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Consider \(n=200\) binomial trials with \(r=80\) successes. (a) Is it appropriate to use a normal distribution to approximate the \(\hat{p}\) distribution? (b) Find a \(95 \%\) confidence interval for the population proportion of successes \(p\). (c) Explain the meaning of the confidence interval you computed.

Short Answer

Expert verified
Yes, a normal distribution is appropriate. The 95% confidence interval for the population proportion is [0.3322, 0.4678]. The interval suggests that we are 95% confident the true proportion of successes lies within this range.

Step by step solution

01

Determine Appropriateness for Normal Approximation

To determine if it's appropriate to use a normal distribution to approximate the distribution of \(\hat{p}\), we need to check the condition \(np \geq 5\) and \(n(1-p) \geq 5\). Here, \(\hat{p} = \frac{r}{n} = \frac{80}{200} = 0.4\). So, \(np = 200 \times 0.4 = 80\) and \(n(1-p) = 200 \times 0.6 = 120\), both of which are greater than 5. Therefore, it's appropriate to use a normal approximation.
02

Calculate the Standard Error

The standard error for the sample proportion is given by the formula \(SE_{\hat{p}} = \sqrt{\frac{\hat{p}(1-\hat{p})}{n}}\). Substituting \(\hat{p} = 0.4\) and \(n = 200\), we get: \(SE_{\hat{p}} = \sqrt{\frac{0.4 \cdot 0.6}{200}} = \sqrt{0.0012} \approx 0.0346\).
03

Find the 95% Confidence Interval

To find the 95% confidence interval, we use the formula \(\hat{p} \pm Z_{\frac{\alpha}{2}} \times SE_{\hat{p}}\), where \(Z_{\frac{\alpha}{2}}\) is the critical value for a 95% confidence level, typically 1.96. So, we calculate: \[\hat{p} \pm 1.96 \times 0.0346 = 0.4 \pm 1.96 \times 0.0346\]. This results in the interval \([0.4 - 0.0678, 0.4 + 0.0678]\), or \([0.3322, 0.4678]\).
04

Interpret the Confidence Interval

The 95% confidence interval \([0.3322, 0.4678]\) means that we can be 95% confident that the true population proportion of successes \(p\) lies within this interval. This does not mean that 95% of the population proportions will lie inside this interval, rather that if we were to take many samples and build a confidence interval from each, 95% of those intervals would contain the true population proportion.

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Key Concepts

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

Normal Distribution
The normal distribution is a fundamental concept in statistics. It is a continuous probability distribution often used to model real-valued random variables. Imagine a symmetrical bell-shaped curve, where the highest point is the mean of the data and represents the average outcome. The spread of the curve is determined by the standard deviation.Why is the normal distribution so important? If a dataset follows a normal distribution pattern, we can make predictions and calculate probabilities. It's used in hypothesis testing and constructing confidence intervals. In the context of binomial trials, when the conditions are met, the distribution of sample proportions can be approximated to a normal distribution, making complex calculations easier.In the exercise, we checked if the sample proportion \(\hat{p}\) could be modeled using a normal distribution. Because both \(np\) and \(n(1-p)\) are greater than 5, our sample is large enough to use this approximation reliably.
Binomial Trials
When we refer to binomial trials, we are talking about experiments or processes that have two possible outcomes, often called "success" and "failure." A simple example is flipping a coin, where outcomes are heads or tails.In mathematical terms, a binomial trial is repeated under identical conditions and each trial has the same probability of success. The number of successes in a given set of trials follows a binomial distribution.For a set of \(n\) binomial trials, each with probability \(p\) of success, the binomial distribution provides the likelihood of different possible numbers of successes. In our example, with \(n=200\) trials and \(r=80\) successes, we calculated the sample proportion \(\hat{p} = \frac{80}{200} = 0.4\) to aid in determining if and how we could apply normal approximation.
Population Proportion
The population proportion is an essential measure in statistics, representing the fraction of a population that has a certain attribute, often designated as \(p\).In sampling, we estimate this value using a sample proportion \(\hat{p}\), which is calculated as the number of observed "successes" divided by the total number of trials. - In the exercise, our \(\hat{p} \) came out to be \(0.4\), derived from 80 successes out of 200 trials.- The close link between \(p\) and \(\hat{p}\) is crucial when constructing a confidence interval: it allows us to estimate the range within which the true population proportion is likely to fall.The confidence interval provides insight into population characteristics without needing direct observation of every member.
Standard Error
The standard error is a statistical term that represents the accuracy with which a sample parameter estimates a population parameter. Essentially, it's the measure of how much \(\hat{p}\), our sample proportion, would vary if we took different samples of the same population size.The formula for standard error of a sample proportion is:\[SE_{\hat{p}} = \sqrt{\frac{\hat{p}(1-\hat{p})}{n}}\]In the exercise, we calculated \(SE_{\hat{p}}\) as approximately \(0.0346\). This value tells us the expected level of deviation of our sample proportion from the true population proportion.By understanding standard error, we can better construct our confidence intervals, quantifying our certainty and paving the way for informed decision-making based on sample data.

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Most popular questions from this chapter

Finance: \(\mathrm{P} / \mathrm{E}\) Ratio The price of a share of stock divided by a company's estimated future earnings per share is called the P/E ratio. High P/E ratios usually indicate "growth" stocks, or maybe stocks that are simply overpriced. Low P/E ratios indicate "value" stocks or bargain stocks. A random sample of 51 of the largest companies in the United States gave the following \(P / E\) ratios (Reference: Forbes). (a) Use a calculator with mean and sample standard deviation keys to verify that \(\bar{x} \approx 25.2\) and \(s \approx 15.5\) (b) Find a \(90 \%\) confidence interval for the \(\mathrm{P} / \mathrm{E}\) population mean \(\mu\) of all large U.S. companies. (c) Find a \(99 \%\) confidence interval for the \(\mathrm{P} / \mathrm{E}\) population mean \(\mu\) of all large U.S. companies. (d) Interpretation Bank One (now merged with J.P. Morgan) had a P/E of \(12,\) AT\&T Wireless had a \(\mathrm{P} / \mathrm{E}\) of \(72,\) and Disney had a \(\mathrm{P} / \mathrm{E}\) of 24 Examine the confidence intervals in parts (b) and (c). How would you describe these stocks at the time the sample was taken? (e) Check Requirements In previous problems, we assumed the \(x\) distribution was normal or approximately normal. Do we need to make such an assumption in this problem? Why or why not? Hint: See the central limit theorem in Section 6.5.

Assume that the population of \(x\) values has an approximately normal distribution. Archaeology: Tree Rings At Burnt Mesa Pueblo, the method of tree-ring dating gave the following years A.D. for an archacological excavation site (Bandelier Archaeological Excavation Project: Summer 1990 Excavations at Burnt Mesa Pueblo, edited by Kohler, Washington State University): \(1268 \quad 1316\) \(\begin{array}{cccc}1189 & 1271 & 1267 & 1272\end{array}\) 1275 (a) Use a calculator with mean and standard deviation keys to verify that the sample mean year is \(\bar{x}=1272,\) with sample standard deviation \(s \approx 37\) years. (b) Find a \(90 \%\) confidence interval for the mean of all tree-ring dates from this archaeological site. (c) Interpretation What does the confidence interval mean in the context of this problem?

Josh and Kendra each calculated a \(90 \%\) confidence interval for the difference of means using a Student's \(t\) distribution for random samples of size \(n_{1}=20\) and \(n_{2}=31 .\) Kendra followed the convention of using the smaller sample size to compute \(d . f .=19 .\) Josh used his calculator and Satterthwaite's approximation and obtained \(d, f . \approx 36.3 .\) Which confidence interval is shorter? Which confidence interval is more conservative in the sense that the margin of error is larger?

What percentage of your campus student body is female? Let \(p\) be the proportion of women students on your campus. (a) If no preliminary study is made to estimate \(p,\) how large a sample is needed to be \(99 \%\) sure that a point estimate \(\hat{p}\) will be within a distance of 0.05 from \(p ?\) (b) The Statistical Abstract of the United States, 1 12th edition, indicates that approximately \(54 \%\) of college students are female. Answer part (a) using this estimate for \(p\).

A requirement for using the normal distribution to approximate the \(\hat{p}\) distribution is that both \(n p > 5\) and \(n q > 5\) since we usually do not know \(p,\) we estimate \(p\) by \(\hat{p}\) and \(q\) by \(\hat{q}=1-\hat{p}\) Then we require that \(n \hat{p}>5\) and \(n \hat{q}>5 .\) Show that the conditions \(n \hat{p}>5\) and \(n \hat{q}>5\) are equivalent to the condition that out of \(n\) binomial trials, both the number of successes \(r\) and the number of failures \(n-r\) must exceed 5 Hint: In the inequality \(n \hat{p}>5,\) replace \(\hat{p}\) by \(r / n\) and solve for \(r .\) In the inequality \(n \hat{q}>5,\) replace \(\hat{q}\) by \((n-r) / n\) and solve for \(n-r\).

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