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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.

Short Answer

Expert verified
90% CI: (21.56, 28.84); 99% CI: (19.38, 31.02). Bank One is a "value" stock; AT&T Wireless is "overpriced"; Disney is typical. Central limit theorem supports our method.

Step by step solution

01

Verify Given Statistics

We are given that the sample mean \( \bar{x} \approx 25.2 \) and the sample standard deviation \( s \approx 15.5 \). Using a calculator or relevant software, you input the 51 data points to calculate these statistics. Ensure they match approximately before proceeding with the confidence intervals.
02

Find 90% Confidence Interval

To find the 90% confidence interval for the mean \( \mu \), use the formula: \[ CI = \bar{x} \pm t \frac{s}{\sqrt{n}} \] where \( \bar{x} = 25.2 \), \( s = 15.5 \), and \( n = 51 \). The critical value \( t \) for 90% confidence for \( n-1 = 50 \) degrees of freedom can be found using t-distribution tables or software, yielding approximately \( t \approx 1.676 \). Substitute these into the formula to get the interval.
03

Calculate 90% Confidence Interval

Using the values from Step 2, the 90% confidence interval is calculated as follows: \[ CI = 25.2 \pm 1.676 \frac{15.5}{\sqrt{51}} \approx 25.2 \pm 3.64 \] This results in a confidence interval of approximately (21.56, 28.84).
04

Find 99% Confidence Interval

For the 99% confidence interval, use the similar formula: \( CI = \bar{x} \pm t \frac{s}{\sqrt{n}} \). The critical value \( t \) for 99% confidence with 50 degrees of freedom is approximately \( t \approx 2.678 \). Substitute these values into the formula.
05

Calculate 99% Confidence Interval

Using the values from Step 4, the 99% confidence interval is calculated as follows: \[ CI = 25.2 \pm 2.678 \frac{15.5}{\sqrt{51}} \approx 25.2 \pm 5.82 \] This results in a confidence interval of approximately (19.38, 31.02).
06

Interpretation of Confidence Intervals

Comparing the P/E ratios: Bank One with 12 is well below both intervals, suggesting it is likely a "value" stock. AT&T Wireless with 72 is well above both intervals, indicating it could be considered "overpriced." Disney with 24 falls within both intervals, suggesting it's typical relative to this sample of companies.
07

Check Requirements

According to the central limit theorem, since our sample size is greater than 30 (\( n = 51 \)), the sample mean will approximately follow a normal distribution regardless of the original distribution. Therefore, we do not need to assume a normal distribution of the \( P/E \) ratios.

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Key Concepts

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

Confidence Intervals
Confidence intervals offer a range of values within which we expect the population parameter, like a mean or proportion, to lie. They provide a way to understand the uncertainty surrounding our sample estimates.
  • To create a confidence interval, we use the sample data mean and a margin of error.
  • The margin of error is calculated using the sample standard deviation and a critical value from a statistical distribution like the t-distribution.
For example, in our exercise, we calculated a 90% confidence interval for the mean P/E ratio using sample mean 25.2 and critical value approximately 1.676. This interval indicates that we are 90% confident that the true mean P/E ratio lies within the interval calculated.
Central Limit Theorem
The Central Limit Theorem (CLT) is a fundamental principle in statistics. It states that the sampling distribution of the sample mean will be approximately normally distributed, provided the sample size is large enough, generally more than 30.
  • This theorem holds true regardless of the population's distribution.
  • As the sample size increases, the distribution of the mean approaches a normal distribution.
In the context of our exercise, the sample size is 51. Thanks to the CLT, we can assume that the distribution of the sample mean P/E ratio approximates a normal distribution. Thus, allowing us to calculate confidence intervals safely without assuming that the P/E ratios themselves are normally distributed.
T-distribution
The t-distribution is a probability distribution that's used when working with small sample sizes or when the population standard deviation is unknown. In many ways, it's similar to a normal distribution but with heavier tails.
  • The t-distribution becomes closer to a normal distribution as sample size increases.
  • Critical values from the t-distribution are used in calculating confidence intervals when the sample size is small or the population standard deviation is unknown.
In our exercise, because the sample size is reasonable (51), we use the t-distribution instead of the normal distribution to determine the critical values. This provides more accurate intervals given that exact population parameters are unknown.
P/E Ratio
The P/E ratio, or Price-to-Earnings ratio, is a key financial metric used to evaluate the value of a company. It measures the current share price relative to its per-share earnings.
  • Higher P/E ratios can indicate growth potential or possibly overvaluation.
  • Conversely, lower ratios may suggest value stocks which are bargains according to market perceptions.
Our job as analysts is to interpret where a company stands relative to these ratios. In the case provided in the exercise, companies like Bank One and AT&T Wireless have significantly different P/E ratios, illustrating their perceived market valuations at the time.
Sample Standard Deviation
Sample standard deviation is a measure of the amount of variability or dispersion in a set of sample values. It gives us an idea of how much individual data points differ from the sample mean.
  • A smaller standard deviation indicates data points are closer to the mean.
  • Larger standard deviation suggests greater variability.
In our exercise, the P/E ratios showed a sample standard deviation of 15.5, reflecting the diversity in company valuations within the sample of 51 U.S. companies. This value is crucial when calculating confidence intervals as it determines the margin of error, helping us understand how varied the P/E ratios are from the mean.

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Most popular questions from this chapter

What price do farmers get for their watermelon crops? In the third week of July, a random sample of 40 farming regions gave a sample mean of \(\bar{x}=\6.88\) per 100 pounds of watermelon. Assume that \(\sigma\) is known to be \(1.92\) per 100 pounds (Reference: Agricultural Statistics\(,\) U.S. Department of Agriculture). (a) Find a \(90 \%\) confidence interval for the population mean price (per 100 pounds) that farmers in this region get for their watermelon crop. What is the margin of error? (b) Sample Size Find the sample size necessary for a \(90 \%\) confidence level with maximal margin of error \(E=0.3\) for the mean price per 100 pounds of watermelon. (c) A farm brings 15 tons of watermelon to market. Find a \(90 \%\) confidence interval for the population mean cash value of this crop. What is the margin of error? Hint: 1 ton is 2000 pounds.

Lorraine was in a hurry when she computed a confidence interval for \(\mu\). Because \(\sigma\) was not known, she used a Student's \(t\) distribution. However, she accidentally used degrees of freedom \(n\) instead of \(n-1 .\) Was her confidence interval longer or shorter than one found using the correct degrees of freedom \(n-1 ?\) Explain.

Suppose \(x\) has a normal distribution with \(\sigma=6 .\) A random sample of size 16 has sample mean \(50 .\) (a) Is it appropriate to use a normal distribution to compute a confidence interval for the population mean \(\mu ?\) Explain. (b) Find a \(90 \%\) confidence interval for \(\mu\) (c) Explain the meaning of the confidence interval you computed.

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.

Isabel Myers was a pioneer in the study of personality types. She identified four basic personality preferences, which are described at length in the book \(A\) Guide to the Development and Use of the Myers-Briggs Type Indicator by Myers and McCaulley (Consulting Psychologists Press). Marriage counselors know that couples who have none of the four preferences in common may have a stormy marriage. Myers took a random sample of 375 married couples and found that 289 had two or more personality preferences in common. In another random sample of 571 married couples, it was found that only 23 had no preferences in common. Let \(p_{1}\) be the population proportion of all married couples who have two or more personality preferences in common. Let \(p_{2}\) be the population proportion of all married couples who have no personality preferences in common. (a) Can a normal distribution be used to approximate the \(\hat{p}_{1}-\hat{p}_{2}\) distribution? Explain. (b) Find a \(99 \%\) confidence interval for \(p_{1}-p_{2}\) (c) Explain the meaning of the confidence interval in part (a) in the context of this problem. Does the confidence interval contain all positive, all negative, or both positive and negative numbers? What does this tell you (at the \(99 \%\) confidence level) about the proportion of married couples with two or more personality preferences in common compared with the proportion of married couples sharing no personality preferences in common?

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