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Do You Eat Red Meat? About \(60 \%\) of American adults include beef and other red meat in their diets. \(-6\) A plant-based meat company contacts an SRS of 1500 American adults and calculates the proportion \(\widehat{p}\) in this sample who eat red meat. a. What is the approximate distribution of \(\widehat{p}\) ? b. If the sample size were 6000 rather than 1500 , what would be the approximate distribution of \(\widehat{p}\) ?

Short Answer

Expert verified
For 1500: Normal with mean 0.60 and SD 0.0126; for 6000: Normal with mean 0.60 and SD 0.0063.

Step by step solution

01

Understand the Problem

We need to find the approximate distribution of the sample proportion, \(\widehat{p}\), of American adults who eat red meat. We are given that about \(60\%\) of the population eat red meat and we have samples of two different sizes (1500 and 6000).
02

Determine the Parameters for 1500 Sample Size

Given \(p = 0.60\) and \(n = 1500\). The mean of the distribution of \(\widehat{p}\) is \(\mu_{\widehat{p}} = p = 0.60\). The standard deviation is given by \(\sigma_{\widehat{p}} = \sqrt{\frac{p(1-p)}{n}} = \sqrt{\frac{0.60 \times 0.40}{1500}}\).
03

Calculate Standard Deviation for 1500 Sample Size

The standard deviation of \(\widehat{p}\) for a sample size of 1500 is \(\sqrt{\frac{0.24}{1500}} = \sqrt{0.00016} \approx 0.0126\).
04

Determine Distribution for 1500 Sample Size

The distribution of \(\widehat{p}\) is approximately normal with mean \(0.60\) and standard deviation \(0.0126\) due to the Central Limit Theorem.
05

Determine the Parameters for 6000 Sample Size

For \(n = 6000\), the mean remains \(\mu_{\widehat{p}} = p = 0.60\). The standard deviation is now \(\sigma_{\widehat{p}} = \sqrt{\frac{0.60 \times 0.40}{6000}}\).
06

Calculate Standard Deviation for 6000 Sample Size

The standard deviation of \(\widehat{p}\) for a sample size of 6000 is \(\sqrt{\frac{0.24}{6000}} = \sqrt{0.00004} \approx 0.0063\).
07

Determine Distribution for 6000 Sample Size

The distribution of \(\widehat{p}\) is approximately normal with mean \(0.60\) and standard deviation \(0.0063\).
08

Conclusion

For a sample size of 1500, \(\widehat{p}\) is normally distributed with mean 0.60 and standard deviation 0.0126. For a sample size of 6000, \(\widehat{p}\) is normally distributed with mean 0.60 and standard deviation 0.0063.

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Key Concepts

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

Central Limit Theorem
The Central Limit Theorem (CLT) is a fundamental concept in statistics that helps us understand how we can make inferences about a population from a sample. It states that when you take a sufficiently large sample from any population, with a known average and standard deviation, the distribution of the sample mean (or proportion) will tend to be normal (bell-shaped), even if the original population distribution is not.

This is particularly useful because it allows statisticians to make predictions about population parameters, even when the population is not normally distributed. In our example, the proportion of American adults eating red meat is taken from a sample. We are interested in the distribution of this sample proportion, represented as \( \hat{p} \).
  • For a sample size of 1500, the distribution of \( \hat{p} \) becomes approximately normal due to the CLT.
  • As the sample size increases to 6000, the redistribution of \( \hat{p} \) gets even closer to the traditional bell curve shape.
This central concept helps us approximate the probability distribution of the sample proportion, thereby modeling the real-world scenario even when dealing with incomplete data from a subset of the population.
Normal Distribution
The normal distribution is a continuous probability distribution characterized by its bell-shaped curve, also known as the Gaussian distribution. It is symmetrical around its mean, indicating that data near the mean are more frequent in occurrence than data far from the mean. This makes it a powerful tool in statistics for simplifying and understanding complex data.

In our scenario, the normal distribution plays a crucial role in understanding the behavior of the sample proportion \( \hat{p} \). Based on the Central Limit Theorem, \( \hat{p} \) approaches normal distribution as the sample size increases.
  • With a sample size of 1500, \( \hat{p} \) follows a normal distribution with a mean \( \mu_{\hat{p}} = 0.60 \) and standard deviation \( \sigma_{\hat{p}} = 0.0126 \).
  • For an increased sample size of 6000, \( \hat{p} \) remains normally distributed with a smaller standard deviation \( \sigma_{\hat{p}} = 0.0063 \).
The key takeaway is that normal distributions allow us to predict the behavior of data with known probabilities, effectively modeling real-world phenomena such as the inclination to consume red meat among a population.
Standard Deviation Calculation
Calculating the standard deviation is an important step in understanding the spread of data around the mean in any probability distribution. In our case, it helps quantify the variability of the sample proportion \( \hat{p} \) of American adults who eat red meat.

Standard deviation is derived using the formula for a sample proportion \( \sigma_{\hat{p}} = \sqrt{\frac{p(1-p)}{n}} \), where \( p \) represents the population proportion (0.60 for red meat consumption), and \( n \) is the sample size.
  • For a sample size of 1500, the standard deviation calculation is \( \sqrt{\frac{0.24}{1500}} \approx 0.0126 \).
  • When increased to 6000, the standard deviation becomes \( \sqrt{\frac{0.24}{6000}} \approx 0.0063 \).
A smaller standard deviation indicates less variability in the sample proportion, meaning the data points are more tightly clustered around the mean. This insight helps statisticians make more precise predictions about the population, illustrating the value of larger sample sizes in reducing uncertainty in statistical analysis.

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Most popular questions from this chapter

Surveying Students. You are planning a survey of students at a large university to determine what proportion favor an increase in student fees to support an expansion of the student newspaper. Using records provided by the registrar, you can select a random sample of students. You will ask each student in the sample whether he or she is in favor of the proposed increase. Your budget will allow a sample of 100 students. a. For a sample of size 100, construct a table of the margins of error for \(95 \%\) confidence intervals when \(\widehat{p}\) takes the values \(0.1,0.3,0.5,0.7\), and \(0.9\). b. A former editor of the student newspaper offers to provide funds for a sample of size 500 . Repeat the margin of error calculations in part (a) for the larger sample size. Then write a short thank-you note to the former editor, describing how the larger sample size will improve the results of the survey. In responding to \(\underline{\text { Exercises } 22.35}\) through \(\underline{22.41}\), follow the Plan, Solve, and Conclude steps of the four- step process.

Harris Announces a Margin of Error. Exercise \(22.25\) describes a Harris Poll survey of smokers in which 848 of a sample of 1010 smokers agreed that smoking would probably shorten their lives. Harris announces a margin of error of \(\pm 3\) percentage points for all samples of about this size. Opinion polls announce the margin of error for \(95 \%\) confidence. a. What is the actual margin of error (in percent) for the large-sample confidence interval from this sample? b. The margin of error is largest when \(\hat{p}=0.5\). What would the margin of error (in percent) be if the sample had resulted in \(\widehat{p}=0.5\) ? c. Why do you think that Harris announces a \(\pm 3 \%\) margin of error for all samples of about this size?

Usage of the Olympic National Park. U.S. National Parks that contain designated wilderness areas are required by law to develop and maintain a wilderness stewardship plan. The Olympic National Park, containing some of the most biologically diverse wilderness in the United States, had a survey conducted in 2012 to collect information relevant to the development of such a plan. National Park Service staff visited 30 wilderness trailheads in moderate-to high-use areas over a 60-day period and asked visitors as they completed their hike to complete a questionnaire. The 1019 completed questionaires, giving a response rate of \(50.4 \%\), provided each subject's opinions on the use and management of wilderness. In particular, there were 694 day users and 325 overnight users in the sample. a. Why do you think the National Park staff only visited trailheads in moderate-to high-use areas to obtain the sample? b. Assuming that the 1019 subjects represent a random sample of users of the wilderness areas in Olympic National Park, give a \(90 \%\) confidence interval for the proportion of day users. c. The response rate was \(49 \%\) for day users and \(52 \%\) for overnight users. Does this lessen any concerns you might have regarding the effect of nonresponse on the interval you obtained in part (b)? Explain briefly. d. Do you think it would be better to refer to the interval in part (b) as a confidence interval for the proportion of day users or the proportion of day users on the most popular trails in the park? Explain briefly.

Do You Listen to Podcasts? In January and February 2019, Edison Research conducted a national telephone survey of 1500 Americans aged 12 and older, using random digit dialing techniques to both cell phones and landlines. The survey included questions about the use of mobile devices, Internet audio, podcasting, social media, smart speakers, and more. Of the 1500 people surveyed, 480 said they had listened to a podcast in the past month. 12 a. What is the margin of error of the large-sample \(95 \%\) confidence interval for the proportion of Americans aged 12 and older who had listened to a podcast in the past month? b. How large a sample is needed to get the common \(\pm 3\) percentage point margin of error? Use the January/February survey of 1500 as a pilot study to get \(p^{*}\)

The value of the \(z\) statistic for the Question \(22.22\) is \(2.53\). This test is a. not significant at either \(\alpha=0.05\) or \(\alpha=0.01\). b. significant at \(\alpha=0.05\) but not at \(\alpha=0.01\). c. significant at both \(\alpha=0.05\) and \(\alpha=0.01\).

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