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In a random sample of 519 judges, it was found that 285 were introverts (see reference of Problem 5). (a) Let \(p\) represent the proportion of all judges who are introverts. Find a point estimate for \(p\). (b) Find a \(99 \%\) 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
(a) Point estimate \( \hat{p} \approx 0.5491 \). (b) 99% CI: \( (0.4929, 0.6053) \). (c) Yes, conditions are satisfied.

Step by step solution

01

Calculate the Point Estimate for p

The point estimate for the proportion \( p \) is calculated by dividing the number of introverted judges by the total number of judges. So, \( \hat{p} = \frac{285}{519} \approx 0.5491 \).
02

Determine the Sample's Standard Deviation

To find the confidence interval, we first determine the standard deviation using the formula \( \sqrt{\frac{\hat{p}(1-\hat{p})}{n}} \), where \( n = 519 \). Thus, \( \sqrt{\frac{0.5491(1-0.5491)}{519}} \approx 0.02183 \).
03

Define the Z-value for the Confidence Level

For a 99% confidence interval, the Z-value associated with the normal distribution is approximately 2.576 (from Z-tables).
04

Calculate the Margin of Error

The margin of error (ME) is calculated as Z-value multiplied by the sample's standard deviation, i.e., \( 2.576 \times 0.02183 \approx 0.0562 \).
05

Compute the Confidence Interval

The confidence interval is given by \( \hat{p} \pm \text{ME} \). So, the 99% confidence interval is \( 0.5491 \pm 0.0562 \), which results in an interval of \( (0.4929, 0.6053) \).
06

Verify the Conditions np > 5 and nq > 5

We verify that both \( np = 285 \) and \( nq = 234 \) are greater than 5, where \( q = 1-p = 0.4509 \). Therefore, both conditions are satisfied.
07

Interpret the Confidence Interval

Since the interval is \( (0.4929, 0.6053) \), we say that we are 99% confident that the true proportion of all judges who are introverts is between 49.29% and 60.53%.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Point Estimate
Understanding the concept of a point estimate is crucial for interpreting sample data. A point estimate is a single value that serves as a best guess of a population parameter.In the context of the exercise, the point estimate represents the proportion of all judges who are introverts. This is computed by taking the number of introverted judges in the sample and dividing it by the total number of judges surveyed.
For example, our calculation was \( \hat{p} = \frac{285}{519} \approx 0.5491 \). This result means that from this sample, we estimate approximately 54.91% of judges are introverts.
Being a singular value, a point estimate doesn't account for variability within the population but gives us an initial idea about our parameter of interest.
Margin of Error
The margin of error (ME) in statistics quantifies the range of uncertainty around a point estimate.
It tells us how much the estimated parameter might deviate from the actual population value due to the randomness of sampling. To calculate the margin of error, we multiply the sample's standard deviation by the Z-value corresponding to the desired confidence level.In our step-by-step solution, the ME was found by evaluating \( 2.576 \times 0.02183 \approx 0.0562 \).
This indicates that the proportion of introverted judges could reasonably vary by about 5.62% above or below our point estimate. This gives our estimate a buffer, assuring us that the true population proportion is within this range.The margin of error becomes smaller with larger sample sizes, which indicates greater precision in the population estimate.
Normal Distribution
A normal distribution is a fundamental concept in statistics used to describe data that cluster around a mean or average. This bell-shaped curve is symmetrical, allowing most data points to fall near the mean.
It becomes crucial when calculating confidence intervals. In our context, the normal distribution helps to determine the Z-value for our confidence interval calculation.
For a 99% confidence level, a Z-value of approximately 2.576 is used. This Z-value shows how far the bounds of our confidence interval are from the mean in terms of standard deviation units. Assuming a normal distribution allows us to make probabilistic statements about our data, reinforcing the reliability of inferential statistics like confidence intervals.
Standard Deviation
Standard deviation is a measure of how spread out the numbers in a data set are. In simpler terms, it tells us how much individual data points deviate from the mean.In the exercise, calculating the standard deviation was essential to finding the margin of error and constructing the confidence interval. The formula used was \( \sqrt{\frac{\hat{p}(1-\hat{p})}{n}} \), where \( \hat{p} \) is the point estimate and \( n \) is the sample size.In our case, this resulted in \( \sqrt{\frac{0.5491(1-0.5491)}{519}} \approx 0.02183 \).A smaller standard deviation indicates that our sample results are closely clustered around the mean, enhancing our confidence in the reliability of the estimate. Conversely, a larger standard deviation suggests more variability.

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Most popular questions from this chapter

In a marketing survey, a random sample of 730 women shoppers revealed that 628 remained loyal to their favorite supermarket during the past year (i.e., did not switch stores). (Source: Trends in the United States: Consumer Attitudes and the Supermarket, The Research Department, Food Marketing Institute.) (a) Let \(p\) represent the proportion of all women shoppers who remain loyal to their favorite supermarket. 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 women supermarket shoppers who remained loyal to their favorite supermarket during the past year? What is the margin of error based on a \(95 \%\) confidence interval?

Sam computed a \(90 \%\) confidence interval for \(\mu\) from a specific random sample of size \(n\). He claims that at the \(90 \%\) confidence level, his confidence interval contains \(\mu\). Is this claim correct? Explain.

In order to use a normal distribution to compute confidence intervals for \(p\), what conditions on \(n p\) and \(n q\) need to be satisfied?

Consider college officials in admissions, registration, counseling, financial aid, campus ministry, food services, and so on. How much money do these people make each year? Suppose you read in your local newspaper that 45 officials in student services earned an average of \(\bar{x}=\$ 50,340\) each year (Reference: Cbronicle of Higher Education). (a) Assume that \(\sigma=\$ 16,920\) for salaries of college officials in student services. Find a \(90 \%\) confidence interval for the population mean salaries of such personnel. What is the margin of error? (b) Assume that \(\sigma=\$ 10,780\) for salaries of college officials in student services. Find a \(90 \%\) confidence interval for the population mean salaries of such personnel. What is the margin of error? (c) Assume that \(\sigma=\$ 4830\) for salaries of college officials in student services. Find a \(90 \%\) confidence interval for the population mean salaries of such personnel. What is the margin of error? (d) Compare the margins of error for parts (a) through (c). As the standard deviation decreases, does the margin of error decrease? (e) Compare the lengths of the confidence intervals for parts (a) through (c). As the standard deviation decreases, does the length of a \(90 \%\) confidence interval decrease?

The U.S. Geological Survey compiled historical data about Old Faithful Geyser (Yellowstone National Park) from 1870 to \(1987 .\) Some of these data are published in the book The Story of Old Faithful, by G. D. Marler (Yellowstone Association Press). Let \(x_{1}\) be a random variable that represents the time interval (in minutes) between Old Faithful eruptions for the years 1948 to \(1952 .\) Based on 9340 observations, the sample mean interval was \(\bar{x}_{1}=63.3\) minutes. Let \(x_{2}\) be a random variable that represents the time interval in minutes between Old Faithful eruptions for the years 1983 to \(1987 .\) Based on 25,111 observations, the sample mean time interval was \(\bar{x}_{2}=72.1\) minutes. Historical data suggest that \(\sigma_{1}=9.17\) minutes and \(\sigma_{2}=\) \(12.67\) minutes. Let \(\mu_{1}\) be the population mean of \(x_{1}\) and let \(\mu_{2}\) be the population mean of \(x_{2}\). (a) Compute a \(99 \%\) confidence interval for \(\mu_{1}-\mu_{2}\). (b) Interpretation: Comment on the meaning of the confidence interval in the context of this problem. Does the interval consist of positive numbers only? negative numbers only? a mix of positive and negative numbers? Does it appear (at the \(99 \%\) confidence level) that a change in the interval length between eruptions has occurred? Many geologic experts believe that the distribution of eruption times of Old Faithful changed after the major earthquake that occurred in \(1959 .\)

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