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Chapter 7: Problem 9

Lorraine computed a confidence interval for \(\mu\) based on a sample of size \(41 .\) Since she did not know \(\sigma,\) she used \(s\) in her calculations. Lorraine used the normal distribution for the confidence interval instead of a Student's \(t\) distribution. Was her interval longer or shorter than one obtained by using an appropriate Student's \(t\) distribution? Explain.

Short Answer

Expert verified
Lorraine's interval is shorter because the normal distribution underestimates variability compared to the Student's t distribution.

Step by step solution

01

Identify Distribution Used

Lorraine used the normal distribution to compute the confidence interval, but the sample size is 41 and population standard deviation \( \sigma \) is unknown. This implies that the Student's \( t \) distribution should have been used.
02

Understand Distribution Characteristics

The normal distribution and the Student's \( t \) distribution are both symmetric and bell-shaped, but the Student's \( t \) distribution has heavier tails, especially for smaller sample sizes. This means it accounts for more variability when the sample size is small. As the sample size increases, the Student's \( t \) distribution becomes similar to the normal distribution.
03

Compare Critical Values

For the same confidence level, the critical value from the Student's \( t \) distribution is greater than that from the normal distribution, when the sample size is not large enough for the two distributions to be approximately equal. Since Lorraine's sample size is 41, the critical value from the Student's \( t \) distribution would be slightly larger than that from the normal distribution.
04

Determine Interval Length

The length of the confidence interval is influenced by the critical value; larger critical values result in longer intervals. Since the Student's \( t \) distribution gives a larger critical value than the normal distribution (due to heavier tails), using the Student's \( t \) distribution would result in a longer confidence interval compared to the normal distribution.

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Key Concepts

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

Student's t distribution
When calculating a confidence interval and the population standard deviation \( \sigma \) is unknown, statisticians often use the Student's \( t \) distribution. This is particularly important in situations where the sample size is relatively small. The Student's \( t \) distribution is similar to the normal distribution; however, it has heavier tails. This heavier tail means that it accounts for more variability in data, which is likely when we don't have the full information about the population.The key properties of the Student's \( t \) distribution include:
  • It is symmetric and bell-shaped, just like the normal distribution.
  • It is defined by degrees of freedom, which in simple terms means the size of the sample minus one (\( n-1 \)).
  • As the sample size increases, the Student's \( t \) distribution becomes similar to the normal distribution, meaning its tails grow lighter as more data is collected.
For Lorraine's calculation, not knowing the population standard deviation was crucial. With a sample size of 41 and no \( \sigma \), the more appropriate choice of distribution would have been the Student's \( t \) rather than the normal distribution.
Normal distribution
The normal distribution is a fundamental concept in statistics. It is often visualized as a bell curve and described by its mean and standard deviation. This distribution is applicable under the assumption that data is continuously distributed and ideally when the full population parameters are known.The normal distribution is:
  • Completely described by two parameters: the mean (\( \mu \)) and standard deviation (\( \sigma \)).
  • Symmetric around its mean.
  • Characterized by its lighter tails compared to the Student's \( t \) distribution, which means it assumes less variability.
Since Lorraine used the normal distribution for her confidence interval, and given that she didn't know \( \sigma \), this may have led to a more narrow and possibly misleading interval. This is because the normal distribution assumes known population parameters, which were not available in this case.
Critical values
Critical values are essential when calculating confidence intervals, as they determine how wide these intervals will be. They are the multiplier of the standard error in the confidence interval formula.Here's what you need to know:
  • For a given confidence level, the critical value from the Student's \( t \) distribution will be larger than from the normal distribution if the sample size is not sufficiently large. This is due to the heavier tails of the Student's \( t \) distribution.
  • In a confidence interval, larger critical values yield wider intervals, indicating greater uncertainty about the estimate.
In Lorraine's situation, using the normal distribution would have provided her with a smaller critical value compared to the Student's \( t \), thereby resulting in a shorter confidence interval. Thus, if you're working with smaller sample sizes or unknown population parameters, opting for the Student's \( t \) critical value is usually more appropriate to capture the true variability in your 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.

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 random sample of 5222 permanent dwellings on the entire Navajo Indian Reservation showed that 1619 were traditional Navajo hogans (Navajo Architecture: Forms, History, Distributions by Jett and Spencer, University of Arizona Press). (a) Let \(p\) be the proportion of all permanent dwellings on the entire Navajo Rescrvation that are traditional hogans. Find a point estimate for \(p\). (b) Find a \(99 \%\) confidence interval for \(p\). Give a brief interpretation of the confidence interval. (c) Do you think that \(n p > 5\) and \(n q > 5\) are satisfied for this problem? Explain why this would be an important consideration.

The National Council of Small Businesses is interested in the proportion of small businesses that declared Chapter 11 bankruptcy last year. Since there are so many small businesses, the National Council intends to estimate the proportion from a random sample. Let \(p\) be the proportion of small businesses that declared Chapter 11 bankruptcy last year. (a) If no preliminary sample is taken to estimate \(p,\) how large a sample is necessary to be \(95 \%\) sure that a point estimate \(\hat{p}\) will be within a distance of 0.10 from \(p ?\) (b) In a preliminary random sample of 38 small businesses, it was found that six had declared Chapter 11 bankruptcy. How many more small businesses should be included in the sample to be \(95 \%\) sure that a point estimate \(\hat{p}\) will be within a distance of 0.10 from \(p ?\)

Jobs and productivity! How do retail stores rate? One way to answer this question is to examine annual profits per employee. The following data give annual profits per employee (in units of 1 thousand dollars per employee) for companies in retail sales. (See reference in Problem \(23 .\) ) Companies such as Gap, Nordstrom, Dillards, JCPenney, Sears, Wal-Mart, Office Depot, and Toys " \(\mathrm{A}\) " Us are included. Assume \(\sigma \approx 3.8\) thousand dollars. $$\begin{array}{rrrrrrrrrrrr}4.4 & 6.5 & 4.2 & 8.9 & 8.7 & 8.1 & 6.1 & 6.0 & 2.6 & 2.9 & 8.1 & -1 . \\\11.9 & 8.2 & 6.4 & 4.7 & 5.5 & 4.8 & 3.0 & 4.3 & -6.0 & 1.5 & 2.9 & 4 . \\ -1.7 & 9.4 & 5.5 & 5.8 & 4.7 & 6.2 & 15.0 & 4.1 & 3.7 & 5.1 & 4.2 &\end{array}$$ (a) Use a calculator or appropriate computer software to verify that, for the preceding data, \(\bar{x} \approx 5.1 .\) (b) Let us say that the preceding data are representative of the entire sector of retail sales companies. Find an \(80 \%\) confidence interval for \(\mu,\) the average annual profit per employee for retail sales. (c) Interpretation Let us say that you are the manager of a retail store with a large number of employees. Suppose the annual profits per employee are less than 3 thousand dollars per employee. Do you think this might be low compared with other retail stores? Explain by referring to the confidence interval you computed in part (b). (d) Interpretation Suppose the annual profits are more than 6.5 thousand dollars per employee. As store manager, would you feel somewhat better? Explain by referring to the confidence interval you computed in part (b). (e) Repeat parts \((b),(c),\) and (d) for a \(95 \%\) confidence interval.
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