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Suppose a political consultant is hired to determine if a school bond is likely to pass in a local election. The consultant randomly samples 250 likely voters and finds that \(52 \%\) of the sample supports passing the bond. Construct a \(95 \%\) confidence interval for the proportion of voters who support the bond. Assume the conditions are met. Based on the confidence interval, should the consultant predict the bond will pass? Why or why not?

Short Answer

Expert verified
Once we've accounted for the margin of error, we'll look at the confidence interval. If the entire interval falls above 50%, the consultant could confidently predict that the bond will pass based on the sample statistics. The detailed conclusion will depend on the calculation of the margin of error and the final confidence interval.

Step by step solution

01

Calculating the sample proportion

First, we calculate the sample proportion (\(p\)), which is the number of successes (those who support the bond) over the total sample size. Given the question, that's \(52\%\) of 250, so \(p=\(0.52\).
02

Calculating the margin of error

Next, we determine the margin of error. We'll use the formula for the margin of error of a proportion, which is \(Z * \sqrt{\frac{p(1-p)}{n}}\), where \(Z\) is the z-score (which equals 1.96 for a 95% confidence interval), \(p\) is the sample proportion, and \(n\) is the sample size. Substituting the given values we get: \(1.96*\sqrt{\frac{0.52*(1-0.52)}{250}}\). This gives us the margin of error.
03

Constructing the confidence interval

Now, we can construct the confidence interval. The confidence interval is the range in which we expect the population proportion to fall with 95% confidence. It's calculated by the formula: \(p \pm \) margin of error. Substituting in the values we've calculated, we get: \(0.52 \pm\) margin of error, giving us our 95% confidence interval.
04

Interpreting the confidence interval

The final step is to interpret the confidence interval. If the interval falls above 50%, then we can predict that the bond will pass. If it falls below 50%, we predict that the bond won't pass. The reason behind this is that passing requires more than 50% of the votes. We'll look at the confidence interval we calculated and make a call based on that.

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Key Concepts

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

Sample Proportion
A sample proportion is a fundamental concept in statistics, especially when deriving conclusions from sample data. In the context of this exercise, the sample proportion is a way to estimate the proportion of a population that has a certain characteristic. Here, it estimates the percentage of voters who support the school bond.

To calculate it, you divide the number of positive responses by the total number of surveyed individuals. In the exercise, 52% of 250 voters support the bond, leading to a sample proportion (\( p \)) of 0.52. This value serves as our best estimate of what the true proportion might be among all potential voters.

Using a sample proportion allows us to infer information about the larger population through the process of statistical inference. However, it's essential to remember it reflects only the sample group and may not perfectly represent the true population proportion.
Margin of Error
The margin of error provides a range around the sample proportion, giving a sense of the potential error in sampling. It is a crucial part of constructing confidence intervals, helping to account for the variability inherent in using sample data.

The formula for calculating margin of error in proportion-based studies is \( Z * \sqrt{\frac{p(1-p)}{n}} \). This accounts for the menial variations between the sample proportion and the actual population proportion.

In our exercise, using the z-score for a 95% confidence level (1.96), and our calculated sample proportion (0.52) over a sample size of 250, the margin of error can be computed. This results in a buffer amount around the sample proportion indicating precision level of our estimate and how close our sample proportion might be to the true population proportion.
Z-Score
The z-score plays a key role in determining how confident we are about our sample estimate representing the real population proportion. It represents the number of standard deviations a data point is from the mean.

For confidence intervals, it is used to find standard deviations within a desired confidence level. For example, a z-score of 1.96 is typically used for a 95% confidence interval. This tells us we expect the true population proportion to fall within this range 95% of the time if we repeated sampling.

In constructing confidence intervals, the z-score is multiplied with the standard error (part of the margin of error formula). The choice of z-score can significantly influence the breadth of the confidence interval. Thus, picking the appropriate z-score is essential to creating meaningful inferences from data.
Population Proportion
The population proportion is an unknown fixed number that represents the exact proportion of individuals in an entire population that have a specific characteristic. It is what statisticians aim to estimate using sample proportions.

In the scenario provided, the consultant is interested in estimating the proportion of the entire voter population supporting the school bond. Although the exact population proportion isn't directly observable, consulting the sample data can give a reasonable estimate through statistical inference.

Utilizing the sample proportion within a confidence interval provides a range for this population proportion, offering a more educated guess about the true state of the population characteristics. The true goal is to capture the population proportion within this interval, thus providing a basis for real-world decision making, such as predicting the outcome of an election.

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Most popular questions from this chapter

According to a 2017 survey conducted by Netflix, \(46 \%\) of couples have admitted to "cheating" on their significant other by streaming a TV show ahead of their partner. Suppose a random sample of 80 Netflix subscribers is selected. a. What percentage of the sample would we expect have "cheated" on their partner? b. Verify that the conditions for the Central Limit Theorem are met. c. What is the standard error for this sample proportion? d. Complete the sentence: We expect _____% of streaming couples to admit to Netflix 鈥渃heating,鈥 give or take _____%.

In a 2017 Harris poll conducted for Uber Eats, 438 of 1019 U.S. adults polled said they were "picky eaters." a. What proportion of the respondents said they were picky eaters? b. Find a \(95 \%\) confidence interval for the population proportion of U.S. adults who say they are picky eaters. c. Would a \(90 \%\) confidence interval based on this sample be wider or narrower than the \(95 \%\) interval? Give a reason for your answer. d. Construct the \(90 \%\) confidence interval. Was your conclusion in part c correct?

\(1,3,5,7\), and 9 are odd and 0,2, 4,6, and 8 are even. Consider a 30 -digit line from a random number table. a. How many of the 30 digits would you expect to be odd on average? b. If you actually counted, would you get exactly the number you predicted in part a? Explain.

Suppose it is known that \(20 \%\) of students at a certain college participate in a textbook recycling program each semester. a. If a random sample of 50 students is selected, do we expect that exactly \(20 \%\) of the sample participates in the textbook recycling program? Why or why not? b. Suppose we take a sample of 500 students and find the sample proportion participating in the recycling program. Which sample proportion do you think is more likely to be closer to \(20 \%\) : the proportion from a sample size of 50 or the proportion from a sample size of \(500 ?\) Explain your reasoning.

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