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In a random sample of size 1,100,338 have the characteristic of interest. a. Compute the sample proportion \(\hat{p}\) with the characteristic of interest. b. Verify that the sample is large enough to use it to construct a confidence interval for the population proportion. c. Construct an \(80 \%\) confidence interval for the population proportion \(p\). d. Construct a \(90 \%\) confidence interval for the population proportion \(p\). e. Comment on why one interval is longer than the other.

Short Answer

Expert verified
Sample proportion is 0.307. The 80% interval is (0.289, 0.325), and the 90% interval is (0.283, 0.331). The 90% interval is longer due to higher confidence.

Step by step solution

01

Calculate the Sample Proportion

The sample proportion, denoted as \( \hat{p} \), is calculated by dividing the number of individuals with the characteristic of interest by the total sample size. We have 338 people with the characteristic out of 1,100 total.\[ \hat{p} = \frac{338}{1100} \approx 0.307 \]
02

Check Sample Size Requirement

To use the sample for a confidence interval, we check if both \( n \hat{p} \) and \( n(1-\hat{p}) \) are greater than 5. Here, \[ n\hat{p} = 1100 \times 0.307 = 337.7 \] \[ n(1-\hat{p}) = 1100 \times (1-0.307) = 762.3 \]Both values are greater than 5, so the sample size is adequate.
03

Construct an 80% Confidence Interval

For an 80% confidence interval, we use the critical value \( z_{\alpha/2} \approx 1.28 \). The formula for the confidence interval is:\[ \hat{p} \pm z_{\alpha/2} \sqrt{\frac{\hat{p}(1-\hat{p})}{n}} \]Substituting the known values:\[ 0.307 \pm 1.28 \sqrt{\frac{0.307(1-0.307)}{1100}} \approx 0.307 \pm 0.018 \]The interval is approximately (0.289, 0.325).
04

Construct a 90% Confidence Interval

For a 90% confidence interval, the critical value \( z_{\alpha/2} \approx 1.645 \). Use the same interval formula:\[ 0.307 \pm 1.645 \sqrt{\frac{0.307(1-0.307)}{1100}} \approx 0.307 \pm 0.024 \]The interval is approximately (0.283, 0.331).
05

Compare Lengths of Intervals

The 90% confidence interval is wider than the 80% interval because a higher confidence level increases the margin of error. This reflects greater uncertainty about the population proportion with higher confidence levels.

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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, often denoted as \( \hat{p} \), is a key statistic used when estimating and making predictions about a population based on a sample. To calculate it, you simply divide the number of successful outcomes (or occurrences of the characteristic of interest) by the total number of observations in the sample.
For our exercise, if we have 338 people out of a total of 1,100 who have the characteristic of interest, the sample proportion \( \hat{p} \) would be calculated as follows:
\[ \hat{p} = \frac{338}{1100} \approx 0.307 \]
This means that approximately 30.7% of the sample possesses the feature we're investigating.
  • This value serves as an estimate for the true population proportion.
  • It forms the basis for constructing confidence intervals.
Understanding \( \hat{p} \) is crucial because it allows you to link the sample data to the broader population context.
Critical Value
The critical value z鈧/鈧 is an important component in constructing confidence intervals. It represents the number of standard deviations a data point is from the mean, accounting for a designated confidence level.
But how do you find the critical value? Typically, it's determined based on the desired confidence level using standard normal distribution tables or calculators. For instance:
  • An 80% confidence interval uses a critical value of approximately 1.28.
  • A 90% confidence interval uses a critical value of approximately 1.645.
The larger the critical value, the wider the confidence interval, reflecting increased certainty.
The choice of critical value affects the range within which we expect to find the population proportion, providing a balance between precision and certainty.
Margin of Error
The margin of error quantifies the uncertainty or variability inherent in estimating a population parameter from a sample statistic. It stretches the sample proportion to create a confidence interval around it.
In mathematical terms, it is expressed as:
\[ \text{Margin of Error} = z_{\alpha/2} \sqrt{\frac{\hat{p}(1-\hat{p})}{n}} \]
Here's how it works in our exercise:
  • For an 80% confidence interval, the margin of error is around 0.018.
  • For a 90% confidence interval, it's approximately 0.024.
As the confidence level increases, so does the margin of error, broadening the interval to account for greater confidence.
It's essential because it defines how 'sure' or 'uncertain' we are about the population proportion estimate, allowing us to communicate the probable range where the true population proportion lies.
Population Proportion
The population proportion (p) is a significant aspect of statistical analysis. It represents the ratio of elements in a population that possess a particular characteristic.
While we rarely know the exact population proportion, we estimate it using sample data. The broader questions focus on understanding how well the sample proportion \( \hat{p} \) estimates the unknown p.
In practical terms:
  • The closer the sample size to the entire population, the more accurate \( \hat{p} \) becomes as an estimator for p.
  • Confidence intervals provide a range for the true population proportion based on our sample proportion \( \hat{p} \).
The goal is to create intervals that provide a high probability of containing the actual population proportion, guiding decisions based on statistical evidence rather than assumptions alone.

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Most popular questions from this chapter

A random sample of 185 college soccer players who had suffered injuries that resulted in loss of playing time was made with the results shown in the table. Injuries are classified according to severity of the injury and the condition under which it was sustained. $$ \begin{array}{|c|c|c|c|} \hline & \text { Minor } & \text { Moderate } & \text { Serious } \\ \hline \text { Practice } & 48 & 20 & 6 \\ \hline \text { Game } & 62 & 32 & 17 \\ \hline \end{array} $$ a. Give a point estimate for the proportion \(p\) of all injuries to college soccer players that are sustained in practice. b. Construct a \(95 \%\) confidence interval for the proportion \(p\) of all injuries to college soccer players that are sustained in practice. c. Give a point estimate for the proportion \(p\) of all injuries to college soccer players that are either moderate or serious.

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