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A researcher wanted to know the percentage of judges who are in favor of the death penalty. He took a random sample of 15 judges and asked them whether or not they favor the death penalty. The responses of these judges are given here. \(\begin{array}{lllllll}\text { Yes } & \text { No } & \text { Yes } & \text { Yes } & \text { No } & \text { No } & \text { No } & \text { Yes } \\\ \text { Yes } & \text { No } & \text { Yes } & \text { Yes } & \text { Yes } & \text { No } & \text { Yes }\end{array}\) a. What is the point estimate of the population proportion? b. Make a \(95 \%\) confidence interval for the percentage of all judges who are in favor of the death penalty.

Short Answer

Expert verified
The point estimate of the proportion is the calculated sample proportion from step 1. The 95% confidence interval is given by the range calculated in step 3.

Step by step solution

01

Calculate the Sample Proportion

Count how many judges were in favor (said 'Yes') and how many judges in total were surveyed. Divide the number of 'Yes' by the total number to calculate the sample proportion. This sample proportion will serve as the point estimate of the population proportion. It is denoted as \(\hat{p} = \frac{x}{n}\), where x is the number of 'Yes' responses and n is the total number of responses.
02

Calculate the Confidence Interval

Use the formula for a Confidence Interval for a proportion which is given by \(\hat{p} \pm z\sqrt{\frac{(\hat{p}(1-\hat{p}))}{n}}\). The value of z here is the z-score, which for a 95% confidence interval is approximately 1.96.
03

Calculate Lower and Upper Limits

Use the sample proportion and standard deviation to calculate the lower and upper limits of the confidence interval. The lower limit will be \(\hat{p} - z\sqrt{\frac{(\hat{p}(1-\hat{p}))}{n}}\) and the upper limit will be \(\hat{p} + z\sqrt{\frac{(\hat{p}(1-\hat{p}))}{n}}\). This gives us a range in which we can be 95% confident that the true population proportion lies.

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Key Concepts

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

Sample Proportion
The sample proportion is a way to express how many respondents in a sample group display a certain attribute. It helps us understand the pattern within a specific subset of a larger population. In the context of this exercise, we are looking at judges who favor the death penalty based on their responses.
To calculate the sample proportion, count the number of 'Yes' responses, which indicates favorability of the death penalty, and divide this by the total number of responses. Suppose out of the 15 judges surveyed, 8 said 'Yes'. The sample proportion (denoted as \(\hat{p}\)) is computed as:
  • \(\hat{p} = \frac{8}{15}\)
This figure will give us an insight into the approximate proportion of the entire population (all judges) that might share this view. Notice how easily the sample proportion summarizes the data encountered in the survey and acts as a crucial stepping stone towards understanding larger population characteristics.
Population Proportion
The population proportion is an estimated or true proportion of a characteristic for the entire population, not just those surveyed. It tells us the percentage of the overall group that shares a specific attribute, like judges who favor the death penalty in this case.
In our given exercise, we wish to estimate the population proportion based on our sample proportion. The calculated sample proportion acts as a point estimate for the population proportion. However, since it's derived from a sample, there's always some level of uncertainty.
We use techniques like confidence intervals, explained below, to provide a sense of how close our sample proportion might be to the true population proportion. It's important to remember that while the sample can guide us, it's not a guaranteed measure of the entire population.
Z-Score
A Z-score is a statistical measure that tells us how many standard deviations a data point is from the mean. It's used here to determine a confidence level for our interval estimate of the population proportion.
When creating a confidence interval, the Z-score helps determine the margin of error, allowing us to account for sampling variability. For a 95% confidence interval, which is a common choice, the Z-score is approximately 1.96.
  • This means if we repeated the sampling process many times, about 95% of the calculated intervals would contain the true population proportion.
In simpler terms, the Z-score adds precision to our estimate by quantifying the certainty of our confidence interval. By using the Z-score in the formula \(\hat{p} \pm z\sqrt{\frac{(\hat{p}(1-\hat{p}))}{n}}\), we adjust our interval range, ensuring it accurately reflects the confidence level we've set.
Point Estimate
The point estimate is an approximate value that serves as the best single estimate of an unknown population parameter. In our case, the parameter of interest is the population proportion of judges who support the death penalty.
  • The sample proportion \(\hat{p}\) is used as a point estimate for the population proportion \(p\).
It's crucial to note that a point estimate is simply an educated guess derived from sample data. While it provides a good approximation, it inherently carries uncertainty because it doesn’t account for sampling variability. However, when coupled with confidence intervals, it gives a fuller picture of our estimate's accuracy. In summary, the point estimate gives us a numerically simple and effective measure of central tendency for the characteristic being studied but must always be interpreted within the context of its associated confidence interval.

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Most popular questions from this chapter

A group of veterinarians wants to test a new canine vaccine for Lyme disease. (Lyme disease is transmitted by the bite of an infected deer tick.) In an area that has a high incidence of Lyme disease, 100 dogs are randomly selected (with their owners' permission) to receive the vaccine. Over a 12 -month period, these dogs are periodically examined by veterinarians for symptoms of Lyme disease. At the end of 12 months, 10 of these 100 dogs are diagnosed with the disease. During the same 12 -month period, \(18 \%\) of the unvaccinated dogs in the area have been found to have Lyme disease. Let \(p\) be the proportion of all potential vaccinated dogs who would contract Lyme disease in this area. a. Find a \(95 \%\) confidence interval for \(p\). b. Does \(18 \%\) lie within your confidence interval of part a? Does this suggest the vaccine might or might not be effective to some degree? c. Write a brief critique of this experiment, pointing out anything that may have distorted the results or conclusions.

Suppose, for a sample selected from a normally distributed population, \(\bar{x}=68.50\) and \(s=8.9\). a. Construct a \(95 \%\) confidence interval for \(\mu\) assuming \(n=16\). b. Construct a \(90 \%\) confidence interval for \(\mu\) assuming \(n=16 .\) Is the width of the \(90 \%\) confidence interval smaller than the width of the \(95 \%\) confidence interval calculated in part a? If yes, explain why. c. Find a \(95 \%\) confidence interval for \(\mu\) assuming \(n=25 .\) Is the width of the \(95 \%\) confidence interval for \(\mu\) with \(n=25\) smaller than the width of the \(95 \%\) confidence interval for \(\mu\) with \(n=16\) calculated in part a? If so, why? Explain.

A computer company that recently developed a new software product wanted to estimate the mean time taken to learn how to use this software by people who are somewhat familiar with computers. A random sample of 12 such persons was selected. The following data give the times taken (in hours) by these persons to learn how to use this software. \(\begin{array}{llllll}1.75 & 2.25 & 2.40 & 1.90 & 1.50 & 2.75 \\ 2.15 & 2.25 & 1.80 & 2.20 & 3.25 & 2.60\end{array}\) Construct a \(95 \%\) confidence interval for the population mean. Assume that the times taken by all persons who are somewhat familiar with computers to learn how to use this software are approximately normally distributed.

A bank manager wants to know the mean amount owed on credit card accounts that become delinquent. A random sample of 100 delinquent credit card accounts taken by the manager produced a mean amount owed on these accounts equal to \(\$ 2640\). The population standard deviation was \(\$ 578\). a. What is the point estimate of the mean amount owed on all delinquent credit card accounts at this bank? b. Construct a \(97 \%\) confidence interval for the mean amount owed on all delinquent credit card accounts for this bank.

What assumption(s) must hold true to use the normal distribution to make a confidence interval for the population proportion, \(p\) ?

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