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Chapter 7: Problem 9

Basic Computation: Sample Size What is the minimal sample size needed for a \(95 \%\) confidence interval to have a maximal margin of error of \(0.1\) (a) if a preliminary estimate for \(p\) is \(0.25\) ? (b) if there is no preliminary estimate for \(p ?\)

Short Answer

Expert verified
(a) If \(p = 0.25\), sample size is 145. (b) Without \(p\) estimate, sample size is 385.

Step by step solution

01

Understand Confidence Interval and Margin of Error

In order to construct a confidence interval for a proportion with a specific margin of error, we use the sample size formula. The sample size needed () can be calculated using the formula for margin of error E: \[ E = Z \sqrt{\frac{p(1-p)}{n}} \] where Z is the z-score corresponding to the confidence level (\(Z = 1.96\) for a 95% confidence interval), p is the proportion estimate, and E is the margin of error.
02

Determine Sample Size with Preliminary Estimate

Given a preliminary estimate for sample proportion \(p = 0.25\) and a margin of error \(E = 0.1\), we rearrange the error formula to solve for the required sample size \(n\): \[ n = \left(\frac{Z}{E}\right)^2 \cdot p(1-p) \]. Substitute \(Z = 1.96\), \(E = 0.1\), and \(p = 0.25\): \[ n = \left(\frac{1.96}{0.1}\right)^2 \cdot 0.25 \cdot (1 - 0.25) \]. Simplify this to find \(n\).
03

Compute Sample Size with Preliminary Estimate

Calculate the sample size using the values: \[ n = (19.6)^2 \cdot 0.25 \cdot 0.75 \]. This results in: \[ n \approx 144.06 \] Since the sample size must be a whole number, round up to the nearest whole number, so \(n = 145\).
04

Determine Sample Size without Preliminary Estimate

When no estimate for \(p\) is available, the formula for \(n\) requires us to use the most conservative estimate \(p = 0.5\) (this maximizes the product \(p(1-p)\), leading to the largest possible \(n\) value). Again, use \[ n = \left(\frac{Z}{E}\right)^2 \cdot p(1-p) \] and substitute \(Z = 1.96\), \(E = 0.1\), and \(p = 0.5\).
05

Compute Sample Size without Preliminary Estimate

Substitute the values into the formula: \[ n = \left(\frac{1.96}{0.1}\right)^2 \cdot 0.5 \cdot (1 - 0.5) \]. Simplify the equation to: \[ n = 19.6^2 \cdot 0.25 \]. Calculate to find: \[ n = 384.16 \]. Round up to the nearest whole number, so \(n = 385\).

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Key Concepts

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

Confidence Interval
A confidence interval is a range of values that likely contain a population parameter, such as a mean or proportion. It's a useful tool because it helps us understand how much precision we have in our estimate. The confidence interval is associated with a confidence level, often expressed as a percentage.

Essentially, a 95% confidence interval means that if we were to take 100 different samples and calculate a confidence interval for each one, we would expect 95 of them to contain the true population parameter.
  • Confidence Level: The probability that the interval contains the true parameter.
  • Interval: The range within which we expect the parameter to fall.
Margin of Error
The margin of error is a statistical term that quantifies the amount of random sampling error in a survey's results. It represents how much you expect the survey results to reflect the true population value due to chance.

For example, if a survey result has a margin of error of 5%, and the survey indicates that 60% of the population have a certain characteristic, we can expect the true proportion to be within 55% to 65%. It's calculated using a formula that includes the z-score and the standard deviation of the sample proportion.
  • Determined by Sample Size: Larger samples result in smaller margins of error.
  • Influenced by Confidence Level: Higher confidence levels broaden the margin of error.
Statistical Proportion
A statistical proportion measures the fraction of a given outcome within a total number of observations. In surveys, it represents the ratio of respondents who exhibit a trait of interest. It's a fundamental concept in understanding the distribution of data across different categories.

When estimating a population proportion, it is represented by "p" in formulas where calculations are concerned with probabilities. Knowing the proportion helps in building confidence intervals and calculating sample sizes.
  • Basic Formula: Ratio of favorable responses to total responses.
  • Key Role in Estimation: Integral for constructing predictions and making inferences.
Z-Score
A z-score is a statistical measure that describes a value's relationship to the mean of a group of values. It's measured in terms of standard deviations from the mean. For purposes related to sample size calculation, z-scores are used to ascertain the confidence level.

In a normal distribution, a z-score can indicate where a data point lies in relation to the mean. For instance, a z-score of 1.96 aligns with a 95% confidence level, which is a common standard in most statistical analyses because it strikes a balance between precision and reliability.
  • Normal Distribution: Z-scores assume data follows a bell curve.
  • Used in Calculations: Key in determining sample sizes and confidence intervals.

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Most popular questions from this chapter

If a \(90 \%\) confidence interval for the difference of means \(\mu_{1}-\mu_{2}\) contains all negative values, what can we conclude about the relationship between \(\mu_{1}\) and \(\mu_{2}\) at the \(90 \%\) confidence level?

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.

Physicians: Solo Practice A random sample of 328 medical doctors showed that 171 have a solo practice (Source: Practice Patterns of General Internal Medicine, American Medical Association). (a) Let \(p\) represent the proportion of all medical doctors who have a solo practice. Find a point estimate for \(p\). (b) Find a \(95 \%\) confidence interval for \(p\). Give a brief explanation of the meaning of the interval. (c) Interpretation As a news writer, how would you report the survey results regarding the percentage of medical doctors in solo practice? What is the margin of error based on a \(95 \%\) confidence interval?

Myers-Briggs: Actors Isabel Myers was a pioneer in the study of personality types. The following information is taken from A Guide to the Development and Use of the Myers-Briggs Type Indicator by Myers and McCaulley (Consulting Psychologists Press). In a random sample of 62 professional actors, it was found that 39 were extroverts. (a) Let \(p\) represent the proportion of all actors who are extroverts. Find a point estimate for \(p\). (b) Find a \(95 \%\) confidence interval for \(p .\) Give a brief interpretation of the meaning of the confidence interval you have found. (c) Check Requirements Do you think the conditions \(n p>5\) and \(n q>5\) are satisfied in this problem? Explain why this would be an important consideration.

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 \(6.5\).
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