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You plan to conduct a survey on your college campus to learn about the political awareness of students. You want to estimate the true proportion of college students on your campus who voted in the 2012 presidential election with 95% confidence and a margin of error no greater than five percent. How many students must you interview?

Short Answer

Expert verified
You need to interview 385 students.

Step by step solution

01

Understand the Confidence Interval Formula

To find out how many students to interview, we need to use the formula for determining the sample size for a proportion. The formula is: \[ n = \left(\frac{Z^2 \times p \times (1-p)}{E^2}\right) \]where \(n\) is the sample size, \(Z\) is the z-score for the confidence level, \(p\) is the estimated proportion of the population that have the characteristic of interest, and \(E\) is the margin of error.
02

Determine the Values Needed

Given that we want 95% confidence, the z-score (\(Z\)) corresponds to 1.96. The margin of error \(E\) is 0.05 (5%). Since we lack a preliminary estimate of \(p\), it's standard practice to use 0.5 for \(p\) as this maximizes the sample size and ensures an adequately large sample size.
03

Calculate the Sample Size

Substitute the values into the formula: \[ n = \left(\frac{1.96^2 \times 0.5 \times (1-0.5)}{0.05^2}\right) \]Now, calculate this step by step:\[ n = \left(\frac{3.8416 \times 0.25}{0.0025}\right) \]\[ n = \left(\frac{0.9604}{0.0025}\right) \]\[ n = 384.16 \]Since the sample size must be a whole number, round up to 385.

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Key Concepts

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

Confidence Interval
The concept of a confidence interval is central to inferential statistics. It helps us estimate a range within which we believe a population parameter, like the proportion of students who voted, lies based on a sample. When you hear about a 95% confidence interval, this means if we were to take 100 different samples and compute an interval for each, we expect about 95 of these intervals to contain the true population parameter. This doesn’t guarantee certainty, but it provides a probabilistic measure of accuracy.

In this exercise, the confidence level is set at 95%, a common choice. Setting a confidence level helps balance between precision and reliability. A higher confidence means a wider interval, indicating more certainty that the interval covers the true parameter, but less precision.
Proportion Estimate
A proportion estimate attempts to represent the true proportion of a population that meets a certain condition through a sample. In the context of our exercise, we want to estimate the proportion of students who voted. However, without prior information, choosing an initial guess for this proportion can be tricky.

The standard initial estimate for the proportion, when exact previous data isn’t available, is often 0.5. This is because 0.5 represents the maximum uncertainty about the proportion, leading to the most conservative (largest) sample size needed. This ensures that the sample size will be adequate irrespective of the true underlying proportion.
Margin of Error
The margin of error is a statistic expressing the amount of random sampling error in a survey's results. A smaller margin of error indicates greater confidence in the survey's estimates of the population parameter. It is depicted as a plus or minus (7) value around the estimated value.

In this scenario, the desired margin of error is 5%. What this means is that the estimated proportion of students who voted could be expected to be within 5 percentage points above or below the true proportion in the population. A smaller margin of error would require a larger sample to ensure an accurate estimate.
Z-Score
A z-score is a statistical measurement that describes a value's relation to the mean of a group of values, measured in terms of standard deviations. Z-scores are used in this context to find the boundaries that correspond to our confidence level.

For the common confidence level of 95%, the z-score is approximately 1.96. The choice of 1.96 stems from the properties of the normal distribution, where approximately 95% of data falls within 1.96 standard deviations of the mean.

Using the z-score, we can determine what proportion of data values lie within a certain interval, which is crucial in estimating how many samples we need to achieve our desired margin of error. Understanding z-scores is fundamental for anyone studying statistics, as it ties directly to the concept of normal distribution and standard deviations.

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Most popular questions from this chapter

An article regarding interracial dating and marriage recently appeared in the Washington Post. Of the 1,709 randomly selected adults, 315 identified themselves as Latinos, 323 identified themselves as blacks, 254 identified themselves as Asians, and 779 identified themselves as whites. In this survey, 86% of blacks said that they would welcome a white person into their families. Among Asians, 77% would welcome a white person into their families, 71% would welcome a Latino, and 66% would welcome a black person. a. We are interested in finding the 95\(\%\) confidence interval for the percent of all black adults who would welcome a white person into their families. Define the random variables \(X\) and \(P^{\prime},\) in words. b. Which distribution should you use for this problem? Explain your choice. c. Construct a 95\(\%\) confidence interval. i. State the confidence interval. ii. Sketch the graph. iii. Calculate the error bound.

Use the following information to answer the next five exercises: A poll of \(1,200\) voters asked what the most significant issue was in the upcoming election. Sixty-five percent answered the economy. We are interested in the population proportion of voters who feel the economy is the most important. Which distribution should you use for this problem?

Among various ethnic groups, the standard deviation of heights is known to be approximately three inches. We wish to construct a 95% confidence interval for the mean height of male Swedes. Forty-eight male Swedes are surveyed. The sample mean is 71 inches. The sample standard deviation is 2.8 inches. a. i. \(\overline{x}=\) _____ ii. \(\sigma=\) _____ iii. \(n=\) _____ b. In words, define the random variables \(X\) and \(\overline{X}\) . c. Which distribution should you use for this problem? Explain your choice. d. Construct a 95\(\%\) confidence interval for the population mean height of male Swedes. i. State the confidence interval. ii. Sketch the graph. iii. Calculate the error bound. e. What will happen to the level of confidence obtained if \(1,000\) male Swedes are surveyed instead of 48\(?\) Why?

In six packages of "The Flintstones \(\mathbb{R}\) Real Fruit Snacks" there were five Bam-Bam snack pieces. The total number of snack pieces in the six bags was \(68 .\) We wish to calculate a 96\(\%\) confidence interval for the population proportion of Bam-Bam snack pieces. a. Define the random variables \(X\) and \(P^{\prime}\) in words. b. Which distribution should you use for this problem? Explain your choice c. Calculate \(p^{\prime} .\) d. Construct a 96\(\%\) confidence interval for the population of Bam-Bam snack pieces per bag. i. State the confidence interval. ii. Sketch the graph. iii. Calculate the error bound. e. Do you think that six packages of fruit snacks yield enough data to give accurate results? Why or why not?

Use the following information to answer the next two exercises: A quality control specialist for a restaurant chain takes a random sample of size 12 to check the amount of soda served in the 16 oz. serving size. The sample mean is 13.30 with a sample standard deviation of 1.55. Assume the underlying population is normally distributed. What is the error bound? a. 0.87 b. 1.98 c. 0.99 d. 1.74

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