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Chapter 9: Problem 42

In a survey of 800 college students in the United States, 576 indicated that they believe that a student or faculty member on campus who uses language considered racist, sexist, homophobic, or offensive should be subject to disciplinary action ("Listening to Dissenting Views Part of Civil Debate," USA TODAY, November 17,2015 ). Assuming that the sample is representative of college students in the United States, construct and interpret a \(95 \%\) confidence interval for the proportion of college students who have this belief.

Short Answer

Expert verified
The 95% confidence interval for the proportion of college students in the United States who believe that individuals who use offensive language on campus should face disciplinary action is \(CI =\frac{576}{800} \pm 1.96 \times \sqrt{\frac{\frac{576}{800}(1-\frac{576}{800})}{800}}\). This means that we are 95% confident that the true proportion of college students in the United States who share this belief lies within this calculated range.

Step by step solution

01

Calculate the sample proportion

We first calculate the sample proportion, which is the proportion of college students in the sample who believe that those who use offensive language on campus should face disciplinary action. To do this, we divide the number of students who have this belief (576) by the total number of students in the sample (800): \[\hat{p} = \frac{576}{800}\]
02

Calculate the critical value for a 95% confidence interval

We need to find the z-score (critical value or z) that corresponds to a 95% confidence interval. The rule of thumb is that a 95% confidence interval corresponds to a z-score of approximately 1.96. You can find this value in a z-score table or use software or online resources.
03

Calculate the margin of error

Now, calculate the margin of error using the formula mentioned earlier: \[margin\_of\_error = z\sqrt{\frac{\hat{p}(1-\hat{p})}{n}}\] Plugging in the values, we get: \[margin\_of\_error = 1.96 \times \sqrt{\frac{\frac{576}{800}(1-\frac{576}{800})}{800}}\]
04

Calculate the confidence interval

Finally, calculate the confidence interval using the sample proportion and the margin of error we calculated: \[CI = \hat{p} \pm margin\_of\_error\] Plugging in the values: \[CI =\frac{576}{800} \pm 1.96 \times \sqrt{\frac{\frac{576}{800}(1-\frac{576}{800})}{800}}\]
05

Interpret the confidence interval

The computed 95% confidence interval represents the range within which we are 95% confident that the true proportion of college students in the United States who believe that individuals who use offensive language on campus should face disciplinary action falls. In other words, if we were to sample many different groups of college students, 95% of the time, the true population proportion would fall within this calculated confidence interval.

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Key Concepts

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

Sample Proportion
When we're looking to understand the opinion of a larger group based on a smaller subset, we calculate the sample proportion. It's a statistic that shows what fraction of our sample has a certain trait or opinion. In our case, out of 800 college students, 576 believe that using offensive language should lead to disciplinary action. This gives us a sample proportion (\r\(\r\hat{p}\r\)\r) by dividing 576 by 800.
Critical Value
To construct a confidence interval, we must determine the critical value. This value is tied to the desired level of confidence we want in our interval, typically expressed in terms of a percentage like 95%. The critical value serves as a multiplier to set the range of our confidence interval and is based on a standardized z-score scale. For a 95% confidence level, the critical value is usually around 1.96. This is because the z-score of 1.96 captures the central 95% of the standard normal distribution.
Margin of Error
The margin of error tells us how much we can expect the sample proportion to vary from the true population proportion. We calculate it using our critical value and the standard error, which is the standard deviation of the sample proportion. This margin of error reflects the uncertainty inherent in our estimation process. It effectively says, 'We're confident that the true number is no more than this amount above or below what we've observed in our sample.'
Z-score
A z-score, or standard score, is a way to describe a data point's relationship to the mean of a group of values, measured in terms of standard deviations from the mean. A z-score can be positive or negative, indicating whether it鈥檚 above or below the mean and by how many standard deviations. In confidence interval calculations, we use the z-score to determine how far out from the sample proportion we'll go to ensure we've captured the true population proportion with the confidence level we desire.
Statistical Significance
The concept of statistical significance allows us to determine whether a result is likely due to an actual effect or just by random chance. In constructing confidence intervals, the notion of significance is embedded in our confidence level; a 95% confidence level suggests that if we were to repeat this survey many times, 95% of the computed intervals would contain the true population proportion. This high level of confidence usually indicates a statistically significant result within the context of the given confidence interval.

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Most popular questions from this chapter

The USA Snapshot titled "Social Media Jeopardizing Your Job?" (USA TODAY, November 12,2014\()\) summarized data from a survey of 1855 recruiters and human resource professionals. The Snapshot indicted that \(53 \%\) of the people surveyed had reconsidered a job candidate based on his or her social media profile. Assume that the sample is representative of the population of recruiters and human resource professionals in the United States. a. Use the given information to estimate the proportion of recruiters and human resource professionals who have reconsidered a job candidate based on his or her social media profile using a \(95 \%\) confidence interval. Give an interpretation of the interval in context and an interpretation of the confidence level of \(95 \%\). b. Would a \(90 \%\) confidence interval be wider or narrower than the \(95 \%\) confidence interval from Part (a)?

A consumer group is interested in estimating the proportion of packages of ground beef sold at a particular store that have an actual fat content exceeding the fat content stated on the label. How many packages of ground beef should be tested in order to have a margin of error of \(0.05 ?\)

The formula used to calculate a large-sample confidence interval for \(p\) is $$ \hat{p} \pm(z \text { critical value }) \sqrt{\frac{\hat{p}(1-\hat{p})}{n}} $$ What is the appropriate \(z\) critical value for each of the following confidence levels? a. \(90 \%\) b. \(99 \%\) c. \(80 \%\)

A random sample will be selected from the population of all adult residents of a particular city. The sample proportion \(\hat{p}\) will be used to estimate \(p,\) the proportion of all adult residents who are registered to vote. For which of the following situations will the estimate tend to be closest to the actual value of \(p ?\) I. \(\quad n=1000, p=0.5\) II. \(\quad n=200, p=0.6\) III. \(n=100, p=0.7\)

In spite of the potential safety hazards, some people would like to have an Internet connection in their car. A preliminary survey of adult Americans has estimated the proportion of adult Americans who would like Internet access in their car to be somewhere around 0.30 (USA TODAY, May 1 , 2009). Use the given preliminary estimate to determine the sample size required to estimate this proportion with a margin of error of 0.02
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