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About \(28 \%\) of private companies are owned by women (The Cincinnati Enquirer, January 26,2006)\(.\) Answer the following questions based on a sample of 240 private companies. a. Show the sampling distribution of \(\bar{p},\) the sample proportion of companies that are owned by women. b. What is the probability the sample proportion will be within ±.04 of the population proportion? c. What is the probability the sample proportion will be within ±.02 of the population proportion?

Short Answer

Expert verified
a. Approximately normal with SE of 0.0285. b. Probability ≈ 0.8384. c. Probability ≈ 0.5160.

Step by step solution

01

Define the Parameters of the Sampling Distribution

The population proportion \( p \) (the proportion of companies owned by women) is known to be 0.28. For a sample size \( n \) of 240, the sample proportion \( \bar{p} \) is approximately normally distributed with mean equal to the population proportion \( p \).
02

Calculate the Standard Error of the Sample Proportion

The standard error (SE) of \( \bar{p} \) is given by the formula \( \sqrt{\frac{p(1-p)}{n}} \). Substituting the known values, \( SE = \sqrt{\frac{0.28 \times 0.72}{240}} \approx 0.0285 \).
03

Define the Range for ±0.04 of the Population Proportion

For part (b), we need the range for \( \bar{p} \) to be within \( 0.28 \pm 0.04 \), which is \( (0.24, 0.32) \).
04

Calculate the Z-scores for ±0.04 Range

The z-score is calculated using \( z = \frac{\bar{p} - p}{SE} \). For \( \bar{p} = 0.24 \), \( z_1 = \frac{0.24 - 0.28}{0.0285} \approx -1.40 \). For \( \bar{p} = 0.32 \), \( z_2 = \frac{0.32 - 0.28}{0.0285} \approx 1.40 \).
05

Find the Probability for Z-scores ±0.04

Using standard normal distribution tables, find \( P(z_1 \leq Z \leq z_2) \). \( P(-1.40 < Z < 1.40) \approx 0.9192 - 0.0808 = 0.8384 \).
06

Define the Range for ±0.02 of the Population Proportion

For part (c), we need the range for \( \bar{p} \) to be within \( 0.28 \pm 0.02 \), which is \( (0.26, 0.30) \).
07

Calculate the Z-scores for ±0.02 Range

For \( \bar{p} = 0.26 \), \( z_1 = \frac{0.26 - 0.28}{0.0285} \approx -0.70 \). For \( \bar{p} = 0.30 \), \( z_2 = \frac{0.30 - 0.28}{0.0285} \approx 0.70 \).
08

Find the Probability for Z-scores ±0.02

Using standard normal distribution tables, find \( P(z_1 \leq Z \leq z_2) \). \( P(-0.70 < Z < 0.70) \approx 0.7580 - 0.2420 = 0.5160 \).

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Key Concepts

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

Sample Proportion
The sample proportion is a statistical measure that represents the fraction of the sample that has a particular characteristic. In our case, it's the proportion of companies in the sample that are owned by women. When we're talking about the sample proportion, we're referring to the notation \(\bar{p}\). For example, if out of 240 companies, 70 are woman-owned, then the sample proportion \(\bar{p}\) would be \(\frac{70}{240} = 0.2917\). This number helps us estimate the population proportion, but with any estimate, there is always some degree of variability. In large samples, the sample proportion tends to be a good approximation of the population proportion due to the law of large numbers. As the sample size increases, the variability of \(\bar{p}\) decreases.
Population Proportion
The population proportion \( p \) is a key parameter in statistics. It represents the true proportion of a population that has a particular characteristic, such as the proportion of all private companies owned by women in this context. Here, the population proportion is known to be 0.28 or 28%. This figure is derived from broader data outside the immediate study and serves as a benchmark for studying sample data. In reality, the population proportion is not always known, so we often rely on sample data to estimate it.
  • Knowing the population proportion allows statisticians to understand the target population better and make informed decisions based on statistical analyses.
  • It forms the foundation for calculating other important statistics, such as the standard error and confidence intervals.
Standard Error
The standard error of the sample proportion is a measure of how much the sample proportion \(\bar{p}\) is expected to vary from the population proportion \(p\). Statistically, it is computed using the formula \(\sqrt{\frac{p(1-p)}{n}}\), where \(n\) is the sample size. For the given problem, with \(p = 0.28\) and \(n = 240\), the standard error computes to approximately 0.0285.
  • A smaller standard error indicates a closer match between the sample and population proportions, and it typically decreases as the sample size increases.
  • Standard error is crucial for constructing confidence intervals and hypothesis testing because it reflects the uncertainty attached to our sample estimates.
Therefore, understanding the standard error allows us to gauge how representative our sample is of the population.
Normal Distribution
The normal distribution is a fundamental concept in statistics that describes how data values are distributed for many real-world phenomena. It is often referred to as a bell curve due to its symmetrical shape. In the context of sampling distributions, the normal distribution is vital because it allows us to apply the central limit theorem. This theorem posits that with a sufficiently large sample size, the distribution of the sample proportion \(\bar{p}\) will approximate a normal distribution, even if the underlying population distribution is not normal.
  • In practical terms: when we talk about something like the 68-95-99.7 rule, we're discussing how data falls relative to the mean in a normal distribution.
  • For calculations involving probabilities of the sample proportion being within a certain range of the population proportion, we use Z-scores as a standard measure.
Thus, the normal distribution is a cornerstone in understanding statistical inference, making sense of the variability in data, and calculating probabilities.

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Most popular questions from this chapter

A population has a mean of 200 and a standard deviation of \(50 .\) A simple random sample of size 100 will be taken and the sample mean \(\bar{x}\) will be used to estimate the population mean. a. What is the expected value of \(\bar{x} ?\) b. What is the standard deviation of \(\bar{x} ?\) c. Show the sampling distribution of \(\bar{x}\). d. What does the sampling distribution of \(\bar{x}\) show?

The Grocery Manufacturers of America reported that \(76 \%\) of consumers read the ingredients listed on a product's label. Assume the population proportion is \(p=.76\) and a sample of 400 consumers is selected from the population.a. Show the sampling distribution of the sample proportion \(\bar{p}\) a. where \(\bar{p}\) is the proportion of the sampled consumers who read the ingredients listed on a product's label. b. What is the probability that the sample proportion will be within ±.03 of the population proportion? c. Answer part (b) for a sample of 750 consumers.

The average price of a gallon of unleaded regular gasoline was reported to be \(\$ 2.34\) in northern Kentucky (The Cincinnati Enquirer, January 21,2006 ). Use this price as the population mean, and assume the population standard deviation is \(\$ .20\) a. What is the probability that the mean price for a sample of 30 service stations is within \(\$ .03\) of the population mean? b. What is the probability that the mean price for a sample of 50 service stations is within \(\$ .03\) of the population mean? c. What is the probability that the mean price for a sample of 100 service stations is within \(\$ .03\) of the population mean? d. Which, if any, of the sample sizes in parts (a), (b), and (c) would you recommend to have at least a .95 probability that the sample mean is within \(\$ .03\) of the population mean?

A population proportion is \(.40 .\) A simple random sample of size 200 will be taken and the sample proportion \(\bar{p}\) will be used to estimate the population proportion. a. What is the probability that the sample proportion will be within ±.03 of the population proportion? b. What is the probability that the sample proportion will be within ±.05 of the population proportion?

Time/CNN voter polls monitored public opinion for the presidential candidates during the 2000 presidential election campaign. One Time/CNN poll conducted by Yankelovich Partners, Inc., used a sample of 589 likely voters (Time, June 26,2000 ). Assume the population proportion for a presidential candidate is \(p=.50 .\) Let \(\bar{p}\) be the sample proportion of likely voters favoring the presidential candidate. a. Show the sampling distribution of \(\bar{p}\) b. What is the probability the Time/CNN poll will provide a sample proportion within ±.04 of the population proportion? c. What is the probability the Time/CNN poll will provide a sample proportion within ±.03 of the population proportion? d. What is the probability the Time/CNN poll will provide a sample proportion within ±.02 of the population proportion?

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