/*! 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 Estimate the minimum sample size... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Estimate the minimum sample size needed to form a confidence interval for the mean of a population having the standard deviation shown, meeting the criteria given. a. \(\quad \sigma=30,95 \%\) confidence, \(E=10\) b. \(\quad \sigma=30,99 \%\) confidence, \(E=10\) c. \(\quad \sigma=30,95 \%\) confidence, \(E=5\)

Short Answer

Expert verified
For part a, n = 35; for part b, n = 60; and for part c, n = 139.

Step by step solution

01

Identify the Formula

The formula for determining the minimum sample size required to estimate a population mean with a given level of confidence and error margin is: \[ n = \left( \frac{Z_{\alpha/2} \cdot \sigma}{E} \right)^2 \] where \( n \) is the sample size, \( Z_{\alpha/2} \) is the Z-value for the given confidence level, \( \sigma \) is the standard deviation, and \( E \) is the maximum error allowed.
02

Find Z-values for Confidence Levels

For 95% confidence, \( Z_{\alpha/2} \) is approximately 1.96. For 99% confidence, \( Z_{\alpha/2} \) is approximately 2.576. These values correspond to how many standard deviations you need to include in your confidence interval to reach the desired confidence level.
03

Calculate Sample Size for Part A

Given \( \sigma = 30 \), 95% confidence level, and error \( E = 10 \), plug the values into the formula: \[ n = \left( \frac{1.96 \times 30}{10} \right)^2 = \left( 5.88 \right)^2 = 34.5744 \]Since the sample size must be a whole number, round up to get \( n = 35 \).
04

Calculate Sample Size for Part B

Given \( \sigma = 30 \), 99% confidence level, and error \( E = 10 \), plug the values into the formula: \[ n = \left( \frac{2.576 \times 30}{10} \right)^2 = \left( 7.728 \right)^2 = 59.726 \]Round up to get \( n = 60 \).
05

Calculate Sample Size for Part C

Given \( \sigma = 30 \), 95% confidence level, and error \( E = 5 \), plug the values into the formula: \[ n = \left( \frac{1.96 \times 30}{5} \right)^2 = \left( 11.76 \right)^2 = 138.2976 \]Round up to get \( n = 139 \).

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 provides a range that is expected to contain the population mean with a certain level of certainty. It is a useful tool in statistics that helps us understand the precision of an estimate based on sample data.
The width of the confidence interval depends on several factors, such as the sample size, confidence level, and variability within the population. The larger the sample size, the narrower the confidence interval, which gives a more precise estimate of the population mean.
  • The confidence level (such as 95% or 99%) indicates how sure we are that the interval contains the true population mean. Higher confidence levels lead to wider intervals.
  • The standard deviation reflects the variability in the data. A smaller standard deviation leads to a narrower confidence interval.
Overall, the confidence interval provides a way to quantify our uncertainty about an estimate and is crucial for making informed decisions based on sample data.
Population Mean
The population mean represents the average of all individuals in a population. It's an important statistic because it gives a central point around which the data is distributed. However, calculating the population mean directly is often impractical because it requires data from every individual in the entire population.
Instead, we usually estimate the population mean by using sample data, which is a smaller subset of the population. This estimate helps us make inferences about the population as a whole. The sample mean, calculated from the data, serves as an approximation of the population mean.
In situations where the population mean is needed but not known, confidence intervals are used to estimate it. This estimation provides both the possible range of the population mean and the level of confidence in that range.
Standard Deviation
Standard deviation is a measure of the amount of variation or dispersion in a set of values. It tells us how much individual data points tend to differ from the mean value.
A small standard deviation indicates that the data points are close to the mean, while a large standard deviation suggests the data points are spread out over a wider range of values.
  • Standard deviation is central to calculating confidence intervals and determining minimum sample sizes because it influences how much the sample mean is expected to vary.
  • The formula for standard deviation is: \[\sigma = \sqrt{\frac{1}{N} \sum_{i=1}^{N} (X_i - \mu)^2}\]where \(N\) is the number of data points, \(X_i\) is each data point, and \(\mu\) is the mean.
Understanding standard deviation is key to interpreting the reliability and precision of statistical estimates.
Confidence Level
The confidence level is a probability metric that quantifies the degree of certainty that the true population parameter is within the calculated confidence interval. Common confidence levels used are 95% and 99%.
A 95% confidence level means that if you were to take 100 different samples and compute a confidence interval for each sample, about 95 of the 100 confidence intervals will contain the true population mean. Similarly, a 99% confidence level implies greater certainty but results in a wider interval.
  • The choice of confidence level affects the Z-value used in the sample size formula.
  • Higher confidence levels use larger Z-values, reflecting the need to widen the interval to ensure a higher probability of containing the true mean.
The confidence level is selected based on how much uncertainty is acceptable and can affect the decisions made based on the statistical analysis results.

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 machine for making precision cuts in dimension lumber produces studs with lengths that vary with standard deviation 0.003 inch. Five trial cuts are made to check the machine's calibration. The mean length of the studs produced is 104.998 inches with sample standard deviation 0.004 inch. Construct a \(99.5 \%\) confidence interval for the mean lengths of all studs cut by this machine. Assume lengths are normally distributed. Hint: Not all the numbers given in the problem are used.

In a random sample of 12,447 hip replacement or revision surgery procedures nationwide, 162 patients developed a surgical site infection. a. Give a point estimate of the proportion of all patients undergoing a hip surgery procedure who develop a surgical site infection. b. Verify that the sample is sufficiently large to use it to construct a confidence interval for that proportion. construct a \(95 \%\) confidence interval for the proportion of all patients undergoing a hip surgery procedure who develop a surgical site infection.

An airline wishes to estimate the weight of the paint on a fully painted aircraft of the type it flies. In a sample of four repaintings the average weight of the paint applied was 239 pounds, with sample standard deviation 8 pounds. Assuming that weights of paint on aircraft are normally distributed, construct a \(99.8 \%\) confidence interval for the mean weight of paint on all such aircraft.

In a random sample of 2,300 mortgages taken out in a certain region last year, 187 were adjustable-rate mortgages. a. Give a point estimate of the proportion of all mortgages taken out in this region last year that were adjustable-rate mortgages. b. Assuming that the sample is sufficiently large, construct a \(99.9 \%\) confidence interval for the proportion of all mortgages taken out in this region last year that were adjustable-rate mortgages.

The variation in time for a baked good to go through a conveyor oven at a large scale bakery has standard deviation 0.017 minute at every time setting. To check the bake time of the oven periodically four batches of goods are carefully timed. The recent check gave a mean of 27.2 minutes with sample standard deviation 0.012 minute. Construct a \(99.8 \%\) confidence interval for the mean bake time of all batches baked in this oven. Assume bake times are normally distributed. Hint: Not all the numbers given in the problem are used.

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.