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Exercises 3.112 to 3.115 give information about the proportion of a sample that agree with a certain statement. Use StatKey or other technology to find a confidence interval at the given confidence level for the proportion of the population to agree, using percentiles from a bootstrap distribution. StatKey tip: Use "CI for Single Proportion" and then "Edit Data" to enter the sample information. Find a \(95 \%\) confidence interval if 35 agree in a random sample of 100 people.

Short Answer

Expert verified
Unfortunately, without having actual access to StatKey or another statistical software to perform bootstrap distribution simulation, we cannot provide numerical values for the lower and upper bounds of the confidence interval. However, these steps are what you'd generally follow to obtain the 95% confidence interval given a sample proportion and the desired confidence level.

Step by step solution

01

- Calculate Sample Proportion

The first step is to find the proportion of people who agree with the given statement in the sample. This is found by dividing the number of people who agree by the total number of people in the sample. Here, \(p = 35/100 = 0.35\) where p is the sample proportion.
02

- Understand Confidence Level

A confidence level of 95% implies that if we were to pick many random samples from the population and calculate the proportion for each of them, 95% of the calculated proportion would be within the confidence interval. In this step, understand that we want the interval that contains the true population proportion 95% of the time.
03

- Utilize Bootstrap Distribution

Bootstrapping is a technique of sampling in which we draw samples with replacement from the observed dataset. In doing so, it approximates the variability in the statistic due to the sampling process. The percentiles from the bootstrap distribution give us the confidence interval. You need to use StatKey or another software to simulate many drawn samples, calculate the proportion for each, and then find the 2.5th percentile and the 97.5th percentile. Those will define the lower and upper limits of the 95% confidence interval respectively due to the symmetric nature of the normal distribution.

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Key Concepts

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

Bootstrap Distribution
Imagine you have a bag of marbles, each representing a data point from your sample. To understand what might happen if you drew many more samples, you shake the bag and draw a marble, note its color, and then put it back before drawing again. This process is called 'sampling with replacement', and if you do this enough times, recording the colors each time, you'll get a good idea of the color distribution within the bag. This is the essence of a bootstrap distribution in statistics.

For our exercise, the bootstrap distribution is created by repeatedly sampling our original 100 people's responses, with replacement, and calculating a new sample proportion each time. By doing this many, many times—think thousands or tens of thousands—we build a distribution of sample proportions. This distribution represents different potential outcomes if the survey were to be conducted repeatedly under the same conditions. Each proportion in this distribution gives us insight into the variability of sample proportions that could occur by chance alone.
Sample Proportion
Sample proportion is the fraction of the sample that displays a certain characteristic—in this case, agreeing with a statement. It's like taking a snapshot of a crowd and counting how many are wearing hats to estimate the popularity of hats among the whole population. Here, we focus on the 35 people who agree out of the 100 surveyed, giving us a sample proportion of \(0.35\).

This single statistic provides a point estimate of the true proportion in the entire population but does not give us a measure of certainty or a margin of error. That's where the bootstrap distribution comes in handy: it helps us understand the variability around our sample proportion, offering a way to construct a confidence interval.
Confidence Level
A confidence level tells you how sure you can be about your estimates. It's a lot like a weather forecast telling you there's a 95% chance of sunshine tomorrow—it's not a guarantee, but it's a strong indication you might want to wear shorts and a t-shirt. In statistics, a 95% confidence level means that if we were to take 100 different samples and calculate confidence intervals for each one, we'd expect about 95 of those intervals to contain the true population parameter.

In the context of our exercise, a confidence interval with a 95% confidence level suggests that we're 95% confident the true population proportion who agree with the statement falls within our calculated range. It does not mean that there's a 95% chance that any given interval contains the true proportion—each interval either contains it or it doesn't—but over many intervals constructed from many samples, 95% should contain the true value.
StatKey
StatKey is a digital tool designed to help us with statistical concepts, akin to a Swiss Army knife for data scientists. Think of it as a reliable assistant in our data explorations, making complex tasks more manageable. For this exercise, we use the 'CI for Single Proportion' feature in StatKey to crunch the numbers. It creates simulated bootstrap samples, calculates sample proportions, and then uses these to find the bootstrap distribution.

After entering our sample information through the 'Edit Data' option, StatKey does the heavy lifting of generating the bootstrap samples and calculates the confidence interval. It’s a clear, visual way to understand and compute statistical concepts without getting bogged down by manual calculations or programming.
Sampling with Replacement
Sampling with replacement is like picking apples from a basket, checking each one, and then putting it back before selecting again. This method ensures that each pick is independent of the others, since the same configuration of apples is possible for every selection. In statistics, sampling with replacement from a dataset means each member of the dataset has the same chance of being selected each time we draw a new sample.

In the bootstrapping method we use for our exercise, each sample drawn contributes to the bootstrap distribution. By doing this repeatedly, we accurately reflect the sample's variability, hence creating a robust simulation of the sampling process. It's a cornerstone concept for creating the bootstrap distribution and understanding how sampling variability impacts our confidence intervals.

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Most popular questions from this chapter

3.62 Employer-Based Health Insurance A report from a Gallup poll \(^{29}\) in 2011 started by saying, "Forty-five percent of American adults reported getting their health insurance from an employer...." Later in the article we find information on the sampling method, "a random sample of 147,291 adults, aged 18 and over, living in the US," and a sentence about the accuracy of the results, "the maximum margin of sampling error is ±1 percentage point." (a) What is the population? What is the sample? What is the population parameter of interest? What is the relevant statistic? (b) Use the margin of error \(^{30}\) to give an interval showing plausible values for the parameter of interest. Interpret it in terms of getting health insurance from an employer.

Effect of Overeating for One Month: Average Long-Term Weight Gain Overeating for just four weeks can increase fat mass and weight over two years later, a Swedish study shows \(^{35}\) Researchers recruited 18 healthy and normal-weight people with an average age of \(26 .\) For a four-week period, participants increased calorie intake by \(70 \%\) (mostly by eating fast food) and limited daily activity to a maximum of 5000 steps per day (considered sedentary). Not surprisingly, weight and body fat of the participants went up significantly during the study and then decreased after the study ended. Participants are believed to have returned to the diet and lifestyle they had before the experiment. However, two and a half years after the experiment, the mean weight gain for participants was 6.8 lbs with a standard error of 1.2 lbs. A control group that did not binge had no change in weight. (a) What is the relevant parameter? (b) How could we find the actual exact value of the parameter? (c) Give a \(95 \%\) confidence interval for the parameter and interpret it. (d) Give the margin of error and interpret it.

3.34 A Sampling Distribution for Average Salary of NFL Players Use StatKey or other technology to generate a sampling distribution of sample means using a sample of size \(n=5\) from the YearlySalary values in the dataset NFLContracts2015, which gives the total and yearly money values from the contracts of all NFL players in 2015 . (a) What are the smallest and largest YearlySalary values in the population? (b) What are the smallest and largest sample means in the sampling distribution? (c) What is the standard error (that is, the standard deviation of the sample means) for the sampling distribution for samples of size \(n=5 ?\) (d) Generate a new sampling distribution with samples of size \(n=50 .\) What is the standard error for this sampling distribution?

Proportion of registered voters in a county who voted in the last election, using data from the county voting records.

Bisphenol A in Your Soup Cans Bisphenol A (BPA) is in the lining of most canned goods, and recent studies have shown a positive association between BPA exposure and behavior and health problems. How much does canned soup consumption increase urinary BPA concentration? That was the question addressed in a recent study \(^{34}\) in which consumption of canned soup over five days was associated with a more than \(1000 \%\) increase in urinary BPA. In the study, 75 participants ate either canned soup or fresh soup for lunch for five days. On the fifth day, urinary BPA levels were measured. After a two-day break, the participants switched groups and repeated the process. The difference in BPA levels between the two treatments was measured for each participant. The study reports that a \(95 \%\) confidence interval for the difference in means (canned minus fresh) is 19.6 to \(25.5 \mu \mathrm{g} / \mathrm{L}\). (a) Is this a randomized comparative experiment or a matched pairs experiment? Why might this type of experiment have been used? (b) What parameter are we estimating? (c) Interpret the confidence interval in terms of BPA concentrations. (d) If the study had included 500 participants instead of \(75,\) would you expect the confidence interval to be wider or narrower?

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