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91Ó°ÊÓ

Sam computed a \(90 \%\) confidence interval for \(\mu\) from a specific random sample of size \(n .\) He claims that at the \(90 \%\) confidence level, his confidence interval contains \(\mu\). Is his claim correct? Explain.

Short Answer

Expert verified
No, Sam's claim is not correct; confidence level applies to the method across many samples, not one interval.

Step by step solution

01

Understand Confidence Interval

A confidence interval is a range of values, derived from a sample, within which we expect a population parameter to lie. A 90% confidence interval means that if we were to take 100 different samples and compute a confidence interval for each sample, we would expect about 90 of those intervals to contain the actual population mean \( \mu \).
02

Analyze Sam's Claim

Sam claims that his specific confidence interval at the 90% confidence level contains \( \mu \). This assertion is about a single interval, and it is important to note that any single confidence interval does not guarantee the inclusion of \( \mu \). The 90% confidence level applies across many samples, not to a single instance.
03

Clarify Probability Interpretation

The probability that any specific confidence interval, such as Sam's, contains \( \mu \) is either 0 (\( \mu \) is not in the interval) or 1 (\( \mu \) is in the interval). The 90% figure doesn't describe this probability, but rather the broader reliability of the method to capture \( \mu \) over numerous samples.
04

Conclusion on Claim Validity

Thus, Sam's claim is a misunderstanding of confidence intervals. His claim is not strictly correct because the confidence level pertains to a long-run frequency of many intervals capturing \( \mu \), rather than a certainty about his individual interval.

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Key Concepts

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

Population Mean
The population mean, often denoted by \( \mu \), represents the average of all values in a population. In probability and statistics, it's crucial when trying to understand the overall behavior of a population.Consider the population mean as a point of balance in a dataset. It gives a holistic view, not focused on an individual observation, but rather the average of all data points.When dealing with samples, we estimate the population mean using the sample mean. However, this sample mean is just an estimate, which might vary from sample to sample. This is where confidence intervals come into play, helping to infer the true population mean based on the sample data we have. Remember, our goal is not just to find what the sample tells us but to infer what the entire population would look like if we had complete data.
Probability Interpretation
Probability interpretation in the context of confidence intervals is often misunderstood. It's important to clarify that a specific confidence interval either contains the population mean \( \mu \) or it does not.- When we speak of a 90% confidence interval, we don't mean there's a 90% chance that the interval we calculated from our sample contains the true mean.- Instead, the 90% figure represents a long-term frequency interpretation. If we were to repeatedly take samples and calculate intervals, around 90% of those intervals would contain \( \mu \).This understanding makes confidence intervals useful for gauging how reliable the method is over many samples, rather than providing a probability statement about just one interval.
Sample Size
Sample size, denoted as \( n \), plays a significant role in the estimation of a confidence interval. It affects how well the sample mean approximates the population mean.Here’s why sample size matters:
  • **Accuracy**: Larger samples tend to give a more accurate estimate of the population mean since they reduce the margin of error. Smaller samples might lead to wider intervals, reflecting greater uncertainty.
  • **Variability**: Increased sample size generally leads to less variability in the confidence interval estimates, narrowing them and making them more precise.
When planning data collection, an adequate sample size is essential for reliable inference, ensuring our results are not just statistical flukes.
Confidence Level
The confidence level indicates the degree of certainty we have that our confidence interval contains the true population parameter.- A 90% confidence level means we expect about 90 out of 100 constructed intervals to contain the true population mean \( \mu \). It's about trustworthiness in the interval generation process, rather than the certainty of any one interval.Consider confidence level as your safety net:
  • Higher confidence levels (e.g., 95%, 99%) increase the width of the interval, providing more assurance that it captures the population mean, but potentially at the cost of precision.
  • Lower confidence levels offer narrower intervals but increase the risk of not capturing \( \mu \).
Ultimately, the confidence level chosen should reflect how much uncertainty you're willing to accept, balancing precision against the probability of capturing the true parameter.

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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 \(\mathrm{P} / \mathrm{E}\) 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). $$ \begin{array}{rrrrrrrrrrrr} 11 & 35 & 19 & 13 & 15 & 21 & 40 & 18 & 60 & 72 & 9 & 20 \\ 29 & 53 & 16 & 26 & 21 & 14 & 21 & 27 & 10 & 12 & 47 & 14 \\ 33 & 14 & 18 & 17 & 20 & 19 & 13 & 25 & 23 & 27 & 5 & 16 \\ 8 & 49 & 44 & 20 & 27 & 8 & 19 & 12 & 31 & 67 & 51 & 26 \\ 19 & 18 & 32 & & & & & & & & & \end{array} $$ (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 \(6.5\).

Large U.S. Companies: Foreign Revenue For large U.S. companies, what percentage of their total income comes from foreign sales? A random sample of technology companies (IBM, Hewlett-Packard, Intel, and others) gave the following information. $$ \begin{aligned} &\text { Technology companies, } \% \text { foreign revenue: } x_{1} ; n_{1}=16\\\ &\begin{array}{llllllll} 62.8 & 55.7 & 47.0 & 59.6 & 55.3 & 41.0 & 65.1 & 51.1 \\ 53.4 & 50.8 & 48.5 & 44.6 & 49.4 & 61.2 & 39.3 & 41.8 \end{array} \end{aligned} $$ Another independent random sample of basic consumer product companies (Goodyear, Sarah Lee, H.J. Heinz, Toys " \(q\) "Us) gave the following information. $$ \begin{aligned} &\text { Basic consumer product companies, } \% \text { foreign revenue: } x_{2} ; n_{2}=17\\\ &\begin{array}{llllllll} 28.0 & 30.5 & 34.2 & 50.3 & 11.1 & 28.8 & 40.0 & 44.9 \\ 40.7 & 60.1 & 23.1 & 21.3 & 42.8 & 18.0 & 36.9 & 28.0 \end{array}\\\ &\begin{aligned} &40.7 \\ &32.5 \end{aligned} \end{aligned} $$ (Reference: Forbes Top Companies.) Assume that the distributions of percentage foreign revenue are mound-shaped and symmetric for these two company types. (a) Use a calculator with mean and standard deviation keys to verify that \(\bar{x}_{1} \approx 51.66, s_{1} \approx 7.93, \bar{x}_{2} \approx 33.60\), and \(s_{2} \approx 12.26\). (b) Let \(\mu_{1}\) be the population mean for \(x_{1}\) and let \(\mu_{2}\) be the population mean for \(x_{2}\). Find an \(85 \%\) confidence interval for \(\mu_{1}-\mu_{2}\). (c) Interpretation Examine the confidence interval and explain what it means in the context of this problem. Does the interval consist of numbers that are all positive? all negative? of different signs? At the \(85 \%\) level of confidence, do technology companies have a greater percentage foreign revenue than basic consumer product companies? (d) Check Requirements Which distribution (standard normal or Student's \(t)\) did you use? Why?

Answer true or false. Explain your answer. Every random sample of the same size from a given population will produce exactly the same confidence interval for \(\mu\).

Basic Computation: Confidence Interval for p Consider \(n=200\) binomial trials with \(r=80\) successes. (a) Cbeck Requirements 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) Interpretation Explain the meaning of the confidence interval you computed.

Jerry tested 30 laptop computers owned by classmates enrolled in a large computer science class and discovered that 22 were infected with keystroke- tracking spyware. Is it appropriate for Jerry to use his data to estimate the proportion of all laptops infected with such spyware? Explain.

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