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Chapter 8: Problem 27

Check whether each of the conditions is met for calculating a confidence interval for the population proportion \(\bar{p}\). Latoya wants to estimate what proportion of the seniors at her boarding high school like the cafeteria food. She interviews an SRS of 50 of the 175 seniors living in the dormitory. She finds that 14 think the cafeteria food is good.

Short Answer

Expert verified
All conditions for calculating a confidence interval for the population proportion \( \bar{p} \) are met.

Step by step solution

01

Calculate Sample Proportion

First, calculate the sample proportion \( \hat{p} \) by dividing the number of seniors in the sample who like the cafeteria food by the total number of sampled seniors. Here, \( \hat{p} = \frac{14}{50} = 0.28 \).
02

Check Sample Size Requirements

For the conditions to be met, the sample size must be sufficiently large. Specifically, both \( n \hat{p} \) and \( n(1 - \hat{p}) \) should be greater than 10. Here, \( n \hat{p} = 50 \times 0.28 = 14 \) and \( n(1 - \hat{p}) = 50 \times 0.72 = 36 \). Both values are greater than 10, satisfying the sample size condition.
03

Check Random Sampling Condition

Latoya selected a Simple Random Sample (SRS) of 50 seniors from 175, which satisfies the condition of random sampling necessary for inference on population proportions.
04

Population Size Condition

Often, the population should be at least 10 times the sample size to ensure independence. Here, the population size is 175, which is more than 10 times the sample size of 50; thus, the condition is met.
05

Conclusion about Conditions

Since all conditions are met鈥攍arge enough sample size, random sampling, and appropriate population size鈥攚e can calculate the confidence interval for the population proportion \( \bar{p} \) of seniors who like the cafeteria food.

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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 way to estimate the proportion of a certain trait within a sample taken from a larger population. In the given exercise, Latoya conducted a survey to find out how many seniors at her school like the cafeteria food. The sample proportion is found by taking the number of positive responses (those who liked the food) and dividing it by the total number of respondents in the sample. For Latoya's survey, the sample proportion is calculated as: \( \hat{p} = \frac{14}{50} = 0.28 \).
  • 14 of the 50 surveyed seniors liked the food.
  • This gives us a sample proportion of 0.28, indicating that 28% of the sampled seniors have a favorable opinion of the cafeteria food.
This proportion gives us an estimate for the larger population, helping Latoya make predictions about all 175 seniors.
Random Sampling
Random sampling is a method used to ensure that every individual within a population has an equal chance of being selected for the sample. In Latoya's case, she used a Simple Random Sample (SRS) to choose 50 seniors out of the 175 in her boarding high school. This random selection is crucial because:
  • It reduces bias, ensuring that the sample truly represents the wider population.
  • Makes the results more generalizable to the entire population.
By using random sampling, Latoya can be more confident that the sample proportion is a true reflection of the views of all seniors in her school, not just the subset she happened to survey. This adds validity to her findings and any resulting conclusions or decisions she needs to make.
Sample Size Requirements
Sample size requirements help to ensure that the statistical measures obtained from a sample can reliably be applied to the larger population. There are specific criteria to determine if a sample size is adequate:
  • The product of the sample size \( n \) and the sample proportion \( \hat{p} \) must be greater than 10: \( n \hat{p} > 10 \).
  • The product of the sample size \( n \) and \( (1 - \hat{p}) \) must also be greater than 10: \( n(1-\hat{p}) > 10 \).
In this exercise,:
  • \( 50 \times 0.28 = 14 \) and \( 50 \times 0.72 = 36 \) are both greater than 10,.
These criteria ensure that the sample data has a normal distribution, a key requirement for calculating accurate confidence intervals. Meeting these sample size requirements gives Latoya more reliable results when estimating the true proportion of seniors who enjoy the cafeteria food.
Population Proportion
The population proportion, denoted \( \bar{p} \), is the ratio of members of a population that have a particular attribute or characteristic. For Latoya's survey, she wishes to estimate this proportion based on her sample of 50 seniors. Some considerations when determining population proportion include:
  • Comparing the proportion from the sample (\( \hat{p} \)) to what is expected under normal circumstances.
  • Determining if there are significant differences or if the sample is a reliable estimator.
  • Ensuring the population from which the sample is drawn significantly exceeds the sample size (often by at least 10 times), which holds true as there are 175 seniors.
Latoya uses her calculated \( \hat{p} \) to estimate \( \bar{p} \). Because her sampling, sample size, and other conditions are sound, her estimate for the population proportion is likely accurate. This is important as it guides decision-making, such as menu planning or school policy changes regarding cafeteria offerings.

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Most popular questions from this chapter

A bunion on the big toe is fairly uncommon in youth and often requires surgery. Doctors used X-rays to measure the angle (in degrees) of deformity on the big toe in a random sample of 37 patients under the age of 21 who came to a medical center for surgery to correct a bunion. The angle is a measure of the seriousness of the deformity. For these 37 patients, the mean angle of deformity was 24.76 degrees and the standard deviation was 6.34 degrees. A dotplot of the data revealed no outliers or strong skewness. \({ }^{26}\) (a) Construct and interpret a \(90 \%\) confidence interval for the mean angle of deformity in the population of all such patients. (b) Researchers omitted one patient with a deformity angle of 50 degrees from the analysis due to a measurement issue. What effect would including this outlier have on the confidence interval in part (a)? Justify your answer without doing any calculations.

A researcher plans to use a random sample of families to estimate the mean monthly family income for a large population. The researcher is deciding between a sample of size \(n=500\) and a sample of size \(n=1000 .\) Compared to using a sample size of \(n=500,\) a \(95 \%\) confidence interval based on a sample size of \(n=1000\) will be (a) narrower and would involve a larger risk of being incorrect. (b) wider and would involve a smaller risk of being incorrect. (c) narrower and would involve a smaller risk of being incorrect. (d) wider and would involve a larger risk of being incorrect. (e) narrower and would have the same risk of being incorrect.

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Several years ago, the U.S. Agency for International Development provided 238,300 metric tons of corn-soy blend (CSB) for emergency relief in countries throughout the world. \(\mathrm{CSB}\) is a highly nutritious, low-cost fortified food. As part of a study to evaluate appropriate vitamin C levels in this food, measurements were taken on samples of CSB produced in a factory. \({ }^{28}\) The following data are the amounts of vitamin C, measured in milligrams per 100 grams \((\mathrm{mg} / 100 \mathrm{~g})\) of blend, for a random sample of size 8 from one production run: $$\begin{array}{llllllll}\hline 26 & 31 & 23 & 22 & 11 & 22 & 14 & 31 \\\\\hline\end{array}$$ Construct and interpret a \(95 \%\) confidence interval for the mean amount of vitamin \(\mathrm{C} \mu\) in the \(\mathrm{CSB}\) from this production run.

A study of commuting times reports the travel times to work of a random sample of 20 employed adults in New York State. The mean is \(\bar{x}=31.25\) minutes, and the standard deviation is \(s_{x}=21.88\) minutes. What is the standard error of the mean? Interpret this value in context.
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