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Chapter 18: Problem 18

Junk mail Direct mail advertisers send solicitations (a.k.a. "junk mail") to thousands of potential customers in the hope that some will buy the company's product. The acceptance rate is usually quite low. Suppose a company wants to test the response to a new flyer, and sends it to 1000 people randomly selected from their mailing list of over 200,000 people. They get orders from 123 of the recipients. a) Create a \(90 \%\) confidence interval for the percentage of people the company contacts who may buy something. b) Explain what this interval means. c) Explain what "90\% confidence" means. d) The company must decide whether to now do a mass mailing. The mailing won't be cost-effective unless it produces at least a \(5 \%\) return. What does your confidence interval suggest? Explain.

Short Answer

Expert verified
The mailing is likely cost-effective since the confidence interval is above 5%.

Step by step solution

01

Identify Sample Proportion

The sample proportion \( p \) is the fraction of recipients who placed an order. This is calculated as \( \frac{123}{1000} = 0.123 \).
02

Determine Standard Error

The standard error (SE) is calculated with the formula \( \text{SE} = \sqrt{\frac{p(1-p)}{n}} \), where \( n \) is the sample size. Hence, \( \text{SE} = \sqrt{\frac{0.123(1-0.123)}{1000}} \approx 0.0103 \).
03

Find Confidence Interval Using Z-score

For a \(90\%\) confidence interval, use a Z-score of 1.645. The confidence interval is \( p \pm Z \times \text{SE} \). Thus, \( 0.123 \pm 1.645 \times 0.0103 \), which results in \([0.106, 0.140] \).
04

Interpret the Confidence Interval

The confidence interval of \([0.106, 0.140] \) suggests that with \(90\%\) confidence, between \(10.6\%\) and \(14\%\) of the contacted people may buy something.
05

Explain "90% Confidence"

"90% confidence" means that if we were to take many samples and build a confidence interval from each of them, \(90\%\) of those intervals would contain the true population proportion.
06

Evaluate Cost-effectiveness Requirement

The required \(5\%\) return falls below the confidence interval's lower bound of \(10.6\%\). Therefore, the mailing is likely to be cost-effective if the true proportion is close to our sample result.

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Key Concepts

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

Sample Proportion
In statistics, the sample proportion (\( p \) ) is an estimate of the true proportion of a population based on a sample. It represents the fraction of voters, respondents or participants who exhibit a particular trait or behavior in the sample. For example, in the exercise, the company sent 1,000 flyers and 123 recipients placed an order. Hence, the sample proportion is calculated by the formula:\[ p = \frac{123}{1000} = 0.123 \]This means that 12.3% of the sample responded positively by placing an order. Understanding the sample proportion is crucial because it acts as a preliminary estimate of the overall response in the entire population, giving us initial insights into how effective the flyer might be.
Standard Error
The standard error (SE) measures the variability or dispersion of the sample proportion and reflects how much the sample results could vary by chance. It's like determining the expected margin of error in surveying. To calculate the SE of a sample proportion, you utilize the formula:\[ \text{SE} = \sqrt{\frac{p(1-p)}{n}} \]- \( p \) is the sample proportion- \( n \) is the sample size
Using the numbers from the exercise:\[ \text{SE} = \sqrt{\frac{0.123 \times (1 - 0.123)}{1000}} \approx 0.0103 \]So, the standard error is about 0.0103. This value indicates the expected extent of variation in our sample results from the true population proportion.
Z-score
The Z-score is a statistical measurement that represents the number of standard deviations a data point is from the mean value. When constructing a confidence interval for a given level of confidence, you use the Z-score to indicate how many standard errors to include around the sample proportion. For a 90% confidence interval, the Z-score is approximately 1.645.The confidence interval is calculated with the formula:\[ \text{Confidence Interval} = p \pm Z \times \text{SE} \]Utilizing our data:\[ 0.123 \pm 1.645 \times 0.0103 \]This results in a confidence interval of approximately \([0.106, 0.140]\). This indicates that we are 90% confident that the true proportion of people who would buy the product lies between 10.6% and 14%.
Response Rate
The response rate is a measure of the effectiveness of a survey or advertisement, indicating the percentage of recipients who take a desired action. In this context, it refers to the proportion of individuals who place an order after receiving the flyer. A healthy response rate can imply a successful campaign.In the given problem, the company received orders from 123 out of 1,000 people, which amounts to a response rate of:\[ \text{Response Rate} = \frac{123}{1000} = 0.123 \text{ or } 12.3\% \]Understanding the response rate helps the company decide if the mass mailing will be cost-effective, as only a minimum return (above 5% response rate) is necessary. Given the confidence interval from the previous calculations, the lower bound of 10.6% assures that the campaign would likely meet profit requirements.

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Most popular questions from this chapter

Back to campus In 2004 ACT, Inc., reported that \(74 \%\) of 1644 randomly selected college freshmen returned to college the next year. The study was stratified by type of college - public or private. The retention rates were \(71.9 \%\) among 505 students enrolled in public colleges and \(74.9 \%\) among 1139 students enrolled in private colleges. a) Will the \(95 \%\) confidence interval for the true national retention rate in private colleges be wider or narrower than the \(95 \%\) confidence interval for the retention rate in public colleges? Explain. b) Do you expect the margin of error for the overall

Conditions For each situation described below, identify the population and the sample, explain what \(p\) and \(\hat{p}\) represent, and tell whether the methods of this chapter can be used to create a confidence interval. a) Police set up an auto checkpoint at which drivers are stopped and their cars inspected for safety problems. They find that 14 of the 134 cars stopped have at least one safety violation. They want to estimate the percentage of all cars that may be unsafe, b) A TV talk show asks viewers to register their opinions on prayer in schools by logging on to a website. Of the 602 people who voted, 488 favored prayer in schools. We want to estimate the level of support among the general public. c) A school is considering requiring students to wear uniforms. The PTA surveys parent opinion by sending a questionnaire home with all 1245 students; 380 surveys are returned, with 228 families in favor of the change. d) A college admits 1632 freshmen one year, and four years later 1388 of them graduate on time. The college wants to estimate the percentage of all their freshman enrollees who graduate on time.

Mislabeled seafood In December \(2011,\) Consumer Reports published their study of labeling of seafood sold in New York, New Jersey, and Connecticut. They purchased 190 pieces of seafood from various kinds of food stores and restaurants in the three states and genetically compared the pieces to standard gene fragments that can identify the species. Laboratory results indicated that \(22 \%\) of these packages of seafood were mislabeled, incompletely labeled, or misidentified by store or restaurant employees. a) Construct a \(95 \%\) confidence interval for the proportion of all seafood packages in those three states that are mislabeled or misidentified. b) Explain what your confidence interval says about seafood sold in these three states. c) A 2009 report by the Government Accountability Board says that the Food and Drug Administration has spent very little time recently looking for seafood fraud. Suppose an official said, "That's only 190 packages out of the billions of pieces of seafood sold in a year. With the small number tested, I don't know that one would want to change one's buying habits." (An official was quoted similarly in a different but similar context). Is this argument valid? Explain.

Margin of error A medical researcher estimates the percentage of children exposed to lead-based paint, adding that he believes his estimate has a margin of error of about \(3 \% .\) Explain what the margin of error means.

More conclusions In January \(2002,\) two students made worldwide headlines by spinning a Belgian euro 250 times and getting 140 heads - that's \(56 \% .\) That makes the \(90 \%\) confidence interval \((51 \%, 61 \%) .\) What does this mean? Are these conclusions correct? Explain. a) Between \(51 \%\) and \(61 \%\) of all euros are unfair. b) We are \(90 \%\) sure that in this experiment this euro landed heads on between \(51 \%\) and \(61 \%\) of the spins. c) We are \(90 \%\) sure that spun euros will land heads between \(51 \%\) and \(61 \%\) of the time. d) If you spin a curo many times, you can be \(90 \%\) sure of getting between \(51 \%\) and \(61 \%\) heads. e) \(90 \%\) of all spun euros will land heads between \(51 \%\) and \(61 \%\) of the time.
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