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91Ó°ÊÓ

Reporting cheating. Students are reluctant 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." 21 Give a \(95 \%\) confidence interval for the proportion of all undergraduates at this university who would report cheating.

Short Answer

Expert verified
The 95% confidence interval is (0.0640, 0.1570).

Step by step solution

01

Identify the sample proportion

First, determine the sample proportion of students who would report cheating. We have 19 out of 172 students who answered "yes." Therefore, the sample proportion \( \hat{p} \) can be calculated as \( \hat{p} = \frac{19}{172} \approx 0.1105 \).
02

Calculate the standard error of the proportion

The standard error (SE) of the sample proportion is calculated with the formula: \( SE = \sqrt{\frac{\hat{p}(1 - \hat{p})}{n}} \), where \( n = 172 \). Substituting the values, we find the standard error: \[ SE = \sqrt{\frac{0.1105 (1 - 0.1105)}{172}} \approx 0.0237 \].
03

Determine the z-score for a 95% confidence level

For a 95% confidence interval, the z-score is 1.96. This is a standard value corresponding to a confidence level of 95% in a normal distribution.
04

Calculate the confidence interval

The confidence interval is calculated using the formula: \( \hat{p} \pm z \times SE \). Substituting the values, the interval is: \[ 0.1105 \pm 1.96 \times 0.0237 \]. This results in: \[ (0.0640, 0.1570) \].
05

Interpret the confidence interval

The 95% confidence interval \((0.0640, 0.1570)\) indicates that we are 95% confident the proportion of all undergraduates at this university who would report cheating is between 6.40% and 15.70%.

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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 is a way to estimate what the overall proportion might be in a larger population. It comes from survey data or experiments and represents the proportion of the sample that exhibits a particular attribute. In the given exercise, we had 172 students surveyed, with 19 indicating they would report cheating.

To find the sample proportion, we denote it as \( \hat{p} \), calculated by dividing the number of successes (students who said "yes") by the total number of observations (total students surveyed). That means: \( \hat{p} = \frac{19}{172} \approx 0.1105 \).

This proportion tells us that approximately 11.05% of the sample students would report cheating. It serves as a point estimate for the true proportion in the entire student population.
Standard Error
The standard error (SE) is crucial as it measures the amount of variability in a sample statistic - in this case, the sample proportion. It helps us understand how much the sample proportion might fluctuate by chance from one sample to another. This fluctuation arises because we're only looking at a portion of the population.

The formula for the standard error of a proportion is \( SE = \sqrt{\frac{\hat{p}(1 - \hat{p})}{n}} \), where \( \hat{p} \) is our sample proportion, and \( n \) is the sample size. Applying our numbers gives \( SE = \sqrt{\frac{0.1105(1 - 0.1105)}{172}} \approx 0.0237 \).

This results in an SE of about 0.0237, indicating the possible variation in the sample proportion across different sets of samples of the same size. The smaller the SE, generally, the more reliable your sample proportion is as an estimate of the population proportion.
Z-Score
A z-score is a value that tells us how many standard deviations a point is from the mean. When building confidence intervals, the z-score helps to determine how much we stretch our interval to capture the true population parameter with a certain level of confidence.

For a 95% confidence level, the z-score used is 1.96. This means that for our interval, we are stretching about 1.96 standard errors in both directions (above and below the sample proportion).

The importance of the z-score stems from the properties of the normal distribution. Given a normal distribution, a z-score indicates the percent of data expected to lie below a particular point. At 1.96, approximately 95% of data in a normal distribution lies below this value, making it reliable for constructing confidence intervals.
Statistical Inference
Statistical inference refers to the process of using data from a sample to make conclusions about a population. This is incredibly important in fields like science, marketing, and social research where studying an entire population is impractical.

In this exercise, statistical inference was used to estimate the proportion of students at a university that would report cheating. Through a confidence interval, we drew a range that we believe contains the true population proportion.

The connection between sampling, standard error, and z-scores plays a crucial role in making these estimations. By acknowledging variability (through standard error) and using z-scores to set bounds, statistical inference allows researchers to make informed statements about a population backed by data collected through suitable sampling. Confidence intervals, like the one calculated, are a common tool in statistical inference to give an estimated range for that population parameter.

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Most popular questions from this chapter

Order in Choice. Does the order in which wine is presented make a difference? In this study, subjects were asked to taste two wine samples in sequence. Both samples given to a subject were the same wine, although subjects were expecting to taste two different samples of a particular variety. Of the 32 subjects in the study, 22 selected the wine presented first when presented with two identical wine samples. \(\frac{30}{}\) a. Do the data give good reason to conclude that the subjects are not equally likely to choose either of the two positions when presented with two identical wine samples in sequence? b. The subjects were recruited in Ontario, Canada, via advertisements to participate in a study of "attitudes and values toward wine." Can we generalize our conclusions to all wine tasters? Explain.

How many American adults must be interviewed to estimate the proportion of all American adults who actively try to avoid drinking regular soda or pop within \(\pm 0.01\) with \(99 \%\) confidence using the large-sample confidence interval? Use \(0.5\) as the conservative guess for \(p\). a. \(n=6765\) b. \(n=9604\) c. \(n=16590\)

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Do You Listen to Podcasts? In January and February 2019, Edison Research conducted a national telephone survey of 1500 Americans aged 12 and older, using random digit dialing techniques to both cell phones and landlines. The survey included questions about the use of mobile devices, Internet audio, podcasting, social media, smart speakers, and more. Of the 1500 people surveyed, 480 said they had listened to a podcast in the past month. 12 a. What is the margin of error of the large-sample \(95 \%\) confidence interval for the proportion of Americans aged 12 and older who had listened to a podcast in the past month? b. How large a sample is needed to get the common \(\pm 3\) percentage point margin of error? Use the January/February survey of 1500 as a pilot study to get \(p^{*}\)

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