/*! 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 115 A researcher classified his subj... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 8: Problem 115

A researcher classified his subjects as innately right-handed or lefthanded by comparing thumbnail widths. He took a sample of 400 men and found that 80 men could be classified as left-handed according to his criterion. Estimate the proportion of all males in the population who would test to be left-handed using a \(95 \%\) confidence interval.

Short Answer

Expert verified
Answer: The 95% confidence interval for the proportion of all males in the population who would be classified as left-handed is between 16.08% and 23.92%.

Step by step solution

01

Calculate the proportion of left-handed men in the sample

The proportion of left-handed men in the sample can be calculated by dividing the number of left-handed men by the total number of men in the sample. In this case, there are 80 left-handed men in a sample of 400 men. Proportion = \(\frac{80}{400} = 0.2\) So, the proportion of left-handed men in the sample is \(0.2\) or \(20\%\).
02

Identify the confidence level

The exercise asks for a \(95\%\) confidence interval. This means we want to be \(95\%\) confident that the true proportion of left-handed men in the population falls within our calculated interval.
03

Calculate the standard error

Standard error (SE) can be calculated using the formula: \(SE = \sqrt{\frac{p(1-p)}{n}}\), where \(p\) is the sample proportion, and \(n\) is the sample size. Plugging in our values, \(p = 0.2\) and \(n=400\), we get: \(SE = \sqrt{\frac{0.2(1 - 0.2)}{400}} = \sqrt{\frac{0.16}{400}} = \sqrt{0.0004} = 0.02\)
04

Determine the critical value

For a \(95\%\) confidence interval, we will use a z-score (critical value) of \(1.96\). This value is based on the standard normal distribution table and corresponds to the confidence level of \(95\%\).
05

Calculate the margin of error

The margin of error can be calculated by multiplying the critical value (\(z\)) by the standard error (SE). Margin of Error = \(z \times SE = 1.96 \times 0.02 = 0.0392\)
06

Calculate the confidence interval

The confidence interval is calculated by finding the range within the sample proportion \(\pm\) the margin of error. Lower Bound = \(p - \text{Margin of Error}\) = \(0.2 - 0.0392 = 0.1608\) Upper Bound = \(p + \text{Margin of Error}\) = \(0.2 + 0.0392 = 0.2392\) Therefore, the \(95\%\) confidence interval for the proportion of all males in the population who would be classified as left-handed is between \(16.08\%\) and \(23.92\%\).

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.

Sample Proportion
The sample proportion is a core concept in statistics that helps us estimate what is happening in a larger population. In the context of our exercise, the sample proportion represents the ratio of left-handed men in our sample group to the total number of men surveyed.
To find the sample proportion, simply divide the number of left-handed men, which is 80, by the total sample size, which is 400. This gives us:
  • Sample Proportion = \( \frac{80}{400} = 0.2 \)
  • This value, 0.2, indicates that 20% of the sample is left-handed.
This proportion serves as a critical starting point for creating a confidence interval, providing a snapshot of how many might be left-handed throughout the whole population.
Standard Error
Once we have the sample proportion, calculating the standard error (SE) helps us understand the variability or possible error inherent in our sample statistic. The standard error is vital because it gives us an idea of how much our sample proportion might differ from the actual population proportion.
To calculate it, use the formula:
  • \[ SE = \sqrt{\frac{p(1-p)}{n}} \]
  • Where \(p\) is the sample proportion (0.2 in our case) and \(n\) is the sample size, 400.
Using these values gives us:
  • \[ SE = \sqrt{\frac{0.2 \times 0.8}{400}} = 0.02 \]
  • A standard error of 0.02 indicates a relatively small amount of potential variation from the sample proportion.
This low standard error suggests that our sample proportion could be a close estimate of the true population proportion.
Margin of Error
The margin of error is a crucial component in constructing a confidence interval. It quantifies the range of possible values around the sample proportion, offering a "buffer zone" indicating uncertainty or potential variability in our estimate.
To calculate the margin of error, multiply the critical value (z-score) by the standard error:
  • Margin of Error = \( z \times SE \)
  • For a 95% confidence interval, the z-score is typically 1.96.
  • Given \( SE = 0.02 \), the margin of error is \( 1.96 \times 0.02 = 0.0392 \).
This tells us the interval around the sample proportion where the true population proportion is likely to fall, with a certain level of confidence. It extends 3.92% above and below the sample proportion of 20%.
Z-score
In statistical terms, a z-score represents the number of standard deviations an element is from the mean. It plays a significant role in confidence intervals, which help us estimate a population parameter, like a proportion.
The z-score connects our sample's findings to the standard normal distribution.
For a 95% confidence level, the z-score we use is 1.96.
  • This value comes from standard normal distribution tables, indicating that approximately 95% of the data falls within 1.96 standard deviations of the mean.
  • The z-score helps ensure that our confidence interval captures the true population parameter 95% of the time.
Using the z-score, we can determine the margin of error, allowing us to construct a reliable confidence interval that reflects our level of certainty about the sample 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

Baseball Fans The first day of baseball comes in late March, ending in October with the World Series. Does fan support grow as the season goes on? Two CNN/USA Today/Gallup polls, one conducted in March and one in November, both involved random samples of 1001 adults aged 18 and older. In the March sample, \(45 \%\) of the adults claimed to be fans of professional baseball, while \(51 \%\) of the adults in the November sample claimed to be fans. \({ }^{14}\) a. Construct a \(99 \%\) confidence interval for the difference in the proportion of adults who claim to be fans in March versus November. b. Does the data indicate that the proportion of adults who claim to be fans increases in November, around the time of the World Series? Explain.

Find a \(99 \%\) lower confidence bound for the binomial proportion \(p\) when a random sample of \(n=400\) trials produced \(x=196\) successes.

In an experiment to assess the strength of the hunger drive in rats, 30 previously trained animals were deprived of food for 24 hours. At the end of the 24 -hour period, each animal was put into a cage where food was dispensed if the animal pressed a lever. The length of time the animal continued pressing the bar (although receiving no food) was recorded for each animal. If the data yielded a sample mean of 19.3 minutes with a standard deviation of 5.2 minutes, estimate the true mean time and calculate the margin of error.

A study was conducted to compare the mean numbers of police emergency calls per 8 -hour shift in two districts of a large city. Samples of 100 8-hour shifts were randomly selected from the police records for each of the two regions, and the number of emergency calls was recorded for each shift. The sample statistics are listed here: Find a \(90 \%\) confidence interval for the difference in the mean numbers of police emergency calls per shift between the two districts of the city. Interpret the interval.

Find a \((1-\alpha) 100 \%\) confidence interval for a population mean \(\mu\) for these values: a. \(\alpha=.01, n=38, \bar{x}=34, s^{2}=12\) b. \(\alpha=.10, n=65, \bar{x}=1049, s^{2}=51\) c. \(\alpha=.05, n=89, \bar{x}=66.3, s^{2}=2.48\)
See all solutions

Recommended explanations on Math Textbooks

Calculus

Read Explanation

Pure Maths

Read Explanation

Mechanics Maths

Read Explanation

Logic and Functions

Read Explanation

Applied Mathematics

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.