/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 19 The price of a share of stock di... [FREE SOLUTION] | 91Ó°ÊÓ

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The price of a share of stock divided by the company's estimated future earnings per share is called the \(\mathrm{P} / \mathrm{E}\) ratio. High \(\mathrm{P} / \mathrm{E}\) 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) 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?

Short Answer

Expert verified
Means and CIs confirmed: Bank One value, Disney typical, AT&T Wireless likely growth.

Step by step solution

01

Calculate Mean and Standard Deviation (Verification)

First, we verify the provided mean and standard deviation using the calculator. Using the data provided, calculate:\[ \bar{x} = \frac{1}{51} \sum_{i=1}^{51} x_i = 25.2 \] and the sample standard deviation:\[ s = \sqrt{\frac{1}{n-1} \sum_{i=1}^{n} (x_i - \bar{x})^2} = 15.5 \] where \( n = 51 \), confirming the given values.
02

Determine 90% Confidence Interval

Use the formula for the confidence interval: \[ \bar{x} \pm t_{\frac{\alpha}{2}} \left(\frac{s}{\sqrt{n}}\right) \]With \( \bar{x} = 25.2, s = 15.5, n = 51 \), and a 90% confidence level, the degrees of freedom \(df = n-1 = 50\). Look up the \( t \)-value from the \( t \)-distribution table \( t_{0.05, 50} \approx 1.676 \):\[ CI = 25.2 \pm 1.676 \left(\frac{15.5}{\sqrt{51}}\right) \approx (21.63, 28.77) \]
03

Determine 99% Confidence Interval

Calculate with:\[ \bar{x} \pm t_{\frac{\alpha}{2}} \left(\frac{s}{\sqrt{n}}\right) \]For a 99% confidence level, \( t \) value is \( t_{0.005, 50} \approx 2.678 \):\[ CI = 25.2 \pm 2.678 \left(\frac{15.5}{\sqrt{51}}\right) \approx (19.04, 31.36) \]
04

Analyze Comparisons to Confidence Intervals

Examine Bank One's P/E of 12, Disney's 24, and AT&T Wireless' 72 with the confidence intervals: - Bank One's P/E of 12 is below both CIs, suggesting it was considered a value stock. - Disney's P/E of 24 falls within both intervals, indicating it was typical for large U.S. companies at the time. - AT&T Wireless' P/E of 72 is above both CIs, indicating it may have been a growth or overpriced stock.

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Key Concepts

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

Mean and Standard Deviation
Understanding mean and standard deviation is crucial in statistical analysis as they provide insights into the central tendency and variability of data. The mean, often called the average, is calculated by summing all data points and dividing by the number of points. It provides a snapshot of where the center of the data tends to fall. In our exercise, the mean \( \bar{x} \) of the P/E ratios was confirmed as \( 25.2 \). This tells us that, on average, the P/E ratio of the sampled companies is 25.2.

Standard deviation, on the other hand, measures the spread of the data around the mean. It tells us how much the individual data points typically deviate from the average. A higher standard deviation indicates more variability, while a lower standard deviation indicates data points clustered closely around the mean. For the given P/E ratios, the standard deviation \( s \) is \( 15.5 \), indicating a decent amount of spread around the mean.

Together, these metrics help us understand not just the average position of our data but also the degree of variability, which is important for making further statistical inferences.
P/E Ratio
The Price-to-Earnings (P/E) ratio is a popular metric used to evaluate the valuation of a company's stock. It is determined by dividing the current market price per share by the company's earnings per share (EPS). This ratio helps investors assess if a stock is overvalued, undervalued, or fairly priced.

In the context of our exercise, a high P/E ratio, like that of AT&T Wireless at 72, could suggest a "growth" stock with potential for high future earnings, or it might simply be overpriced relative to its earnings. Conversely, Bank One's lower P/E ratio of 12 suggests it may have been perceived as a "value" stock, potentially undervalued and offering a good buying opportunity.

The P/E ratio is a snapshot, one of several tools investors use to gauge the attractiveness of a stock, and should be considered in combination with other financial metrics.
t-Distribution
When constructing confidence intervals, especially with smaller sample sizes, the t-distribution becomes essential. It is used instead of the normal distribution to account for the additional uncertainty inherent when estimating from a limited data set. The t-distribution is slightly broader, with thicker tails, and provides a more conservative estimate.

For this exercise, when calculating the 90% and 99% confidence intervals for the mean P/E ratio, the t-distribution was used with \( 50 \) degrees of freedom (since \( n-1 \) degrees of freedom apply). The t-values obtained (1.676 for 90% and 2.678 for 99% confidence intervals) were used in the formula to calculate these intervals. This approach ensures that we account for any uncertainties, providing more reliable estimates of where the true mean P/E for large U.S. companies likely falls.

Using the t-distribution is crucial in making sure our estimates can be trusted, especially with the variability present in the data set highlighted by our earlier standard deviation calculation.
Statistical Analysis
Statistical analysis is about making informed inferences from data. In the exercise, we took an important step by calculating the confidence intervals for the population mean P/E ratio. These intervals help to gauge the precision of our mean estimate and the degree of certainty we have regarding the true mean.

The 90% confidence interval, (21.63, 28.77), suggests that there is a 90% probability that the true mean P/E ratio of all large U.S. companies falls within this range. Similarly, the 99% confidence interval (19.04, 31.36) indicates a tighter requirement for accuracy, providing a broader range to ensure greater certainty.

By analyzing how specific companies' P/E ratios, like Disney's and AT&T Wireless's, compare to these intervals, investors can form judgments regarding whether these stocks are typical, under- or overpriced. Such analysis is foundational in making data-driven investment decisions, leveraging statistical methods to reduce uncertainty and highlight potential investments.

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Most popular questions from this chapter

Case studies showed that out of 10,351 convicts who escaped from U.S. prisons, only 7867 were recaptured (The Book of Odds, by Shook and Shook, Signet). (a) Let \(p\) represent the proportion of all escaped convicts who will eventually be recaptured. Find a point estimate for \(p\). (b) Find a \(99 \%\) confidence interval for \(p .\) Give a brief statement of the meaning of the confidence interval. (c) Is use of the normal approximation to the binomial justified in this problem? Explain.

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Over the past several months, an adult patient has been treated for tetany (severe muscle spasms). This condition is associated with an average total calcium level below \(6 \mathrm{mg} / \mathrm{dl}(\) Reference: Manual of Laboratory and Diagnostic Tests, F. Fischbach). Recently, the patient's total calcium tests gave the following readings (in \(\mathrm{mg} / \mathrm{dl}\) ). \(9.3\) \(\begin{array}{llllll}8.8 & 10.1 & 8.9 & 9.4 & 9.8 & 10.0\end{array}\) \(\begin{array}{lll}9.9 & 11.2 & 12.1\end{array}\) (a) Use a calculator to verify that \(\bar{x}=9.95\) and \(s \approx 1.02\). (b) Find a \(99.9 \%\) confidence interval for the population mean of total calcium in this patient's blood. (c) Based on your results in part (b), do you think this patient still has a calcium deficiency? Explain.

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