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91Ó°ÊÓ

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 90% confidence interval for the proportion of customers who would buy something is between 10.59% and 14.01%. This means that we are 90% confident that the actual proportion of customers who will respond with an order lies within this range. Given that the lower limit of this interval exceeds the 5% return rate needed for profitability, it suggests that a mass mailing could be profitable.

Step by step solution

01

Calculate Sample Proportion

The first step is to calculate the sample proportion \((p)\). This is done by dividing the number of successful outcomes (orders) in the sample by the size of the sample. In this case we have \(p = \frac{123}{1000} = 0.123\).
02

Determine the Standard Error

Now, calculate the standard error. The formula for standard error in a proportion is \(\sqrt{\frac{p(1-p)}{n}}\). Plug the values into the equation to get \(\sqrt{\frac{0.123 × (1-0.123)}{1000}}\) which approximately equals 0.0104.
03

Calculate Confidence Interval

To calculate a 90% confidence interval, use the formula \(p \pm Z \times \) standard error where \(Z\) corresponds to the degree of confidence; for a 90% interval, it's 1.645. Plug in the values to get \(0.123 \pm 1.645 \times 0.0104\). The resulting interval is \(0.123 \pm 0.0171\), or between 0.1059 and 0.1401.
04

Interpret Confidence Interval

This interval means that we are 90% confident that the true proportion of customers who would respond with an order lies between 10.59% and 14.01%.
05

Explain 90% Confidence

90% confidence means that if the company were to repeat this process (of sending out 1000 mailers and calculating the confidence interval) many times, approximately 90% of those calculated intervals would contain the true proportion of customers who order.
06

Should the Company do a Mass Mailing?

Since the lower limit of our confidence interval, 10.59%, is above the 5% break-even point, the confidence interval suggests the mass mailing could be profitable, as we could expect at least a 10.59% return rate 90% of the time according to our estimation.

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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 key statistic in determining the behavior of a large population based on a smaller subset. When the company received 123 orders from its mailings to 1000 people, this generated a sample proportion. Mathematically, this is represented as \( p = \frac{123}{1000} = 0.123 \). This means 12.3% of the sampled individuals responded positively.Understanding the sample proportion is essential because it estimates the percentage of the total population that may respond in the same way. By identifying the sample proportion, companies can make initial assumptions about wider customer behaviors. But, remember that this is an average derived from a specific subset and subject to random sampling variations.
Standard Error
Standard Error (SE) quantifies the variability or "spread" of the sample proportion. It essentially indicates the accuracy of the sample proportion when used to estimate the population proportion.The formula to compute the standard error for a proportion is \( \sqrt{\frac{p(1-p)}{n}} \). For the advertisement example, that is:
  • \( p = 0.123 \)
  • \( n = 1000 \)
Insert these values into the formula: \( \sqrt{\frac{0.123 \times (1-0.123)}{1000}} \approx 0.0104 \). Thus, the standard error here is approximately 0.0104, showing the typical amount of variation one might expect in different samples of the same size. With smaller standard errors indicating more reliable estimates, this computation gives us more confidence in the results.
Statistical Inference
Statistical inference is the broader process in which conclusions about a population are drawn from sample data. Using the sample proportion and standard error, the company forms conclusions about possible customer responses across its entire mailing list. One main goal of statistical inference is to estimate population parameters, like the proportion of customers who would order, and determine how confident we can be in these estimates. Actions based on inferences are guided by computed margins of error and confidence intervals, which specify ranges of plausible values for the parameter of interest. It converts the randomness of sampling into meaningful insights for decision-making, such as predicting the effectiveness of a mass mailing.
Z-score
In statistics, a Z-score indicates how many standard deviations a data point is from the mean. When calculating confidence intervals, the Z-score is crucial. It adjusts for the fact that data naturally varies and helps to determine the "confidence" one can have in results.For a 90% confidence interval, as considered for the mail advertisement case, the Z-score used is 1.645. This value corresponds to the level of confidence that we want. Inserting this into the confidence interval formula results in:\[ 0.123 \pm 1.645 \times 0.0104 \]This calculation provides the range \( 0.1059 \text{ to } 0.1401 \), meaning we are 90% confident that the true population proportion lies within 10.59% to 14.01%. The choice of 1.645 reflects the specific confidence level, emphasizing how Z-scores allow us to tailor our inferences to particular assurances over sampling results.

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Most popular questions from this chapter

Hiring In preparing a report on the economy, we need to estimate the percentage of businesses that plan to hire additional employees in the next 60 days. a. How many randomly selected employers must we contact in order to create an estimate in which we are \(98 \%\) confident with a margin of error of \(5 \% ?\) b. Suppose we want to reduce the margin of error to \(3 \%\). What sample size will suffice? C. Why might it not be worth the effort to try to get an interval with a margin of error of only \(1 \% ?\)

Baseball fans In a poll taken in December 2012, Gallup asked 1006 national adults whether they were baseball fans; \(48 \%\) said they were. Almost five years earlier, in February \(2008,\) only \(35 \%\) of a similar-size sample had reported being baseball fans. a. Find the margin of error for the 2012 poll if we want \(90 \%\) confidence in our estimate of the percent of national adults who are baseball fans. b. Explain what that margin of error means. c. If we wanted to be \(99 \%\) confident, would the margin of error be larger or smaller? Explain. d. Find that margin of error. e. In general, if all other aspects of the situation remain the same, will smaller margins of error produce greater or less confidence in the interval?

Rickets Vitamin D, whether ingested as a dietary supplement or produced naturally when sunlight falls on the skin, is essential for strong, healthy bones. The bone disease rickets was largely eliminated in England during the 1950 s, but now there is concern that a generation of children more likely to watch TV or play computer games than spend time outdoors is at increased risk. A recent study of 2700 children randomly selected from all parts of England found \(20 \%\) of them deficient in vitamin D. a. Find a \(98 \%\) confidence interval. b. Explain carefully what your interval means. c. Explain what "98\% confidence" means.

Confidence intervals Several factors are involved in the creation of a confidence interval. Among them are the sample size, the level of confidence, and the margin of error. Which statements are true? a. For a given sample size, higher confidence means a smaller margin of error. b. For a specified confidence level, larger samples provide smaller margins of error. c. For a fixed margin of error, larger samples provide greater confidence. d. For a given confidence level, halving the margin of error requires a sample twice as large.

Teenage drivers An insurance company checks police records on 582 accidents selected at random and notes that teenagers were at the wheel in 91 of them. a. Create a \(95 \%\) confidence interval for the percentage of all auto accidents that involve teenage drivers. b. Explain what your interval means. c. Explain what "95\% confidence" means. d. A politician urging tighter restrictions on drivers' licenses issued to teens says, "In one of every five auto accidents, a teenager is behind the wheel." Does your confidence interval support or contradict this statement? Explain.

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