/*! 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 5 Isabel Myers was a pioneer in th... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Isabel Myers was a pioneer in the study of personality types. The following information is taken from \(A\) Guide to the Development and Use of the Myers- Briggs Type Indicator, by Myers and McCaulley (Consulting Psychologists Press). In a random sample of 62 professional actors, it was found that 39 were extroverts. (a) Let \(p\) represent the proportion of all actors who are extroverts. Find a point estimate for \(p\). (b) Find a \(95 \%\) confidence interval for \(p\). Give a brief interpretation of the meaning of the confidence interval you have found. (c) Do you think the conditions \(n p>5\) and \(n q>5\) are satisfied in this problem? Explain why this would be an important consideration.

Short Answer

Expert verified
The point estimate for \(p\) is approximately 0.629. The \(95\%\) confidence interval is (0.507, 0.751). Conditions \(np>5\) and \(nq>5\) are satisfied, confirming the sample size is adequate for a normal approximation.

Step by step solution

01

Identify Given Values

From the problem, we know that there are 62 professional actors, among whom 39 are extroverts. We let the observed number of extroverts be denoted as \(X=39\), and the total number of actors \(n=62\).
02

Estimate the Proportion

The sample proportion \(\hat{p}\) is calculated by dividing the number of extroverts by the total number of actors. Hence, \(\hat{p} = \frac{X}{n} = \frac{39}{62}\). Calculating this gives \(\hat{p} \approx 0.629\). This is our point estimate for the population proportion \(p\).
03

Find Standard Error

The standard error of the sample proportion is given by the formula \(SE_{p} = \sqrt{\frac{\hat{p}(1-\hat{p})}{n}}\). Substituting \(\hat{p}=0.629\), and \(n=62\), we find \(SE_{p} = \sqrt{\frac{0.629 \times (1-0.629)}{62}}\), which approximately equals \(0.062\).
04

Compute the Confidence Interval

Using a \(95\%\) confidence level, the critical value for a normal distribution is \(z=1.96\). The confidence interval is calculated using \(\hat{p} \pm z \times SE_{p}\), which gives \(0.629 \pm 1.96 \times 0.062\). This calculation results in the interval approximately from \(0.507\) to \(0.751\).
05

Interpret the Confidence Interval

The \(95\%\) confidence interval \((0.507, 0.751)\) means we are 95\% confident that the actual proportion of all professional actors who are extroverts is between these two values.
06

Verify Condition Satisfaction

We check whether \(np > 5\) and \(nq > 5\) are satisfied, where \(\hat{q} = 1-\hat{p}\). So, \(np = 62 \times 0.629 = 39\) and \(nq = 62 \times 0.371 = 23\). Both conditions are satisfied (\(39 > 5\) and \(23 > 5\)). These conditions ensure the sample size is large enough for the normal approximation to be valid.

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.

Point Estimate
A point estimate is a single value that serves as a best guess for an unknown population parameter, such as a proportion or mean. In the context of Isabel Myers' study on actors, the point estimate for the proportion of all actors who are extroverts gives us a snapshot of what is likely true about the entire population based on the sample data. To find the point estimate for the proportion of extroverts, you calculate the sample proportion, denoted as \(\hat{p}\). The sample proportion is determined by dividing the number of extroverts in the sample by the total number of actors sampled. Thus, \(\hat{p} = \frac{39}{62} \approx 0.629\). This means that approximately 62.9% of the sampled actors are extroverts, which serves as our point estimate for the entire population's proportion \(p\).
  • Point estimates provide a quick method to estimate unknown values.
  • They are based solely on sample data and do not account for variability.
Sample Proportion
The sample proportion, often denoted as \(\hat{p}\), is a statistic that gives a measure of how many members of a sample fall into a particular category. It's crucial when you want to infer population-level characteristics from sample data. In this exercise, the sample proportion is calculated by taking the number of extroverts, 39, and dividing it by the total sample size of 62. Hence, the sample proportion is \(\hat{p} = \frac{39}{62} \approx 0.629\).

The sample proportion is pivotal because it helps us construct other statistical tools, like confidence intervals, which provide a range of values within which we expect the actual population parameter to fall. By understanding the sample proportion, one can begin to understand broader patterns about the population.
  • The sample proportion is easy to compute and acts as the foundation for further analyses.
  • It varies from sample to sample, which is why we use confidence intervals to account for this variability.
Standard Error
The standard error is a measure of how much the sample proportion \(\hat{p}\) is expected to vary from the true population proportion \(p\). It gives us an idea of the precision of our sample proportion as an estimate of the population proportion. The formula for standard error of the sample proportion is \(SE_{p} = \sqrt{\frac{\hat{p}(1-\hat{p})}{n}}\).

In this scenario, substituting \(\hat{p} = 0.629\) and \(n = 62\), we calculate \(SE_{p} \approx 0.062\). This tells us that the sample proportion is estimated to have a small deviation around the population proportion, suggesting this sample provides a precise estimate.
  • Standard error decreases with larger sample sizes due to better estimation precision.
  • A smaller standard error indicates more reliable sample proportion estimates.
Normal Approximation Conditions
Normal approximation conditions are criteria used to determine if it is valid to use the normal distribution to approximate the sampling distribution of a proportion. For this approximation to be valid, two conditions must be met: \(np > 5\) and \(nq > 5\), where \(q = 1 - \hat{p}\).

Here, \(np = 62 \times 0.629 = 39\) and \(nq = 62 \times 0.371 = 23\). Both exceed 5, confirming that the sample size is large enough for the normal distribution to be an appropriate approximation. This is critical for constructing confidence intervals and performing hypothesis tests based on the normal distribution.
  • Satisfying these conditions ensures that the approximation leads to accurate and valid statistical interpretations.
  • They help manage the risk of erroneous conclusions stemming from small sample sizes.

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

At wind speeds above 1000 centimeters per second \((\mathrm{cm} / \mathrm{sec})\), significant sand-moving events begin to occur. Wind speeds below \(1000 \mathrm{~cm} / \mathrm{sec}\) deposit sand and wind speeds above \(1000 \mathrm{~cm} / \mathrm{sec}\) move sand to new locations. The cyclic nature of wind and moving sand determines the shape and location of large dunes (Reference: Hydraulic, Geologic, and Biologic Research at Great Sand Dunes National Monument and Vicinity, Colorado, Proceedings of the National Park Service Research Symposium). At a test site, the prevailing direction of the wind did not change noticeably. However, the velocity did change. Sixty wind speed readings gave an average velocity of \(\bar{x}=1075 \mathrm{~cm} / \mathrm{sec}\). Based on long-term experience, \(\sigma\) can be assumed to be \(265 \mathrm{~cm} / \mathrm{sec} .\) (a) Find a \(95 \%\) confidence interval for the population mean wind speed at this site. (b) Does the confidence interval indicate that the population mean wind speed is such that the sand is always moving at this site? Explain.

As the degrees of freedom increase, what distribution does the Student's \(t\) distribution become more like?

What price do farmers get for their watermelon crops? In the third week of July, a random sample of 40 farming regions gave a sample mean of \(\bar{x}=\$ 6.88\) per 100 pounds of watermelon. Assume that \(\sigma\) is known to be \(\$ 1.92\) per 100 pounds (Reference: Agricultural Statistics, U.S. Department of Agriculture). (a) Find a \(90 \%\) confidence interval for the population mean price (per 100 pounds) that farmers in this region get for their watermelon crop. What is the margin of error? (b) Find the sample size necessary for a \(90 \%\) confidence level with maximal error of estimate \(E=0.3\) for the mean price per 100 pounds of watermelon. (c) A farm brings 15 tons of watermelon to market. Find a \(90 \%\) confidence interval for the population mean cash value of this crop. What is the margin of error? Hint: 1 ton is 2000 pounds.

A random sample of 328 medical doctors showed that 171 had a solo practice. (Source: Practice Patterns of General Internal Medicine, American Medical Association.) (a) Let \(p\) represent the proportion of all medical doctors who have a solo practice. Find a point estimate for \(p\). (b) Find a \(95 \%\) confidence interval for \(p .\) Give a brief explanation of the meaning of the interval. (c) As a news writer, how would you report the survey results regarding the percentage of medical doctors in solo practice? What is the margin of error based on a \(95 \%\) confidence interval?

If the original \(x\) distribution has a relatively small standard deviation, the confidence interval for \(\mu\) will be relatively short.

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.