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A random sample of 300 Oberlin students were asked if they regularly smoke marijuana. 60 replied in the affirmative. Find an approximate \(95 \%\) confidence interval on the true proportion, \(p\), of Oberlin marijuana smokers.

Short Answer

Expert verified
The 95% confidence interval for the true proportion of Oberlin students who regularly smoke marijuana lies between approximately \(14.37\%\) and \(25.63\%\).

Step by step solution

01

Calculate the sample proportion (p-hat)

To calculate the sample proportion, we will divide the number of affirmative responses (60) by the total number of students in the sample (300). p-hat = 60 / 300 p-hat = 0.2 The sample proportion of Oberlin students who regularly smoke marijuana is 0.2 or 20%.
02

Determine the z-score for a 95% confidence interval

For a 95% confidence interval, we will use a z-score of 1.96. This value is derived from the standard normal distribution table, where 95% of the area under the curve falls between -1.96 and 1.96.
03

Calculate the margin of error

To calculate the margin of error, we will use the following formula: Margin of Error = z-score * √(p-hat*(1-p-hat) / n) Where: - z-score = 1.96 (from Step 2) - p-hat = 0.2 (from Step 1) - n = 300 (sample size) Margin of Error = 1.96 * √(0.2*(1-0.2) / 300) Margin of Error ≈ 0.0563
04

Find the lower and upper bounds of the confidence interval

Finally, we will find the 95% confidence interval for the true proportion, p, by subtracting and adding the margin of error to the sample proportion (p-hat). Lower Bound = p-hat - Margin of Error = 0.2 - 0.0563 ≈ 0.1437 Upper Bound = p-hat + Margin of Error = 0.2 + 0.0563 ≈ 0.2563
05

Conclusion

We can say with 95% confidence that the true proportion of Oberlin students who regularly smoke marijuana lies between 0.1437 (14.37%) and 0.2563 (25.63%).

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Key Concepts

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

Sample Proportion
When you're trying to understand statistical data, knowing the sample proportion can be incredibly helpful. The sample proportion, often denoted as \(\hat{p}\), is essentially the fraction of the sample that exhibits a particular trait or characteristic. In our exercise about Oberlin students, this represents the number of students who said they regularly smoke marijuana divided by the total number of students sampled. In mathematical terms, it's expressed as:
  • \(\hat{p} = \frac{x}{n} \)
Where \(x\) is the number of individuals with the trait of interest—in this case, those who said they smoke marijuana—and \(n\) is the total number of individuals in the sample. For example, if 60 out of 300 students said "yes," \(\hat{p}\) would be \(\frac{60}{300} = 0.2\), or 20%. This percentage gives us a snapshot of the sample's behavior.
Z-Score
The z-score is a crucial part of figuring out confidence intervals for proportions. It tells us how many standard deviations an element is from the mean, and in the context of a confidence interval, it helps us to determine the boundaries of our interval. Z-scores come from the standard normal distribution, which is a bell-shaped curve that illustrates how values are dispersed. When we're talking about a 95% confidence interval, we use a z-score of 1.96. This value indicates that 95% of the data falls within 1.96 standard deviations from the mean in a normal distribution.
  • At 95%, we believe there's a good chance our true proportion lies within the calculated range.
  • It's vital when estimating the range within which the true proportion of a population lies based on our sample data.
Understanding the z-score helps us translate our initial sample data into more expansive insights about the true population.
Margin of Error
The margin of error provides the range within which we expect the true proportion of the population to fall. It's like a safety net around our sample proportion, and it acknowledges the sample's inherent imprecision when attempting to represent the whole population. We calculate it using the formula:
  • \( \text{Margin of Error} = z \times \sqrt{\frac{\hat{p}(1-\hat{p})}{n}} \)
Here, \(z\) is the z-score for our chosen confidence level, \(\hat{p}\) is the sample proportion, and \(n\) is the total number of samples. In the Oberlin exercise, this was \(1.96 \times \sqrt{\frac{0.2 \cdot 0.8}{300}} \approx 0.0563\). This means we are approximately 5.63% off from the true proportion, reminding us that sample results always have some uncertainty.
True Proportion
The true proportion, symbolized as \(p\), represents the actual fraction of the entire population that exhibits a certain trait, characteristic, or behavior. It's the goal of our estimation efforts because it tells us about the broader population rather than just the sample group.In our exercise, we aimed to estimate the true proportion of Oberlin students who regularly smoke marijuana. However, it's rarely directly measurable—hence, we rely on our sample to make an educated guess. Our confidence interval seeks to bracket this elusive value by suggesting a range around our sample proportion.
  • For instance, with a 95% confidence interval from 0.1437 to 0.2563, we believe with high certainty that the true proportion is likely within this range.
  • We trust this interval because it accounts for sample variability and uncertainty.
Ultimately, the true proportion reflects the larger population, based on insights derived from our sample data.

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Most popular questions from this chapter

In assessing the desirability of windowless schools, officials asked 144 elementary school children whether or not they like windows in their classrooms. Thirty percent of the children preferred windows. Establish a \(0.95\) confidence-interval estimate of the proportion of elementary school children who like windows in their classrooms.

Mr. Greenberg owns a gas station in Philadelphia. His busiest hour is from 11 to noon. On May 18 th, Mr. Greenberg's daughter Beth surveyed 36 customers. She found that the 36 people bought an average of 12 gallons of gasoline with a standard deviation of 4 gallons. (a) Find a point estimate for \(\mu\), the mean number of gallons of gas people buy. (b) Establish a \(.95\) confidence interval for \(\mu\). (c) Establish a 99 confidence interval for \(\mu\).

A random sample of 10 Miss America contestants had a mean age of \(22.6\) years with a standard deviation of 2 years. A random sample of 12 Miss U.S.A. candidates had a mean age of \(19.6\) with a standard deviation of \(1.6 .\) Assume the population variances are equal. Find a \(90 \%\) confidence interval estimate for the difference between the population means.

Derive a method for determining a \(100(1-\alpha) \%\) confidence interval for the unknown variance, \(\sigma^{2}\), of a normal distribution when the mean is also unknown.

A random sample of size \(\mathrm{n}\) is taken from a very large number of army recruits. The average weight for the recruits in the sample \((\underline{X})\) is 160 pounds and the sample standard deviation is 10 pounds. Suppose that we want the \(90 \%\) confidence interval to be equal at most to 5 pounds. What size random sample greater than \(30(\mathrm{n}>30)\) should we take?

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