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Chapter 8: Problem 119

According to a recent survey of \(1,200\) people, 61\(\%\) feel that the president is doing an acceptable job. We are interested in the population of people who feel the president is doing an acceptable job. a. Define the random variables \(X\) and \(P^{\prime}\) in words. b. Which distribution should you use for this problem? Explain your choice. c. Construct a 90\(\%\) confidence interval for the population of people who feel the president is doing an acceptable job. i. State the confidence interval. ii. Sketch the graph. iii. Calculate the error bound.

Short Answer

Expert verified
a. \(X\): Number and \(P'\): Proportion supporting views. b. Binomial/normally approximated due to large sample size. c. 90% CI is (0.5861, 0.6339); error bound is 0.0239.

Step by step solution

01

Define Random Variables

In this problem, we define two random variables: - **\(X\):** The number of people in the sample who feel that the president is doing an acceptable job. - **\(P^{\prime}\):** The proportion of the sample who feel that the president is doing an acceptable job. That is, \(P^{\prime} = \frac{X}{n}\) where \(n\) is the sample size.
02

Choose the Distribution

We should use the binomial proportion distribution because we are dealing with proportions and binary outcomes (i.e., people either feel the president is doing an acceptable job or they do not). Given that our sample size is large, we can approximate this distribution using a normal distribution for constructing confidence intervals.
03

Calculate Sample Proportion

The sample proportion \(P^{\prime}\) is calculated as follows:\[ P^{\prime} = \frac{X}{n} = \frac{61\% \times 1200}{1200} = 0.61 \] This tells us that 61% of the people in our sample have the desired response.
04

Confidence Interval Using Z-distribution

To construct a 90% confidence interval, we use the formula:\[ CI = \left( P^{\prime} - Z \cdot \sqrt{\frac{P^{\prime} (1 - P^{\prime})}{n}}, P^{\prime} + Z \cdot \sqrt{\frac{P^{\prime} (1 - P^{\prime})}{n}} \right) \]The Z-value for a 90% confidence level is approximately 1.645.
05

Calculate Margin of Error (E)

The Margin of Error (E) is calculated using:\[ E = Z \cdot \sqrt{\frac{P^{\prime}(1 - P^{\prime})}{n}} \]Substitute the values: \[ E = 1.645 \cdot \sqrt{\frac{0.61 \times 0.39}{1200}} \approx 0.0239 \]
06

State the Confidence Interval

Apply the margin of error (E) to the sample proportion (\(P^{\prime}\)): \[ CI = (0.61 - 0.0239, 0.61 + 0.0239) = (0.5861, 0.6339) \]This interval represents the range within which we are 90% 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.

Random Variables
In statistics, random variables are used to quantify outcomes of a random process. For this exercise, we define two random variables related to the survey about the president's performance. Random variables can take on different values based on underlying random phenomena.
  • **\(X\)** represents the number of individuals in the sample who feel the president is doing an acceptable job. It is a discrete random variable because it counts individuals in categories.
  • **\(P^{\prime}\)** is the proportion of the sample who share this opinion, given by the formula \(P^{\prime} = \frac{X}{n}\), where \(n\) is the total sample size, which is 1200.
Understanding these variables is crucial as they help estimate broader population characteristics from a sample.
Normal Distribution
The normal distribution is a bell-shaped curve that is symmetrical around its mean. In our exercise, we use the normal distribution to build a confidence interval for the proportion of the population who believe the president is performing acceptably.
  • The choice of normal distribution comes from the large sample size of 1200 people, which makes it reasonable to approximate other distributions (like binomial) with it.
  • Normal distribution allows us to leverage its well-defined properties such as standard deviation and mean to compute probabilities and confidence intervals effectively.
Using a normal distribution, we can apply the **Central Limit Theorem**, which states that our sample proportions will tend to follow a normal distribution, especially with large samples.
Margin of Error
The margin of error is a measure that represents the range of uncertainty around a sample statistic. It helps us understand how much our sample result (the proportion) might differ from the true population proportion.
  • To find the margin of error in this context, we use the formula: \[ E = Z \cdot \sqrt{\frac{P^{\prime}(1 - P^{\prime})}{n}} \] where \(Z\) is the z-score corresponding to the desired confidence level.
  • For a 90% confidence level, the z-score is approximately 1.645, and by substituting the values, we found the margin of error to be about 0.0239.
The margin of error reflects the maximum expected difference between the sample proportion and the actual population proportion with a given level of confidence.
Binomial Proportion Distribution
A binomial proportion distribution is particularly useful when outcomes are binary, such as yes/no decisions, and when we're interested in the proportion of outcomes in one category.
  • In this problem, each participant can either feel the president is doing an acceptable job or not, making it a binary outcome.
  • The sample proportion \(P^{\prime}\) gives us an estimated proportion of people sharing the opinion in the total population.
  • Although the primary distribution is binomial due to the binary outcome, with a large sample size, a normal approximation is used for constructing the confidence interval.
This approximation simplifies calculations and offers a practical way to interpret and convey statistical conclusions.

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Most popular questions from this chapter

Use the following information to answer the next two exercises: A quality control specialist for a restaurant chain takes a random sample of size 12 to check the amount of soda served in the 16 oz. serving size. The sample mean is 13.30 with a sample standard deviation of 1.55. Assume the underlying population is normally distributed. Find the 95\(\%\) Confidence Interval for the true population mean for the amount of soda served. a. \((12.42,14.18)\) b. \((12.32,14.29)\) C. \((12.50,14.10)\) d. Impossible to determine

Suppose that a committee is studying whether or not there is waste of time in our judicial system. It is interested in the mean amount of time individuals waste at the courthouse waiting to be called for jury duty. The committee randomly surveyed 81 people who recently served as jurors. The sample mean wait time was eight hours with a sample standard deviation of four hours. a. i. \(\overline{x}=\) _____ ii. \(s_{x}=\) _____ iii. \(n=\) _____ iv. \(n-1=\) _____ b. Define the random variables \(X\) and \(\overline{X}\) in words. c. Which distribution should you use for this problem? Explain your choice. d. Construct a 95\(\%\) confidence interval for the population mean time wasted. i. State the confidence interval. ii. Sketch the graph. iii. Calculate the error bound. e. Explain in a complete sentence what the confidence interval means.

Use the following information to answer the next six exercises: One hundred eight Americans were surveyed to determine the number of hours they spend watching television each month. It was revealed that they watched an average of 151 hours each month with a standard deviation of 32 hours. Assume that the underlying population distribution is normal. Why would the error bound change if the confidence level were lowered to 95%?

Use the following information to answer the next two exercises: Marketing companies are interested in knowing the population percent of women who make the majority of household purchasing decisions. When designing a study to determine this population, what is the minimum number you would need to survey to be 90\(\%\) confident that the population proportion is estimated to within 0.05\(?\)

Use the following information to answer the next 16 exercises: The Ice Chalet offers dozens of different beginning ice-skating classes. All of the class names are put into a bucket. The 5 P.M., Monday night, ages 8 to 12, beginning ice-skating class was picked. In that class were 64 girls and 16 boys. Suppose that we are interested in the true proportion of girls, ages 8 to 12, in all beginning ice-skating classes at the Ice Chalet. Assume that the children in the selected class are a random sample of the population. What is being counted?
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