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Chapter 9: Problem 14

In a random sample of 40 felons convicted of aggravated assault, it was determined that the mean length of sentencing was 54 months, with a standard deviation of 8 months. Construct and interpret a \(95 \%\) confidence interval for the mean length of sentence for an aggravated assault conviction.

Short Answer

Expert verified
The 95% confidence interval for the mean sentence length is approximately (51.52, 56.48) months.

Step by step solution

01

Identify the Given Information

Extract the given values from the problem: the sample size (=40), the sample mean (=54 months), and the sample standard deviation (=8 months).
02

Determine the Confidence Level

Identify the confidence level, which is 95%. This implies we are using a z-score corresponding to 95% confidence, which is 1.96.
03

Calculate the Standard Error

The formula for the standard error (SE) is given by = \(\frac{蟽}{\sqrt{n}}\). Plugging the known values, we get \(\frac{8}{\sqrt{40}}\). Simplify this to find the SE.
04

Apply the Confidence Interval Formula

Use the confidence interval formula: = \(\overline{x} \pm Z* \cdot SE\). Here, \(\overline{x}\) is the sample mean, Z* is the z-score (1.96 for 95% confidence), and SE is the standard error calculated in the previous step.
05

Calculate the Margin of Error

Use the formula for the margin of error: \(Z* \cdot SE = 1.96 \cdot 1.2649 鈮� 2.4780\).
06

Find the Confidence Interval

Calculate the interval: \(54 \pm 2.4780\), yielding (54 - 2.4780, 54 + 2.4780) 鈮� (51.522, 56.478).
07

Interpret the Confidence Interval

We can say with 95% confidence that the true mean sentence length for aggravated assault convictions lies between approximately 51.52 and 56.48 months.

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Key Concepts

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

Confidence Level
First things first, let's talk about the confidence level. This is a key concept when constructing a confidence interval. The confidence level tells you how sure you can be about the results of your sample. For instance, a 95% confidence level means that if you were to take 100 different samples, the true population mean would lie within the calculated confidence interval in 95 of those samples. In our exercise example, a 95% confidence level was used, represented by a z-score of 1.96. This level of certainty is often used in statistics to balance accuracy and precision.
Sample Mean
The sample mean is another fundamental concept. It refers to the average value calculated from a sample of data. Think of it as the central point of the sample data. In this exercise, the sample mean sentence length for aggravated assault convictions was 54 months. This mean is crucial because it serves as the starting point for constructing the confidence interval. Since it represents the average based on the sample, it provides a useful estimate of the population mean.
Standard Error
Next up, the standard error (SE). This measures the variability of the sample mean. It's crucial for understanding how much the sample mean is expected to fluctuate due to random sampling. The smaller the standard error, the more representative the sample mean is of the population mean. The formula for SE is \(\frac{蟽}{\text{sqrt}(n)}\), where 蟽 is the sample standard deviation, and n is the sample size. For our exercise, the standard deviation is 8 months, and the sample size is 40. Plugging these values in, we get\(\frac{8}{\text{sqrt}(40)}\), which results in the SE. This value helps to compute the margin of error and, eventually, the confidence interval.
Margin of Error
Finally, let's talk about the margin of error. This value gives you an idea of how much the sample estimate (sample mean) can deviate from the true population mean. It's calculated using the z-score and the standard error. The formula for the margin of error is \(\text{Z*} \times SE\), which in this case translates to \1.96 \times 1.2649 \. The result? Approximately 2.4780. This margin of error is then used to create the confidence interval, which gives us a range where the true mean is likely to be. It essentially tells us how 'off' our sample mean could be from the actual mean due to sample variability.

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Most popular questions from this chapter

In March 2014, Harris Interactive conducted a poll of a random sample of 2234 adult Americans 18 years of age or older and asked, "Which is more annoying to you, tailgaters or slow drivers who stay in the passing lane?" Among those surveyed, 1184 were more annoyed by tailgaters. (a) Explain why the variable of interest is qualitative with two possible outcomes. What are the two outcomes? (b) Verify the requirements for constructing a \(90 \%\) confidence interval for the population proportion of all adult Americans who are more annoyed by tailgaters than slow drivers in the passing lane. (c) Construct a \(90 \%\) confidence interval for the population proportion of all adult Americans who are more annoyed by tailgaters than slow drivers in the passing lane.

Construct the appropriate confidence interval. A simple random sample of size \(n=40\) is drawn from a population. The sample mean is found to be \(\bar{x}=120.5,\) and the sample standard deviation is found to be \(s=12.9\). Construct a \(99 \%\) confidence interval for the population mean.

The American Society for Microbiology (ASM) and the Soap and Detergent Association (SDA) jointly commissioned two separate studies, both of which were conducted by Harris Interactive. In one of the studies, 1001 adults were interviewed by telephone and asked about their handwashing habits. In the telephone interviews, 921 of the adults said they always wash their hands in public restrooms. In the other study, the hand-washing behavior of 6076 adults was inconspicuously observed within public restrooms in four U.S. cities and 4679 of the 6076 adults were observed washing their hands. (a) In the telephone survey, what is the variable of interest? Is it qualitative or quantitative? (b) What is the sample in the telephone survey? What is the population to which this study applies? (c) Verify that the requirements for constructing a confidence interval for the population proportion of adults who say they always wash their hands in public restrooms are satisfied. (d) Using the results from the telephone interviews, construct a \(95 \%\) confidence interval for the proportion of adults who say they always wash their hands in public restrooms. (e) In the study where hand-washing behavior was observed, what is the variable of interest? Is it qualitative or quantitative? (f) We are told that 6076 adults were inconspicuously observed, but were not told how these adults were selected. We know randomness is a key ingredient in statistical studies that allows us to generalize results from a sample to a population. Suggest some ways randomness might have been used to select the individuals in this study. (g) Verify the requirements for constructing a confidence interval for the population proportion of adults who actually washed their hands while in a public restroom. (h) Using the results from the observational study, construct a \(95 \%\) confidence interval for the proportion of adults who wash their hands in public restrooms. (i) Based on your findings in parts (a) through (h), what might you conclude about the proportion of adults who say they always wash their hands versus the proportion of adults who actually wash their hands in public restrooms? (j) Cite some sources of variability in both studies.

Construct the appropriate confidence interval. A simple random sample of size \(n=12\) is drawn from a population that is normally distributed. The sample mean is found to be \(\bar{x}=45,\) and the sample standard deviation is found to be \(s=14\). Construct a \(90 \%\) confidence interval for the population mean.

The Sullivan Statistics Survey I asks, "Would you be willing to pay higher taxes if the tax revenue went directly toward deficit reduction?" Treat the survey respondents as a random sample of adult Americans. Go to www.pearsonhighered.com/sullivanstats to obtain the data file SullivanSurveyI using the file format of your choice for the version of the text you are using. The column "Deficit" has survey responses. Construct and interpret a \(90 \%\) confidence interval for the proportion of adult Americans who would be willing to pay higher taxes if the revenue went directly toward deficit reduction.
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