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Use the information given to find the necessary confidence interval for the binomial proportion \(p .\) Interpret the interval that you have constructed. A \(90 \%\) confidence interval for \(p\), based on a random sample of \(n=300\) observations from a binomial population with \(x=263\) successes.

Short Answer

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A) (0.8362, 0.9172) B) (0.8421, 0.9263) C) (0.8235, 0.9099) D) (0.8564, 0.9325) Answer: A) (0.8362, 0.9172) Explanation: By calculating the sample proportion, critical z-value, margin of error, and determining the 90% confidence interval, we find that the true proportion of success in the population lies between 83.62% and 91.72%.

Step by step solution

01

Find the sample proportion

To find sample proportion (\(\hat{p}\)), we need to divide the number of successes (x) by the number of observations (n): $$ \hat{p} = \frac{x}{n} = \frac{263}{300} $$ Calculate the sample proportion: $$ \hat{p} = \frac{263}{300} = 0.8767 $$
02

Find the critical z-value

To find the critical z-value corresponding to a 90% confidence interval, we can use a z-table or calculator. The z-value corresponds to the probability in both tails. Since this is a two-tailed test, we will use \(\frac{1 - 0.9}{2} = 0.05\) for the z-value: $$ Z_{\frac{\alpha}{2}} = Z_{0.05} = 1.645 $$
03

Calculate the margin of error

Using the critical z-value and sample proportion, we will calculate the margin of error: $$ \text{Margin of Error} = Z_\frac{\alpha}{2} * \sqrt{\frac{\hat{p}(1-\hat{p})}{n}} = 1.645 * \sqrt{\frac{0.8767(1-0.8767)}{300}} $$ Calculate the margin of error: $$ \text{Margin of Error} \approx 0.0405 $$
04

Determine the confidence interval

Now that we have calculated both the sample proportion and margin of error, we can determine the 90% confidence interval for the proportion p: $$ \text{Confidence Interval} = \hat{p} \pm \text{Margin of Error} = 0.8767 \pm 0.0405 $$ Calculate the interval: $$ \text{Lower Bound} = 0.8767 - 0.0405 = 0.8362 \\ \text{Upper Bound} = 0.8767 + 0.0405 = 0.9172 $$ So, the 90% confidence interval for the binomial proportion is \((0.8362, 0.9172)\).
05

Interpret the confidence interval

The confidence interval of \((0.8362, 0.9172)\) means that we are 90% confident that the true proportion of success in the population lies between 83.62% and 91.72%.

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Key Concepts

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

Sample Proportion
Understanding the sample proportion is crucial in statistics, especially when working with binomial distributions. The sample proportion, denoted as \( \hat{p} \), is a ratio representing the number of successes to the total number of trials in a sample. To calculate \( \hat{p} \), you divide the count of observed successes, \( x \), by the overall sample size, \( n \).

In our exercise, with 263 successes out of 300 trials, the sample proportion is \( 0.8767 \). This figure acts as an estimate of the true proportion in the entire population, assuming that the sample is representative.

Why is it Important?

  • It is a critical value for hypothesis testing of proportions.
  • It helps in constructing confidence intervals for the population proportion.
  • It can determine if results from the sample can be generalized for the population.
Knowing the sample proportion is the first step in gauging the occurrence rate of an event of interest within the population.
Critical Z-value
The critical z-value plays a key role when building confidence intervals for a binomial proportion. This value is derived from the standard normal distribution and corresponds to a specific confidence level – the percentage of all possible samples that include the true population parameter.

For a 90% confidence level, we aim to find the z-value that leaves 5% in each tail (since it's two-tailed, equating to 10% total in the tails). Using statistical tables or software, we find that the critical z-value is \( 1.645 \).

Choice of Confidence Level

  • A higher confidence level means a larger z-value.
  • The critical z-value is instrumental in calculating the margin of error for the interval.
  • Understanding this concept is essential for accurate interval estimation.
Grasping how the critical z-value affects your confidence interval gives insight into the reliability and precision of your statistics.
Margin of Error
Lastly, the margin of error quantifies the level of uncertainty in estimating the population proportion. It is affected by both the sample proportion and the critical z-value. To calculate the margin of error, you multiply the critical z-value by the standard error of the sample proportion.

In mathematical terms, for our case this would be \( 1.645 \times \sqrt{(0.8767 \times (1 - 0.8767) / 300)} \approx 0.0405 \).

Significance of the Margin of Error

  • It indicates the range within which the true population proportion is likely to be found.
  • A smaller margin of error means a more precise estimate.
  • This metric is pivotal for decision-making based on the interval estimate.
Recognizing the variables that influence the margin of error help students make sense of the interval they compute. For example, a larger sample would typically lead to a smaller margin of error, suggesting a tighter confidence interval. This detail illustrates the trade-off between the size of the sample and the precision of the estimate.

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Most popular questions from this chapter

Auto Accidents A recent year's records of auto accidents occurring on a given section of highway were classified according to whether the resulting damage was \(\$ 3000\) or more and to whether a physical injury resulted from the accident. The data follows: $$\begin{array}{lcc}\hline & \text { Under } \$ 3000 & \$ 3000 \text { or More } \\\\\hline \text { Number of Accidents } & 32 & 41 \\\\\text { Number Involving Injuries } & 10 & 23\end{array}$$ a. Estimate the true proportion of accidents involving injuries when the damage was \(\$ 3000\) or more for similar sections of highway and find the margin of error. b. Estimate the true difference in the proportion of accidents involving injuries for accidents with damage under \(\$ 3000\) and those with damage of \(\$ 3000\) or more. Use a \(95 \%\) confidence interval.

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Using the sample information given in Exercises \(20-21\), give the best point estimate for the population mean \(\mu\) and calculate the margin of error: A random sample of \(n=75\) observations from a quantitative population produced \(\bar{x}=29.7\) and \(s^{2}=10.8\).

Confidence An increase in the rate of consumer savings is frequently tied to a lack of confidence in the economy. A random sample of \(n=200\) savings accounts in a local community showed a mean increase in savings account values of \(7.2 \%\) over the past 12 months, with a standard deviation of 5.6\%. Estimate the mean percent increase in savings account values over the past 12 months for depositors in the community. Find the margin of error for your estimate.

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