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What proportion of students are willing to report cheating by other students? A student project put this question to an SRS of 172 undergraduates at a large university: "You witness two students cheating on a quiz. Do you go to the professor?" Only 19 answered "Yes." \({ }^{12}\) (a) Identify the population and parameter of interest. (b) Check conditions for constructing a confidence interval for the parameter. (c) Construct a \(99 \%\) confidence interval for \(p .\) Show your method. (d) Interpret the interval in context.

Short Answer

Expert verified
The proportion lies between 4.83% and 17.27% with 99% confidence.

Step by step solution

01

Identify the population and parameter

The population is all undergraduates at the large university. The parameter of interest is the proportion of these students who are willing to report cheating by answering 'Yes' to the question.
02

Check conditions for confidence interval

To construct a valid confidence interval for the proportion, we need to check: 1) Randomness: The sample should be a simple random sample (SRS) which is satisfied here. 2) Normality: Both \(np\) and \(n(1-p)\) must be at least 10. Using \(\hat{p} = \frac{19}{172}\), check \( np = 19 \) and \( n(1-p) = 153 \), both of which are ≥ 10.
03

Calculate sample proportion

The sample proportion \( \hat{p} \) is calculated as the number of 'Yes' responses divided by the total number of students surveyed. Therefore, \( \hat{p} = \frac{19}{172} \approx 0.1105 \).
04

Find the standard error

The standard error (SE) of the sample proportion is calculated using the formula: \[ SE = \sqrt{ \frac{\hat{p}(1-\hat{p})}{n} } \]. This gives \[ SE = \sqrt{ \frac{0.1105 \times 0.8895}{172} } \approx 0.0241 \].
05

Determine the critical value

For a 99% confidence interval, the critical value \(z^*\) can be found using the standard normal distribution table or calculator. For 99%, \(z^* \approx 2.576\).
06

Construct the confidence interval

Using the formula for the confidence interval \( \hat{p} \pm z^* \times SE \), the interval is calculated as follows: \( 0.1105 \pm 2.576 \times 0.0241 \). This results in an interval of \( [0.0483, 0.1727] \).
07

Interpret the confidence interval

With 99% confidence, we can say that the proportion of all undergraduates at this university willing to report cheating lies between 4.83% and 17.27%.

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Key Concepts

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

Confidence Interval
A confidence interval provides a range of values within which we can expect the population parameter to fall, with a certain level of confidence. In statistical practice, this is usually expressed as a percentage, such as a 95% or 99% confidence interval, which implies that if we were to take repeated samples of the same size and construct a confidence interval for each sample, then a certain percentage of those intervals would contain the true parameter value.
To calculate a confidence interval for a proportion, like the one discussed in our example, you need the sample proportion, the standard error of the sample proportion, and a critical value from the standard normal distribution. The critical value depends on the desired confidence level. In our exercise, we established a 99% confidence interval using a critical value of approximately 2.576.
The final step in constructing the confidence interval is to apply the formula: \( \hat{p} \pm z^* \times SE \). This gives us an interval estimate for the true proportion of students willing to report cheating. This range allows us a level of certainty about the population proportion based on the data from our sample.
Sample Proportion
The sample proportion, denoted as \( \hat{p} \), is a statistic used to estimate the population proportion. It represents the fraction of the sample with a particular attribute, serving as an approximation for the population characteristic. The sample proportion is essential because it forms the basis of our confidence interval estimation.
In our exercise, the sample proportion was determined by dividing the number of students who answered "Yes" by the total number of students surveyed. Specifically, \( \hat{p} = \frac{19}{172} \approx 0.1105 \). This means approximately 11.05% of the sample students indicated they would report cheating. This value provides a point estimate around which our confidence interval is centered.
  • The sample proportion provides an immediate snapshot of the sample, offering insights before relating it to the broader population.
  • It is fundamental in calculating other statistical measures, such as the standard error, which are necessary for constructing the confidence interval.
Population and Parameter
In statistical studies, understanding the difference between population and parameter is crucial.
A population refers to the entire group about which researchers want to draw conclusions. It encompasses all individuals or elements relevant to the study. In our example, the population is all undergraduates at the university. The parameter is a summary measure that describes a particular characteristic of the population, such as the proportion of all undergraduates willing to report cheating.
The parameter of interest in this context is the proportion of undergraduates who would answer "Yes" to reporting cheating. We want to estimate this value based on our sample data. However, it's important to remember that the true population parameter is often unknown and needs to be estimated through statistical methods. The processes of sampling and subsequent statistical analysis aim to provide the best possible approximation of this parameter.
Standard Error
Standard error (SE) quantifies the variability of a sample statistic. It provides an estimate of how much the sample proportion may differ from the true population proportion due to sampling variability.
To calculate the standard error for a sample proportion, use the formula: \[ SE = \sqrt{ \frac{\hat{p}(1-\hat{p})}{n} } \]. This computation ensures that as the sample size increases, the standard error decreases, reflecting more precise estimates.
In our exercise, the standard error was calculated using the sample proportion \( \hat{p} = 0.1105 \) and sample size \( n = 172 \). The result was approximately 0.0241, indicating the extent to which the sample proportion could fluctuate by chance. An important aspect of standard error is that it plays a vital role in constructing confidence intervals. It helps determine the margin of error, illustrating how sure we are about our sample estimate compared to the true population parameter. By understanding standard error, one gains deeper insight into the reliability of a sample statistic.

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Most popular questions from this chapter

Does living near power lines cause leukemia in children? The National Cancer Institute spent 5 years and \(\$ 5\) million gathering data on this question. The researchers compared 638 children who had leukemia with 620 who did not. They went into the homes and measured the magnetic fields in children's bedrooms, in other rooms, and at the front door. They recorded facts about power lines near the family home and also near the mother's residence when she was pregnant. Result: no connection between leukemia and exposure to magnetic fields of the kind produced by power lines was found. (a) Was this an observational study or an experiment? Justify your answer. (b) Does this study show that living near power lines doesn't cause cancer? Explain.

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Young people have a better chance of full-time employment and good wages if they are good with numbers. How strong are the quantitative skills of young Americans of working age? One source of data is the National Assessment of Educational Progress (NAEP) Young Adult Literacy Assessment Survey, which is based on a nationwide probability sample of households. The NAEP survey includes a short test of quantitative skills, coveringmainly basic arithmetic and the ability to apply it to realistic problems. Scores on the test range from 0 to \(500 .\) For example, a person who scores \(233 \mathrm{can}\) add the amounts of two checks appearing on a bank deposit slip; someone scoring 325 can determine the price of a meal from a menu; a person scoring 375 can transform a price in cents per ounce into dollars per pound. Suppose that you give the NAEP test to an SRS of 840 people from a large population in which the scores have mean 280 and standard deviation \(\sigma=60\). The mean \(\bar{x}\) of the 840 scores will vary if you take repeated samples. (a) Describe the shape, center, and spread of the sampling distribution of \(\bar{x}\). (b) Sketch the sampling distribution of \(\bar{x}\). Mark its mean and the values \(1,2,\) and 3 standard deviations on either side of the mean. (c) According to the \(68-95-99.7\) rule, about \(95 \%\) of all values of \(\bar{x}\) lie within a distance \(m\) of the mean of the sampling distribution. What is \(m ?\) Shade the region on the axis of your sketch that is within \(m\) of the mean. (d) Whenever \(\bar{x}\) falls in the region you shaded, the population mean \(\mu\) lies in the confidence interval \(\bar{x} \pm m\). For what percent of all possible samples does the interval capture \(\mu ?\)

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