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Chapter 8: Problem 69

What is the point estimator of the population proportion, \(p ?\)

Short Answer

Expert verified
The point estimator of the population proportion, \( p \), is the sample proportion represented by \( \hat{p} \), which is calculated by dividing the number of successes by the total number of trials in the sample.

Step by step solution

01

Definition of Point Estimator for Population Proportion

The point estimate of a population proportion is simply the sample proportion. It's given by \( \hat{p} = \frac{x}{n} \) where:- \( \hat{p} \) is the sample proportion (decimal form)- \( x \) is the number of successes- \( n \) is the total number of trials
02

Calculate Point Estimator

To calculate the point estimator, count the number of successes in the sample and divide by the total number of trials in the sample. For instance, if you have a sample of 100 people and 25 of them have the trait you're interested in, then the point estimate of the population proportion is \( \hat{p} = \frac{25}{100} = 0.25 \)
03

Interpret the Point Estimator

The point estimator gives the best estimate of the population proportion based on the sample data. For instance, if the sample proportion, \( \hat{p} \), is 0.25, this would suggest that the best estimate for the population proportion is 25%.

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Key Concepts

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

Population Proportion
The population proportion is a fundamental concept in statistics. It represents the fraction of a population that possesses a particular attribute. For example, in a city, the population proportion might refer to the percentage of people who own a pet.

Understanding the population proportion is crucial in various fields, such as market research, public health, and sociology. However, the true population proportion is often unknown. Hence, we need statistical methods to estimate it, relying on sample data to give us an insight into the larger population.
Sample Proportion
The sample proportion is used as an estimate of the population proportion. It's calculated from a smaller, manageable group drawn from the population, which represents the larger whole.

The formula for the sample proportion is simple: \( \hat{p} = \frac{x}{n} \). Here, \( x \) is the number of favorable outcomes or successes, and \( n \) is the total number of observations in the sample.
  • For example, if out of 200 surveyed people, 60 say they ride a bicycle to work, the sample proportion is \( \hat{p} = \frac{60}{200} = 0.30 \) or 30%.
  • It provides a snapshot of the "population" through this sample lens.
Bear in mind that the sample proportion is subject to variability because different samples can yield different proportions. The larger the sample, the closer the estimate may be to the actual population proportion.
Statistical Estimation
Statistical estimation is the process of making inferences about a population based on sample data. This involves techniques like point estimation, which seeks to provide the best "single" value estimate of a population parameter (like population proportion) based on the sample data.

The main tools in statistical estimation include:
  • Point estimation: Provides a single number as the estimate without considering the potential variability of this estimate.
  • Interval estimation: Offers a range (interval) of values within which the population parameter is expected to lie.
Point estimation is often preferred for its simplicity. However, to get a clearer picture, it is commonly supplemented with interval estimation to assess the degree of uncertainty involved in the estimate.

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Most popular questions from this chapter

For each of the following, find the area in the appropriate tail of the \(t\) distribution. a. \(t=2.467\) and \(d f=28\) b. \(t=-1.672\) and \(d f=58\) c. \(t=-2.670\) and \(n=55\) d. \(t=2.819\) and \(n=23\)

A local gasoline dealership in a small town wants to estimate the average amount of gasoline that people in that town use in a 1-week period. The dealer asked 44 randomly selected customers to keep a diary of their gasoline usage, and this information produced the following data on gas used (in gallons) by these people during a 1-week period. The population standard deviation is not known. \(\begin{array}{rrrrrrrrrrr}23.1 & 13.6 & 25.8 & 10.0 & 7.6 & 18.9 & 26.6 & 23.8 & 12.3 & 15.8 & 21.0 \\ 26.9 & 22.9 & 18.3 & 23.5 & 21.6 & 15.5 & 23.5 & 11.8 & 15.3 & 11.9 & 19.2 \\ 14.5 & 9.6 & 12.1 & 18.0 & 20.6 & 14.2 & 7.1 & 13.2 & 5.3 & 13.1 & 10.9 \\ 10.5 & 5.1 & 5.2 & 6.5 & 8.3 & 10.5 & 7.4 & 7.4 & 5.3 & 10.6 & 13.0\end{array}\) Construct a \(95 \%\) confidence interval for the average weekly gas usage by people in this town.

In a Time/Money Magazine poll of Americans of age 18 years and older, \(65 \%\) agreed with the statement, "We are less sure our children will achieve the American Dream" (Time, October 10,2011 ). Assume that this poll was based on a random sample of 1600 Americans. a. Construct a \(95 \%\) confidence interval for the proportion of all Americans of age 18 years and older who will agree with the aforementioned statement. b. Explain why we need to construct a confidence interval. Why can we not simply say that \(65 \%\) of all Americans of age 18 years and older agree with the aforementioned statement?

A sample of 18 observations taken from a normally distributed population produced the following data: \(\begin{array}{lllllllll}28.4 & 27.3 & 25.5 & 25.5 & 31.1 & 23.0 & 26.3 & 24.6 & 28.4\end{array}\) \(\begin{array}{llllllllll}37.2 & 23.9 & 28.7 & 27.9 & 25.1 & 27.2 & 25.3 & 22.6 & 22.7\end{array}\) a. What is the point estimate of \(\mu\) ? b. Make a \(99 \%\) confidence interval for \(\mu .\) c. What is the margin of error of estimate for \(\mu\) in part b?

At Farmer's Dairy, a machine is set to fill 32 -ounce milk cartons. However, this machine does not put exactly 32 ounces of milk into each carton; the amount varies slightly from carton to carton. It is known that when the machine is working properly, the mean net weight of these cartons is 32 ounces. The standard deviation of the amounts of milk in all such cartons is always equal to \(.15\) ounce. The quality control department takes a sample of 25 such cartons every week, calculates the mean net weight of these cartons, and makes a \(99 \%\) confidence interval for the population mean. If either the upper limit of this confidence interval is greater than \(32.15\) ounces or the lower limit of this confidence interval is less than \(31.85\) ounces, the machine is stopped and adjusted. A recent sample of 25 such cartons produced a mean net weight of \(31.94\) ounces. Based on this sample, will you conclude that the machine needs an adjustment? Assume that the amounts of milk put in all such cartons have a normal distribution.
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