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Chapter 9: Problem 13

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 is (51.44, 56.56) months.

Step by step solution

01

- Identify the Given Information

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

- Find the Critical Value

Since the sample size is less than 30, use the t-distribution. For a 95% confidence level with 39 degrees of freedom ( = n - 1), look up the critical value (t_{crit}), which is approximately 2.022.
03

- Calculate the Standard Error

The standard error of the mean (SE) is calculated as follows: SE = s / N . Therefore, SE = 8 / 40 鈮� 1.265.
04

- Determine the Margin of Error

The margin of error (ME) is calculated by multiplying the t-critical value by the standard error: ME = t_{crit} SE 鈮� 2.022 1.265 鈮� 2.56.
05

- Construct the Confidence Interval

Add and subtract the margin of error from the sample mean to construct the confidence interval: CI = ( - ME, + ME) 鈮� (54 - 2.56, 54 + 2.56) 鈮� (51.44, 56.56).
06

- Interpret the Confidence Interval

With 95% confidence, the true mean length of sentence for aggravated assault convictions falls between 51.44 months and 56.56 months.

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Key Concepts

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

t-distribution
The t-distribution is an essential concept when dealing with small sample sizes, usually less than 30. Unlike the normal distribution, which is symmetrical and bell-shaped, the t-distribution is slightly broader with heavier tails. This characteristic accounts for additional uncertainty with smaller samples. When sample sizes are large, the t-distribution closely resembles the normal distribution. In our exercise, because the sample size is 40, we use the t-distribution to find the critical value for constructing the confidence interval. This ensures our interval accurately reflects the increased variability observed in smaller samples.
sample mean
The sample mean is a crucial step in estimating the population mean. It is the average of all sampled data points. The sample mean serves as our best estimate for the population mean, albeit with some uncertainty. In our case, the sample mean length of sentencing for 40 felons is 54 months. While this value gives us an idea, it is just an estimate, and there's variability involved. This is why we construct confidence intervals 鈥� to provide a range that likely contains the true population mean.
standard error
Standard error (SE) measures the variability of the sample mean from the population mean. It decreases with larger sample sizes and gives us an idea of how precise our sample mean is as an estimate of the population mean. The formula for calculating SE is s divided by the square root of the sample size (N), where s is the sample standard deviation. For our felon sentence length example, SE = 8 / 鈭�40 鈮� 1.265. This slight value measures the dispersion of our sample mean from the actual population mean.
margin of error
The margin of error (ME) provides a range around the sample mean that is likely to contain the population mean. It is calculated by multiplying the critical value from the t-distribution by the standard error: ME = t_{crit} * SE. For a 95% confidence level with 39 degrees of freedom in our example, the critical value (t_{crit}) is approximately 2.022. Thus, ME 鈮� 2.022 * 1.265 鈮� 2.56. This margin of error is then used to extend on both sides of the sample mean to create the confidence interval.
critical value
The critical value serves as a multiplier in confidence interval calculations. It derives from the t-distribution table based on the chosen confidence level and degrees of freedom鈥攊n our case, 95% confidence and 39 degrees of freedom. The critical value for these parameters is about 2.022. A higher confidence level would yield a larger critical value, broadening the confidence interval. In using the t-distribution, the critical value adjusts for the additional uncertainty due to smaller sample sizes, ensuring a more accurate confidence interval.

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Most popular questions from this chapter

A simple random sample of size \(n<30\) for \(a\) quantitative variable has been obtained. Using the normal probability plot, the correlation between the variable and expected z-score, and the boxplot, judge whether a t-interval should be constructed. $$ n=13 ; \text { Correlation }=0.966 $$

Explain why the \(t\) -distribution has less spread as the number of degrees of freedom increases.

Jane wants to estimate the proportion of students on her campus who eat cauliflower. After surveying 20 students, she finds 2 who eat cauliflower. Obtain and interpret a \(95 \%\) confidence interval for the proportion of students who eat cauliflower on Jane's campus using Agresti and Coull's method.

How much time do Americans spend eating or drinking? Suppose for a random sample of 1001 Americans age 15 or older, the mean amount of time spent eating or drinking per day is 1.22 hours with a standard deviation of 0.65 hour. Source: American Time Use Survey conducted by the Bureau of Labor Statistics (a) A histogram of time spent eating and drinking each day is skewed right. Use this result to explain why a large sample size is needed to construct a confidence interval for the mean time spent eating and drinking each day. (b) There are over 200 million Americans age 15 or older. Explain why this, along with the fact that the data were obtained using a random sample, satisfies the requirements for constructing a confidence interval. (c) Determine and interpret a \(95 \%\) confidence interval for the mean amount of time Americans age 15 or older spend eating and drinking each day. (d) Could the interval be used to estimate the mean amount of time a 9-year- old American spends eating and drinking each day? Explain.

Gallup polled 982 likely voters immediately preceding the 2014 North Carolina senate race. The results of the survey indicated that incumbent Kay Hagan had the support of \(47 \%\) of respondents, while challenger Thom Tillis had support of \(46 \% .\) The poll's margin of error was \(3 \% .\) Gallup suggested the race was too close to call. Use the concept of a confidence interval to explain what this means.
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