/*! 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 1 A sample of 49 observations is t... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

A sample of 49 observations is taken from a normal population. The sample mean is 55 , and the sample standard deviation is \(10 .\) Determine the 99 percent confidence interval for the population mean.

Short Answer

Expert verified
The 99% confidence interval for the population mean is (51.322, 58.678).

Step by step solution

01

Identify the Known Values

We are given that the sample size \( n = 49 \), the sample mean \( \bar{x} = 55 \), and the sample standard deviation \( s = 10 \). We are asked to find a 99% confidence interval for the population mean.
02

Understand Confidence Interval Formula

The confidence interval for the population mean when the sample standard deviation is known and the sample comes from a normal distribution is given by \( \bar{x} \pm z_{\alpha/2} \times \frac{s}{\sqrt{n}} \). Here, \( \bar{x} \) is the sample mean, \( z_{\alpha/2} \) is the critical value for the standard normal distribution, \( s \) is the sample standard deviation, and \( n \) is the sample size.
03

Find the Critical Value

For a 99% confidence interval, \( \alpha = 0.01 \). Thus for \( z_{\alpha/2} \) we find the value where the cumulative probability is \( 1 - \alpha/2 = 0.995 \). Corresponding to this cumulative probability, the critical value \( z_{0.005} \approx 2.576 \).
04

Calculate the Standard Error

The standard error (SE) of the sample mean is calculated as \( \frac{s}{\sqrt{n}} \). Substituting the given values, we get \( SE = \frac{10}{\sqrt{49}} = \frac{10}{7} \approx 1.4286 \).
05

Calculate the Margin of Error

The margin of error is found by multiplying the critical value by the standard error: \( 2.576 \times 1.4286 \approx 3.678 \).
06

Find the Confidence Interval

The confidence interval is given by \( \bar{x} \pm \text{Margin of Error} \). Therefore, the 99% confidence interval is \( 55 \pm 3.678 \), which yields the interval \( (51.322, 58.678) \).

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.

Population Mean
A population mean is a crucial measure in statistics. It represents the average value of a characteristic in an entire population. The population mean gives us insight into the overall tendency of the dataset. For instance, if you're studying the heights of adult males in a country, the population mean would be the average height if we could measure every adult male in that country.

Often, we can't measure every single individual due to constraints like time and resources, so we sample a smaller group that represents the population. This sample provides an estimate of the population mean. In this exercise, we used a sample size of 49 with a sample mean (\(ar{x}\)) of 55. Understanding the population mean helps us to make predictions and decisions based on statistical data.
Sample Standard Deviation
The sample standard deviation (\(s\)) is a measure of how spread out the values in a sample are around the sample mean. It provides a quantifiable measure of variability, or dispersion, of the sample data points. A small standard deviation signifies that the data points are close to the sample mean, while a large standard deviation indicates the opposite.

In our case, the sample standard deviation is 10. This implies that the individual observations vary, on average, by 10 units from the sample mean of 55. Using the standard deviation, statisticians can determine the reliability of the sample mean as an estimator of the population mean. It's a fundamental component in calculating the confidence interval, reflecting the degree of uncertainty associated with the estimate of the mean.
Critical Value
The critical value is an essential element in determining the confidence interval. It represents the threshold or boundary at which the probability of the observed statistic being a certain value is equal to a specified confidence level. It helps in defining the range within which the population parameter is expected to lie.

For a 99% confidence interval, we select a critical value (\(z_{\alpha/2}\)) corresponding to the tail probabilities of the standard normal distribution. In this exercise, the critical value is approximately 2.576, which aligns with the requirement that the total area under the tails outside the confidence interval adds up to 1% (0.5% for each tail). This ensures that 99% of the distribution falls within the calculated interval, providing a high level of confidence that the true population mean lies within this range.
Standard Error
The standard error quantifies the amount of variability one can expect in the sample mean as an estimator of the population mean. It is a direct function of the sample standard deviation and the square root of the sample size. The standard error tells us how much the sample mean would vary if we repeated the sample multiple times.

Mathematically, it is calculated as \( \frac{s}{\sqrt{n}} \) where \(s\) is the sample standard deviation and \(n\) is the sample size. In our example, the standard error is roughly 1.4286, showing how the mean of our sample (49 observations) could vary. The smaller the standard error, the closer the sample mean is to the population mean. By combining the standard error with the critical value, we can calculate the margin of error, providing insights into the reliability of the sample mean as an estimate of the population mean.

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 manufacturer of a new line of ink jet printers would like to include as part of their advertising the number of pages a user can expect from a print cartridge. A sample of 10 cartridges revealed the following number of pages printed. $$\begin{array}{|lllllllll|}\hline 2,698 & 2,028 & 2,474 & 2,395 & 2,372 & 2,475 & 1,927 & 3,006 & 2,334 & 2,379 \\\\\hline\end{array}$$ a. What is the point estimate of the population mean? b. Develop a 95 percent confidence interval for the population mean.

A state meat inspector in lowa has been given the assignment of estimating the mean net weight of packages of ground chuck labeled "3 pounds." Of course, he realizes that the weights cannot be precisely 3 pounds. A sample of 36 packages reveals the mean weight to be 3.01 pounds, with a standard deviation of 0.03 pounds. a. What is the estimated population mean? b. Determine a 95 percent confidence interval for the population mean.

The estimate of the population proportion is to be within plus or minus \(.10,\) with a 99 percent level of confidence. The best estimate of the population proportion is .45. How large a sample is required?

There are 20,000 eligible voters in York County, South Carolina. A random sample of 500 York County voters revealed 350 plan to vote to return Louella Miller to the state senate. Construct a 99 percent confidence interval for the proportion of voters in the county who plan to vote for Ms. Miller. From this sample information, can you confirm she will be reelected?

A recent study by the American Automobile Dealers Association revealed the mean amount of profit per car sold for a sample of 20 dealers was \(\$ 290,\) with a standard deviation of \(\$ 125 .\) Develop a 95 percent confidence interval for the population mean.

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.