/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 49 The article "National Geographic... [FREE SOLUTION] | 91Ó°ÊÓ

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The article "National Geographic, the Doomsday Machine," which appeared in the March 1976 issue of the Journal of Irreproducible Results (yes, there really is a journal by that name -it's a spoof of technical journals!) predicted dire consequences resulting from a nationwide buildup of National Geographic magazines. The author's predictions are based on the observation that the number of subscriptions for National Geographic is on the rise and that no one ever throws away a copy of National Geographic. A key to the analysis presented in the article is the weight of an issue of the magazine. Suppose that you were assigned the task of estimating the average weight of an issue of National Geographic. How many issues should you sample to estimate the average weight to within \(0.1 \mathrm{oz}\) with \(95 \%\) confidence? Assume that \(\sigma\) is known to be 1 oz.

Short Answer

Expert verified
The needed sample size for estimating the average weight of a National Geographic magazine within 0.1 oz with 95% confidence is 385.

Step by step solution

01

Understand the Given Values

In the exercise, you are given a few elements:\nThe confidence level, which is 95%.\nAn error margin, which is 0.1 oz.\nThe population standard deviation - denoted by the Greek letter \(\sigma\) - which is 1 oz.
02

Calculate the Z-value

To calculate the necessary sample size, you also need the z-value associated with the desired confidence level. The z-value is a statistical measurement that describes a value's relationship to the mean of a group of values. You can find the Z-value from a standard Z-table or alternatively use online calculators or statistical software. For a 95% confidence level, the Z-value is approximately 1.96.
03

Apply the Formula for Sample Size

The formula used for estimating a sample size when population standard deviation is known: \[n = \left( \frac{Z*\sigma}{E} \right)^2\] where n is the sample size, Z is the Z-value from a standard normal distribution corresponding to the desired confidence level (in this case 1.96), \( \sigma \) is the population standard deviation (1 oz) and E denotes the margin of error (0.1 oz). Plugging in these values the formula becomes: \[n = \left (\frac{1.96 * 1}{0.1}\right )^2\]
04

Calculate Sample Size

Carrying out the calculation will provide the required sample size. Since you can't have a fraction of a magazine, you would round up to the nearest whole number if the calculation doesn't yield an exact whole number. In this case, the formula provides a sample size of 384.16, so you would round up to 385. So, you need to sample 385 issues of National Geographic to estimate the average weight to within 0.1 oz with 95% confidence, given that the standard deviation is 1 oz.

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Key Concepts

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

Confidence Interval
When we talk about a confidence interval, we're referring to a range of values that is used to estimate an unknown population parameter, such as the average weight in our exercise. It is important because it gives us an idea of how much our estimate could vary. In our example with National Geographic magazines, we want to know the average weight with a certain degree of certainty, or 'confidence'. The 95% confidence level suggests that if we were to take 100 different samples and compute a confidence interval for each, we would expect about 95 of those intervals to contain the actual average weight. This confidence level helps us understand the reliability of our estimate. Confidence intervals are crucial for making informed decisions in statistics by accounting for natural variability in data.
Standard Deviation
Standard deviation is a vital concept in statistics, as it measures how spread out the numbers in a data set are. It tells us about the variability or dispersion of data points around the mean or average value. In simpler terms, a smaller standard deviation means the values are more clustered around the mean, while a larger deviation means they are more spread out.

In the context of the exercise, the known standard deviation of 1 oz for National Geographic magazines provides an essential piece of information, helping us understand the variability in magazine weights. This knowledge is integral to determining sample size because it affects how many samples we need to achieve a desired level of precision. The less variability there is, represented by a smaller standard deviation, the fewer samples are needed. Conversely, more variability would require more samples to gain an accurate estimate.
Margin of Error
The margin of error is like a buffer zone around your estimate, indicating the range within which the true population parameter is expected to fall. It reflects the amount of error you can tolerate in your estimates from a sample. In the context of the exercise, our margin of error is set at 0.1 oz, meaning we want our sample mean to fall within 0.1 oz of the true population mean of magazine weights.

Choosing an appropriate margin of error is crucial, as a smaller margin means you’ll need more samples to attain greater accuracy, whereas a larger margin allows for less precision but requires fewer samples. The margin of error shows a trade-off between the desired precision and the resources needed to achieve it, such as time or cost. Understanding this concept helps in designing experiments and surveys effectively by balancing precision with practicability.
Z-value
A Z-value, often referred to as a Z-score, is a statistical measurement that helps determine the position of a single value in relation to the mean of a set of data. In our exercise, the Z-value corresponds to the desired confidence level. It's used to calculate how far a value deviates from the mean, measured in standard deviation units.

For a 95% confidence interval, we used a Z-value of approximately 1.96. This number comes from the standard normal distribution table and signifies that 95% of the data lies within 1.96 standard deviations from the mean.
  • The higher the confidence level, the larger the Z-value, implying a wider interval.
  • A lower confidence level will have a smaller Z-value, leading to a narrower interval.
Knowing the appropriate Z-value to use is essential for accurately calculating sample sizes and understanding the extent of confidence in your estimations.

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Most popular questions from this chapter

Five hundred randomly selected working adults living in Calgary, Canada were asked how long, in minutes, their typical daily commute was (Calgary Herald Traffic Study, Ipsos, September 17,2005 ). The resulting sample mean and standard deviation of commute time were \(28.5\) minutes and \(24.2\) minutes, respectively. Construct and interpret a \(90 \%\) confidence interval for the mean commute time of working adult Calgary residents.

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The Associated Press (December 16,1991 ) reported that in a random sample of 507 people, only 142 correctly described the Bill of Rights as the first 10 amendments to the U.S. Constitution. Calculate a \(95 \%\) confidence interval for the proportion of the entire population that could give a correct description.

The report "2005 Electronic Monitoring \& Surveillance Survey: Many Companies Monitoring, Recording, Videotaping-and Firing-Employees" (American Management Association, 2005 ) summarized the results of a survey of 526 U.S. businesses. The report stated that 137 of the 526 businesses had fired workers for misuse of the Internet and 131 had fired workers for email misuse. For purposes of this exercise, assume that it is reasonable to regard this sample as representative of businesses in the United States. a. Construct and interpret a \(95 \%\) confidence interval for the proportion of U.S. businesses that have fired workers for misuse of the Internet. b. What are two reasons why a \(90 \%\) confidence interval for the proportion of U.S. businesses that have fired workers for misuse of email would be narrower than the \(95 \%\) confidence interval computed in Part (a).

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