum them up, then: \[ s^2 = \frac{1}{18} \sum (x_i - \bar{x})^2 = 41.7 \]\[ s = \sqrt{41.7} = 6.46 \]
04

Find the Critical Value

Since the sample size is less than 30, we use the \(t\)-distribution. With 18 degrees of freedom and 99% confidence, use a \(t\)-table or calculator to find the critical value \(t^*\). For 18 degrees of freedom, \(t^* \approx 2.878\).
05

Calculate the Confidence Interval

The formula for a confidence interval is:\[ \bar{x} \pm t^* \frac{s}{\sqrt{n}} \]Substituting the values we have:\[ 24.79 \pm 2.878 \frac{6.46}{\sqrt{19}} \approx 24.79 \pm 4.27 \]This gives the interval:\[ (20.52, 29.06) \]
06

Interpret the Confidence Interval

With 99% confidence, the true mean size of a university dance company is between 20.52 and 29.06 students.

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.

Normal Distribution
Many natural phenomena, when analyzed, tend to form a pattern that is symmetric and bell-shaped. This is what we call a normal distribution. In statistics, data that follow a normal distribution are often described by their mean and standard deviation. The majority of values cluster around the mean, creating the bell shape.

Features of a normal distribution include:
  • A central peak at the mean.
  • Symmetry around the mean.
  • A predictable spread of data around the mean, dictated by the standard deviation.
Assuming normal distribution is crucial because it allows us to use specific methods to estimate parameters like confidence intervals. In our case, since the number of students is assumed to follow a normal distribution, it means we can rely on methods like the t-distribution for our calculations.
Sample Mean
The sample mean is an estimate of the true population mean. It's a crucial part of inferential statistics because it provides a single value representing the center of the sampled data.

To calculate the sample mean:
  • Add up all the individual data points (here: 21, 25, 32, ..., 40).
  • Divide the sum by the number of data points.
For our example, the sum of the sample data was 471, and the number of data points was 19, giving a sample mean, \(\bar{x} = 24.79\). This value serves as our best estimate of the average size of a university dance company from the provided sample.
Sample Standard Deviation
The sample standard deviation is a measure of how spread out the sample data are from the sample mean. It indicates the amount of variation or dispersion in a set of values. A larger standard deviation means more variability in the data.

To find the sample standard deviation:
  • Subtract the sample mean from each data point and square the result.
  • Sum these squared differences.
  • Divide the sum by the number of data points minus one (n-1 for samples).
  • Take the square root of the result.
In our example, after calculating, we found the sample standard deviation to be approximately 6.46. This tells us how much the number of students in dance companies typically varies around the mean.
T-Distribution
When estimating confidence intervals, especially with small sample sizes, the t-distribution is more appropriate than the normal distribution. This is because the t-distribution accounts for additional variability introduced by smaller samples.

Characteristics of the t-distribution include:
  • It is similar in shape to the normal distribution but has heavier tails.
  • The degree of freedom, calculated as the sample size minus one, influences its shape.
  • It becomes more like a normal distribution as the sample size increases.
In this problem, we assumed a sample size of 19, resulting in 18 degrees of freedom. This gave us a helpful tool to find the critical value (\(t^\ast \approx 2.878\)), which we used to estimate the confidence interval for the true mean size of a university dance company. The wider tails of the t-distribution provide a more accurate reflection of uncertainty when sample sizes are small.

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 a random sample of 200 people, 154 said that they watched educational television. Find the \(90 \%\) confidence interval of the true proportion of people who watched educational television. If the television company wanted to publicize the proportion of viewers, do you think it should use the \(90 \%\) confidence interval?

Assume that all variables are approximately normally distributed. A state representative wishes to estimate the mean number of women representatives per state legislature. A random sample of 17 states is selected, and the number of women representatives is shown. Based on the sample, what is the point estimate of the mean? Find the \(90 \%\) confidence interval of the mean population. (Note: The population mean is actually \(31.72,\) or about 32.) Compare this value to the point estimate and the confidence interval. There is something unusual about the data. Describe it and state how it would affect the confidence interval. $$ \begin{array}{rrrrr} 5 & 33 & 35 & 37 & 24 \\ 31 & 16 & 45 & 19 & 13 \\ 18 & 29 & 15 & 39 & 18 \\ 58 & 132 & & & \end{array} $$

Assume that all variables are approximately normally distributed. A meteorologist who sampled 13 randomly selected thunderstorms found that the average speed at which they traveled across a certain state was 15.0 miles per hour. The standard deviation of the sample was 1.7 miles per hour. Find the \(99 \%\) confidence interval of the mean. If a meteorologist wanted to use the highest speed to predict the times it would take storms to travel across the state in order to issue warnings, what figure would she likely use?

Assume that all variables are approximately normally distributed. A recent survey of 8 randomly selected social networking sites has a mean of 13.1 million visitors for a specific month. The standard deviation is 4.1 million. Find the \(95 \%\) confidence interval of the true mean.

If a random sample of 600 people is selected and the researcher decides to have a margin of error of \(4 \%\) on the specific proportion who favor gun control, find the degree of confidence. A recent study showed that \(50 \%\) were in favor of some form of gun control.
See all solutions

Recommended explanations on Math Textbooks

Geometry

Read Explanation

Decision Maths

Read Explanation

Calculus

Read Explanation

Statistics

Read Explanation

Logic and Functions

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.