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Chapter 9: Problem 63

An Associated Press article on potential violent behavior reported the results of a survey of 750 workers who were employed full time (San Luis Obispo Tribune, September 7. 1999). Of those surveyed, 125 indicated that they were so angered by a coworker during the past year that they felt like hitting the coworker (but didn't). Assuming that it is reasonable to regard this sample of 750 as a random sample from the population of full-time workers, use this information to construct and interpret a \(90 \%\) confidence interval estimate of \(p\), the true proportion of full-time workers so angered in the last year that they wanted to hit a colleague.

Short Answer

Expert verified
Once above calculations are completed, both upper and lower boundaries of the 90% confidence interval can be calculated. This interval will represent the estimated range of true proportion of workers who felt so angered that they wanted to hit a coworker with 90% confidence.

Step by step solution

01

Calculation of Sample Proportion

Firstly, we calculate the sample proportion (\(p虃\)) which is given by the ratio of the number of positive outcomes to the total outcomes in the sample. In this case, it's \(\frac{125}{750}\). So, \(p虃=\frac{125}{750}=0.167\)
02

Calculation of Standard Error

Next, we calculate the standard error. The formula for standard error (\(SE\)) is \(\sqrt{\frac{p虃(1-p虃)}{n}}\) where \(p虃\) is the sample proportion and \(n\) is the sample size. Here, this is \(\sqrt{\frac{0.167(1-0.167)}{750}}.\)
03

Determining Z-Score

The z-score corresponding to a 90% confidence interval is 1.645 (you can find this in any standard normal distribution table). So, let's denote our z-value as \(z=1.645\).
04

Constructing the Confidence Interval

The formula for constructing the confidence interval for a proportion is given by \(p虃 \pm z*SE\). This means we have to calculate \(0.167 \pm 1.645*SE\). Once we substitute the value of \(SE\) calculated in step 2, we have our lower and upper bounds for the 90% confidence interval. These values represent the estimated range wherein the true population proportion lies with 90% confidence.

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Key Concepts

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

Sample Proportion
Understanding the sample proportion is crucial in estimating the population proportion. In our exercise, the sample proportion is the ratio of workers who felt like hitting a coworker to the total number surveyed. Here, it's calculated as \[\hat{p} = \frac{125}{750} = 0.167.\] This result means 16.7% of the surveyed workers were so angered that they contemplated hitting a coworker. The sample proportion \(\hat{p}\) serves as an estimate of the true proportion of all full-time workers who felt similarly.
Standard Error
The standard error (SE) gives us an idea of the variability in our estimate of the population proportion. It's like a measure of how much the sample proportion might differ from the actual population proportion. For standard error calculation, use the formula:\[SE = \sqrt{\frac{\hat{p}(1-\hat{p})}{n}}.\]In this case:\[SE = \sqrt{\frac{0.167 \times 0.833}{750}}.\]By applying this, we get a clearer understanding of the precision of our estimate. A smaller SE indicates a more reliable estimate.
Z-score
A z-score represents the number of standard deviations a data point is from the mean. In the context of confidence intervals, it helps define the confidence level we wish to establish for our estimate. For a 90% confidence interval, the z-score is 1.645. This value is sourced from standard normal distribution tables and signifies that we want our interval to capture the true proportion 90% of the time. A higher z-score would result in a wider interval offering more confidence.
Population Proportion Estimation
Estimating the population proportion involves creating a confidence interval around the sample proportion. This interval estimates where the true population proportion likely lies.Using the formula:\[\hat{p} \pm z \times SE,\]we compute:\[0.167 \pm 1.645 \times SE.\]The result provides an interval, suggesting that the true percentage of full-time workers who felt like hitting a coworker is within this range with 90% confidence. This method allows researchers to generalize findings from a sample to a broader population effectively.

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Most popular questions from this chapter

The study 鈥淒igital Footprints" (Pew Internet \& American Life Project, www.pewinternet.org, 2007) reported that \(47 \%\) of Internet users have searched for information about themselves online. The \(47 \%\) figure was based on a random sample of Internet users. For purposes of this exercise, suppose that the sample size was \(n=300\) (the actual sample size was much larger). Construct and interpret a \(90 \%\) confidence interval for the proportion of Internet users who have searched online for information about themselves.

The authors of the paper "Deception and Design: The Impact of Communication Technology on Lying Behavior" (Proceedings of Computer Human Interaction [2004]) asked 30 students in an upper division communications course at a large university to keep a journal for 7 days, recording each social interaction and whether or not they told any lies during that interaction. A lie was defined as "any time you intentionally try to mislead someone." The paper reported that the mean number of lies per day for the 30 students was \(1.58\) and the standard deviation of number of lies per day was \(1.02 .\) a. What assumption must be made in order for the \(t\) confidence interval of this section to be an appropriate method for estimating \(\mu\), the mean number of lies per day for all students at this university? b. Would you recommend using the \(t\) confidence interval to construct an estimate of \(\mu\) as defined in Part (a)? Explain why or why not.

A random sample of 10 houses heated with natural gas in a particular area, is selected, and the amount of gas (in therms) used during the month of January is determined for each house. The resulting observations are as follows: \(\begin{array}{llllllllll}103 & 156 & 118 & 89 & 125 & 147 & 122 & 109 & 138 & 99\end{array}\) a. Let \(\mu_{J}\) denote the average gas usage during January by all houses in this area. Compute a point estimate of \(\mu_{J}\) b. Suppose that 10,000 houses in this area use natural gas for heating. Let \(\tau\) denote the total amount of gas used by all of these houses during January. Estimate \(\tau\) using the given data. What statistic did you use in computing your estimate? c. Use the data in Part (a) to estimate \(p\), the proportion of all houses that used at least 100 therms. d. Give a point estimate of the population median us- age based on the sample of Part (a). Which statistic did you use?

The eating habits of 12 bats were examined in the article "Foraging Behavior of the Indian False Vampire Bat" (Biotropica [1991]: 63-67). These bats consume insects and frogs. For these 12 bats, the mean time to consume a frog was \(\bar{x}=21.9\) minutes. Suppose that the standard deviation was \(s=7.7\) minutes. Construct and interpret a \(90 \%\) confidence interval for the mean suppertime of a vampire bat whose meal consists of a frog. What assumptions must be reasonable for the one-sample \(t\) interval to be appropriate?

USA Today (October 14, 2002) reported that \(36 \%\) of adult drivers admit that they often or sometimes talk on a cell phone when driving. This estimate was based on data from a sample of 1004 adult drivers, and a bound on the error of estimation of \(3.1 \%\) was reported. Assuming a \(95 \%\) confidence level, do you agree with the reported bound on the error? Explain.
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