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The proportion of individuals insured by the All-Driver Automobile Insurance Company who received at least one traffic ticket during a five-year period is .15 a. Show the sampling distribution of \(\bar{p}\) if a random sample of 150 insured individuals is used to estimate the proportion having received at least one ticket. b. What is the probability that the sample proportion will be within ±.03 of the population proportion?

Short Answer

Expert verified
The probability is approximately 0.7154.

Step by step solution

01

Understand the Given Information

We need to find the sampling distribution of \( \bar{p} \), where \( p = 0.15 \), the population proportion. The sample size \( n = 150 \). We also need to calculate the probability that \( \bar{p} \) is within ±0.03 of \( p \).
02

Calculate the Standard Error of \( \bar{p} \)

The standard error (SE) of the sample proportion \( \bar{p} \) is calculated using the formula \( SE = \sqrt{\frac{p(1-p)}{n}} \). Substitute \( p = 0.15 \) and \( n = 150 \) into the formula: \[ SE = \sqrt{\frac{0.15 \times 0.85}{150}} = \sqrt{\frac{0.1275}{150}} \approx 0.028 \]
03

Describe the Sampling Distribution

According to the Central Limit Theorem, since \( n \) is large, \( \bar{p} \) follows a normal distribution with mean \( \mu_{\bar{p}} = p = 0.15 \) and standard deviation equal to the standard error calculated as \( 0.028 \). Thus, \( \bar{p} \sim N(0.15, 0.028^2) \).
04

Define the Range for the Probability Calculation

We need to find the probability that \( \bar{p} \) is within ±0.03 of \( p \). This implies \( 0.12 \leq \bar{p} \leq 0.18 \).
05

Standardize to find Z-scores

Convert \( \bar{p} = 0.12 \) and \( \bar{p} = 0.18 \) to Z-scores using the formula \( Z = \frac{\bar{p} - p}{SE} \). For \( 0.12 \): \[ Z = \frac{0.12 - 0.15}{0.028} \approx -1.07 \] For \( 0.18 \): \[ Z = \frac{0.18 - 0.15}{0.028} \approx 1.07 \]
06

Calculate the Probability using the Z-table

Using the standard normal table, find the probabilities: - \( P(Z < 1.07) \approx 0.8577 \) - \( P(Z < -1.07) \approx 0.1423 \) Thus, the probability that \( 0.12 \leq \bar{p} \leq 0.18 \) is \[ P(-1.07 < Z < 1.07) = P(Z < 1.07) - P(Z < -1.07) \approx 0.8577 - 0.1423 = 0.7154 \]
07

Conclusion

Therefore, the sampling distribution of \( \bar{p} \) is normally distributed with \( \mu = 0.15 \) and \( \sigma = 0.028 \). The probability that the sample proportion is within ±0.03 of the population proportion is approximately 0.7154.

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Key Concepts

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

Central Limit Theorem
The Central Limit Theorem (CLT) is a fundamental principle in statistics that describes the behavior of sampling distributions. It states that when you take a sufficiently large sample size from a population, the distribution of the sample mean will approximate a normal distribution, regardless of the shape of the population distribution.

This theorem is particularly useful when dealing with proportions, as it allows us to make predictions about a sample proportion using the normal distribution. In the context of our example, because we have a large sample size of 150, we apply the CLT to conclude that the sample proportion \( \bar{p} \) is normally distributed.
  • It doesn't matter if the original population distribution is skewed; as long as the sample size is large, the mean of the sample distribution will be normally distributed.
  • The mean of the sampling distribution of the sample proportion \( \bar{p} \) will equal the population proportion \( p \).
Understanding the CLT helps us calculate the probability of the sample proportion \( \bar{p} \) falling within a specified range, such as within ±0.03 of the population proportion of 0.15.
Standard Error
Standard Error (SE) is a key concept in statistics used to quantify the variation of a sample statistic from the population parameter. For the sample proportion, SE measures how much the sample proportion \( \bar{p} \) is expected to vary around the population proportion \( p \).

The formula for the Standard Error of the sample proportion is \[ SE = \sqrt{\frac{p(1-p)}{n}} \] where \( p \) is the population proportion, and \( n \) is the sample size. In our example with \( p = 0.15 \) and \( n = 150 \), substituting these values gives:
\[ SE = \sqrt{\frac{0.15 \times 0.85}{150}} \approx 0.028 \]
  • Smaller SE indicates that the sample proportion is more consistent and closely clusters around the population proportion.
  • In practical terms, a small SE means more precise estimates of the population proportion from our sample data.
Understanding SE allows us to make more informed predictions about the population based on our sample data, enhancing confidence in our estimates.
Z-scores
Z-scores are a statistical measure that standardizes an observation within a distribution in terms of standard deviations from the mean. They are particularly useful in determining the probability of an observation occurring within a standard normal distribution.

In our context, we use Z-scores to standardize the sample proportion values to determine probabilities. Calculate the Z-score with the formula: \[ Z = \frac{\bar{p} - p}{SE} \]
Here are the steps we used to calculate the Z-scores for the sample proportions 0.12 and 0.18:
  • For \( \bar{p} = 0.12 \): \( Z = \frac{0.12 - 0.15}{0.028} \approx -1.07 \)
  • For \( \bar{p} = 0.18 \): \( Z = \frac{0.18 - 0.15}{0.028} \approx 1.07 \)

  • The Z-score tells us how far, in standard deviations, an observation is from the mean.
  • You can use Z-scores and the standard normal distribution table to find probabilities associated with specific sample proportion ranges.
Understanding and calculating Z-scores enable us to translate sample data into more general insights about the population.
Population Proportion
Population proportion, usually denoted by \( p \), is a measure that indicates the fraction of the total population that possesses a particular characteristic. In our example, \( p = 0.15 \), meaning 15% of the individuals in the population have received at least one traffic ticket.

It serves as a foundational parameter in statistical analysis, particularly when dealing with sampling distributions. To estimate the population proportion from a sample proportion reliably, you need to consider:
  • The sample size, which should be large enough to ensure that the Central Limit Theorem applies, making the sampling distribution approximately normal.
  • The population proportion and its complement \( 1-p \) to compute the standard error.

  • Understanding the population proportion is crucial when interpreting the results of a study, as it provides context for how the sample relates to the entire population.
  • Knowing the proportion helps in setting up confidence intervals and hypothesis tests to make data-driven decisions.
Recognizing the role of population proportion also aides in drawing more significant inferences from sampled data to total the population.

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Most popular questions from this chapter

Lori Jeffrey is a successful sales representative for a major publisher of college textbooks. Historically, Lori obtains a book adoption on \(25 \%\) of her sales calls. Viewing her sales calls for one month as a sample of all possible sales calls, assume that a statistical analysis of the data yields a standard error of the proportion of .0625 a. How large was the sample used in this analysis? That is, how many sales calls did Lori make during the month? b. Let \(\bar{p}\) indicate the sample proportion of book adoptions obtained during the month. Show the sampling distribution of \(\bar{p}\) c. Using the sampling distribution of \(\bar{p},\) compute the probability that Lori will obtain book adoptions on \(30 \%\) or more of her sales calls during a one-month period.

BusinessWeek surveyed MBA alumni 10 years after graduation \((\text {Business Week} \text { , September } 22,\) 2003 ). One finding was that alumni spend an average of \(\$ 115.50\) per week eating out socially. You have been asked to conduct a follow-up study by taking a sample of 40 of these MBA alumni. Assume the population standard deviation is \(\$ 35\) a. Show the sampling distribution of \(\bar{x}\), the sample mean weekly expenditure for the \(40 \mathrm{MBA}\) alumni. b. What is the probability that the sample mean will be within \(\$ 10\) of the population mean? c. Suppose you find a sample mean of \(\$ 100 .\) What is the probability of finding a sample mean of \(\$ 100\) or less? Would you consider this sample to be an unusually low spending group of alumni? Why or why not?

The American Association of Individual Investors (AAII) polls its subscribers on a weekly basis to determine the number who are bullish, bearish, or neutral on the short-term prospects for the stock market. Their findings for the week ending March \(2,2006,\) are consistent with the following sample results (AAII website, March 7, 2006). Bullish \(409 \quad\) Neutral \(299 \quad\) Bearish \(291\) Develop a point estimate of the following population parameters. a. The proportion of all AAII subscribers who are bullish on the stock market. b. The proportion of all AAII subscribers who are neutral on the stock market. c. The proportion of all AAII subscribers who are bearish on the stock market.

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 \(\bar{p}\), 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.

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