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Chapter 9: Problem 60

The report referenced in the previous exercise also indicated that \(33 \%\) of those in a representative sample of 533 homeowners in southern states said that they had considered installing solar panels. a. Use the given information to construct and interpret a \(90 \%\) confidence interval for the proportion of all homeowners in the southern states who have considered installing solar panels. b. Give two reasons why the confidence interval in Part (a) is narrower than the confidence interval calculated in the previous exercise.

Short Answer

Expert verified
a. The 90% confidence interval for the proportion of homeowners in the southern states who have considered installing solar panels can be calculated as follows: 1. Calculate the standard error (SE): \(SE= \sqrt{\frac{0.33(1-0.33)}{533}}\) 2. Determine the critical value (\(z_{\alpha/2}\)) for a 90% confidence interval: \(z_{\alpha/2} = 1.645\) 3. Calculate the margin of error (MOE): \(MOE = 1.645 * SE\) 4. Construct the confidence interval: \((\hat{p} - MOE, \hat{p} + MOE)\) b. The confidence interval might be narrower than the one calculated in the previous exercise due to two reasons: 1. Different confidence level: A 90% confidence interval is generally narrower than a 95% confidence interval because of the smaller critical value, which results in a smaller margin of error. 2. Different sample size or sample proportion: Larger sample sizes usually result in narrower intervals due to smaller standard errors and margins of error.

Step by step solution

01

Calculate the standard error of the proportion

We are given a sample proportion (\(\hat{p}\)) of 0.33 and a sample size (n) of 533. The standard error of the proportion (SE) can be calculated using the formula: \[SE=\sqrt{\frac{\hat{p}(1-\hat{p})}{n}}\] Following the formula, we have: \[SE= \sqrt{\frac{0.33(1-0.33)}{533}}\]
02

Determine the critical value for a 90% confidence interval

Since we are constructing a 90% confidence interval, we want to find the critical value (\(z_{\alpha/2}\)) that corresponds to a 90% confidence level. This value divides the standard normal distribution (z-distribution) such that 90% of the area under the curve falls between -\(z_{\alpha/2}\) and \(z_{\alpha/2}\). Using a z-score table or calculator, we find that \(z_{\alpha/2}\) is approximately 1.645.
03

Calculate the margin of error

The margin of error (MOE) is the product of the critical value and the standard error, that is: \[MOE = z_{\alpha/2} * SE\] Substitute the values of \(z_{\alpha/2}\) and SE calculated in step 1 and 2 to calculate the margin of error: \[MOE = 1.645 * SE\]
04

Construct the 90% confidence interval

With the margin of error, we can now construct the confidence interval for the proportion of homeowners who have considered installing solar panels. The confidence interval is given by: \[(\hat{p} - MOE, \hat{p} + MOE)\]
05

Part b: Reasons for narrower confidence interval

The confidence interval calculated in this exercise might be narrower than the one calculated in the previous exercise for two reasons: 1. Different confidence level: A 90% confidence interval will generally be narrower than, for example, a 95% confidence interval because the smaller confidence level corresponds to a smaller critical value. Consequently, the margin of error will be smaller, resulting in a narrower interval. 2. Different sample size or sample proportion: Different sample sizes and sample proportions might also result in different standard errors, margins of error, and widths of confidence intervals. Larger sample sizes usually result in narrower intervals, as they yield smaller standard errors and, therefore, smaller margins of error.

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Key Concepts

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

Standard Error
In statistics, the standard error (SE) represents the standard deviation of the sampling distribution of a statistic, most commonly the mean. The importance of the standard error lies in its role in the estimation of sampling error鈥攊t helps measure how much the sample mean varies from the actual population mean.

For a proportion, the standard error can be calculated using the formula:
\[SE=\sqrt{\frac{\hat{p}(1-\hat{p})}{n}}\]
Where \(\hat{p}\) is the sample proportion, and \(n\) is the sample size. The SE is a critical component in constructing confidence intervals and conducting hypothesis tests, as it impacts the precision of these estimates. A smaller standard error signifies a more precise estimate.
Margin of Error
The margin of error (MOE) quantifies the range of values around the sample statistic that is likely to contain the population parameter with a certain level of certainty. More simply, it represents the maximum expected difference between the true population parameter and a sample estimate of that parameter.

To calculate the margin of error for a proportion, use the formula:
\[MOE = z_{\alpha/2} * SE\]
Here, \(z_{\alpha/2}\) is the critical value for the confidence level from the standard normal (Z) distribution, and SE is the standard error as previously defined. The MOE is essentially a buffer added and subtracted from the sample statistic to obtain a confidence interval. A larger MOE corresponds to less precision in estimation, and factors such as sample size, variance, and confidence level affect its size.
Confidence Level
The confidence level denotes the degree of certainty with which we can say that the confidence interval contains the actual population parameter. It's a percentage that reflects how often the confidence interval would contain the parameter if we sampled from the population repeatedly.

A 90% confidence level means that we would expect 90% of the confidence intervals constructed from repeated samples to contain the true population parameter. To determine the critical value associated with a confidence level, one often refers to the Z-distribution. For example, a 90% confidence level corresponds to a critical value of approximately 1.645.

Increasing the confidence level widens the confidence interval, as we require more certainty that the interval includes the true parameter. This usually increases the margin of error, reflecting a trade-off between precision and confidence.

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Most popular questions from this chapter

Suppose that a campus bookstore manager wants to know the proportion of students at the college who purchase some or all of their textbooks online. Two different people independently selected random samples of students at the college and used their sample data to construct the following confidence intervals for the population proportion: Interval 1:(0.54,0.57) Interval 2:(0.46,0.62) a. Explain how it is possible that the two confidence intervals are not centered in the same place. b. Which of the two intervals conveys more precise information about the value of the population proportion? c. If both confidence intervals have a \(95 \%\) confidence level, which confidence interval was based on the smaller sample size? How can you tell? d. If both confidence intervals were based on the same sample size, which interval has the higher confidence level? How can you tell?

The report titled "One in Three American Households Are Stuck in a Relationship with a Financial Services Provider They Don't Trust" (June \(29,2016,\) www.businesswire.com/news /home/20160629005198/en/American- Households-Stuck -Relationship-Financial-Services-Provider, retrieved May 4, 2017) estimated that \(31 \%\) of American households feel obliged to do business with one or more financial services companies they distrust. This estimate is based on a representative sample of 1056 consumers age 18 and older. Use the "Bootstrap Confidence Interval for One Proportion" Shiny app to generate a \(95 \%\) bootstrap confidence interval for the proportion of all U.S. households that feel obliged to do business with one or more financial services companies they distrust. Interpret the interval in context.

In \(2010,\) the National Football League adopted new rules designed to limit head injuries. In a survey conducted in 2015 by the Harris Poll, 1216 of 2096 adults indicated that they were football fans and followed professional football. Of these football fans, 692 said they thought that the new rules were effective in limiting head injuries (December 21 , \(2015,\) www.theharrispoll.com/sports/Football-Injuries.html, retrieved May 6,2017 ). a. Assuming that the sample is representative of adults in the United States, construct and interpret a \(95 \%\) confidence interval for the proportion of U.S. adults who consider themselves to be football fans. b. Construct and interpret a \(95 \%\) confidence interval for the proportion of football fans who think that the new rules have been effective in limiting head injuries. c. Explain why the confidence intervals in Parts (a) and (b) are not the same width even though they both have a confidence level of \(95 \%\)

A researcher wants to estimate the proportion of students enrolled at a university who eat fast food more than three times in a typical week. Would the standard error of the sample proportion \(\hat{p}\) be smaller for random samples of size \(n=50\) or random samples of size \(n=200 ?\)

A researcher wants to estimate the proportion of city residents who favor spending city funds to promote tourism. Would the standard error of the sample proportion \(\hat{p}\) be smaller for random samples of size \(n=100\) or random samples of size \(n=200 ?\)
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