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Chapter 8: Problem 3

Determine the point estimator you would use and calculate the value of the point estimate. Tonya wants to estimate what proportion of the seniors in her school plan to attend the prom. She interviews an SRS of 50 of the 750 seniors in her school and finds that 36 plan to go to the prom.

Short Answer

Expert verified
The point estimator is the sample proportion; the point estimate is 0.72.

Step by step solution

01

Understanding the Problem

Tonya wants to estimate the proportion of seniors planning to attend the prom. She has conducted a Simple Random Sample (SRS) of 50 seniors out of 750, of which 36 plan to attend the prom. We need to determine a suitable point estimator and calculate the point estimate.
02

Selecting the Point Estimator

In this scenario, we want to estimate a population proportion. The appropriate point estimator for a population proportion is the sample proportion, often denoted as \( \hat{p} \). It is calculated by dividing the number of favorable outcomes by the total number of observations in the sample.
03

Calculating the Sample Proportion

To find the sample proportion \( \hat{p} \), use the formula: \[ \hat{p} = \frac{x}{n} \] where \( x = 36 \) (the number of seniors who plan to attend) and \( n = 50 \) (total number of seniors surveyed).
04

Substituting Values

Substitute \( x = 36 \) and \( n = 50 \) into the formula: \[ \hat{p} = \frac{36}{50} \].
05

Simplifying the Expression

Calculate the division in the expression to find the value of \( \hat{p} \): \[ \hat{p} = 0.72 \]. This means the point estimate for the proportion of seniors planning to attend is 0.72 or 72%.

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Key Concepts

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

Point Estimator
A point estimator is a statistical tool used to provide a single, most likely value for a particular parameter from a population, based on sample data. Visualize it as a best guess or estimation of what the real value might be. Here, we want to estimate the proportion of seniors planning to attend the prom.
  • For proportions, the sample proportion, denoted as \( \hat{p} \), is used as the point estimator.
  • It is calculated by dividing the number of favorable outcomes by the total sample size.
  • This value represents the most probable estimate of the actual, but unknown, population proportion.
In the exercise, Tonya's sample provided a sample proportion of 0.72. This means her point estimator suggests that about 72% of seniors plan to attend the prom.
Population Proportion
Population proportion is a term used to indicate the fraction of individuals in a population who exhibit a particular attribute of interest. In Tonya's case, it's the proportion of seniors who plan to go to the prom.
  • The population proportion is denoted as \( p \).
  • However, \( p \) is generally unknown because it's infeasible to survey every member of a large population.
  • Instead, we rely on sampling and the sample proportion \( \hat{p} \) as our estimate to infer what the actual proportion might be.
Thus, Tonya uses the proportion from her Simple Random Sample as a practical approximation of the true population proportion.
Simple Random Sample (SRS)
A Simple Random Sample (SRS) is a foundational concept in statistics wherein each member of a population is equally likely to be chosen. This method helps ensure the randomness and representativeness of the sample.
  • Every individual in the population has an equal chance of being selected.
  • This mitigates biases that could skew the results of the study.
  • Tonya surveyed 50 seniors out of a total of 750. By doing so, she executed a SRS.
SRS is fundamental because it increases the likelihood that the sample data accurately reflects the population's characteristics, offering more credible estimates such as the point estimator of the population proportion.

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Most popular questions from this chapter

PTC is a substance that has a strong bitter taste for some people and is tasteless for others. The ability to taste \(\mathrm{PTC}\) is inherited. About \(75 \%\) of Italians can taste \(\mathrm{PTC}\), for example. You want to estimate the proportion of Americans who have at least one Italian grandparent and who can taste PTC. (a) How large a sample must you test to estimate the proportion of PTC tasters within 0.04 with \(90 \%\) confidence? Answer this question using the \(75 \%\) estimate as the guessed value for \(\hat{p}\). (b) Answer the question in part (a) again, but this time use the conservative guess \(\hat{p}=0.5 .\) By how much do the two sample sizes differ?

A national opinion poll found that \(44 \%\) of all American adults agree that parents should be given vouchers that are good for education at any public or private school of their choice. The result was based on a small sample. (a) How large an SRS is required to obtain a margin of error of 0.03 (that is, \(\pm 3 \%\) ) in a \(99 \%\) confidence interval? Answer this question using the previous poll's result as the guessed value for \(\hat{p}\). (b) Answer the question in part (a) again, but this time use the conservative guess \(\hat{p}=0.5 .\) By how much do the two sample sizes differ?

The admissions director from Big City University found that (107.8,116.2) is a \(95 \%\) confidence interval for the mean IQ score of all freshmen. Discuss whether each of the following explanations is correct. (a) There is a \(95 \%\) probability that the interval from 107.8 to 116.2 contains \(\mu\) (b) There is a \(95 \%\) chance that the interval (107.8, 116.2 ) contains \(\bar{x}\) (c) This interval was constructed using a method that produces intervals that capture the true mean in \(95 \%\) of all possible samples. (d) If we take many samples, about \(95 \%\) of them will contain the interval (107.8,116.2) (e) The probability that the interval (107.8,116.2) captures \(\mu\) is either 0 or 1 , but we don't know which.

The Trial Urban District Assessment (TUDA) is a government-sponsored study of student achievement in large urban school districts. TUDA gives a reading test scored from 0 to \(500 . \mathrm{A}\) score of 243 is a "basic" reading level and a score of 281 is "proficient." Scores for a random sample of 1470 eighth- graders in Atlanta had \(\bar{x}=240\) with standard deviation \(42.17 .^{24}\) (a) Calculate and interpret a \(99 \%\) confidence interval for the mean score of all Atlanta eighth-graders. (b) Based on your interval from part (a), is there good evidence that the mean for all Atlanta eighth-graders is less than the basic level? Explain.

Most people can roll their tongues, but many can't. The ability to roll the tongue is genetically determined. Suppose we are interested in determining what proportion of students can roll their tongues. We test a simple random sample of 400 students and find that 317 can roll their tongues. The margin of error for a \(95 \%\) confidence interval for the true proportion of tongue rollers among students is closest to (a) 0.0008 . (c) 0.03 . (c) 0.05 . (b) 0.02 (d) 0.04
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