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Chapter 6: Problem 6

A random sample of 100 cable TV movies contained an average of \(3.38\) acts of physical violence per program with a standard deviation of \(0.30 .\) At the \(99 \%\) level, what is your estimate of the population value?

Short Answer

Expert verified
The 99% confidence interval is from 3.30272 to 3.45728.

Step by step solution

01

- Identify all given values

We are given the sample mean \(\bar{x} = 3.38\), the sample standard deviation \(s = 0.30\), the sample size \(n = 100\), and the confidence level (99%).
02

- Find the critical value

For a 99% confidence level, the critical value \(z\) can be found using standard normal distribution tables. The critical value for 99% confidence level is approximately 2.576.
03

- Calculate the standard error

The standard error (SE) of the mean is given by the formula \(\text{SE} = \frac{s}{\root n}\). Substituting the given values: \(\text{SE} = \frac{0.30}{\root{100}} = 0.03\).
04

- Compute the margin of error

The margin of error (ME) is calculated by multiplying the critical value \(z\) by the standard error \(\text{SE}\). Thus, \(\text{ME} = z \times \text{SE} = 2.576 \times 0.03 = 0.07728\).
05

- Determine the confidence interval

Add and subtract the margin of error from the sample mean to find the confidence interval. Lower limit = \(\bar{x} - \text{ME} = 3.38 - 0.07728 = 3.30272\), Upper limit = \(\bar{x} + \text{ME} = 3.38 + 0.07728 = 3.45728\).

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Key Concepts

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

Sample Mean
The sample mean, often denoted as \(\bar{x}\), is the average of a set of observations from a sample. It's a crucial statistic because it gives us a central value of the data we're examining. In this exercise, the sample mean is 3.38 acts of physical violence per TV movie. This value is calculated by summing up all acts of violence in all sample movies and then dividing by the number of movies (100 in this case). The sample mean provides a point estimate of the population mean.
Standard Error
The standard error (SE) measures the spread of the sample mean from the population mean. It is calculated using the sample standard deviation (\text{s}) and the sample size (\text{n}). The formula for SE is \( SE = \frac{s}{\root n} \). In our example, we have \( s = 0.30 \) and \( n = 100 \), so \( SE = \frac{0.30}{\root{100}} = 0.03 \). The smaller the SE, the more precise our estimate of the population mean will be.
Critical Value
The critical value is a factor used to compute the margin of error and depends on the desired confidence level. For a 99% confidence level, we utilize the z-score from the standard normal distribution, which is approximately 2.576. This value marks the boundary within which the true population mean lies 99% of the time when the sample is repeatedly taken from the population.
Margin of Error
The margin of error (ME) quantifies the range within which the population parameter (mean, proportion, etc.) is likely to fall. It combines the critical value and the standard error through the formula \( ME = z \times SE \). For this exercise, \( ME = 2.576 \times 0.03 = 0.07728 \). The margin of error provides the extent of unknown errors due to sampling variability. It is added and subtracted from the sample mean to create the confidence interval.
99% Confidence Level
Confidence levels indicate how confident we are that our interval estimate will contain the true population parameter. A 99% confidence level means that if we were to take 100 different samples and compute a confidence interval for each sample, we would expect approximately 99 of those intervals to contain the population mean. This high confidence level results in a broader confidence interval to ensure the true mean is captured. For this problem, we calculated a 99% confidence interval as \(3.30272 \leq \mu \leq 3.45728 \). This interval suggests that the true average acts of physical violence per TV movie in the population is between 3.30272 and 3.45728 with 99% confidence.

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Most popular questions from this chapter

A random sample of 100 patients treated in a program for alcoholism and drug dependency over the past 10 years was selected. It was determined that 53 of the patients had been readmitted to the program at least once. At the \(95 \%\) level, construct an estimate of the population proportion.

For each of the following three sample sizes, construct the \(95 \%\) confidence interval for the population proportion. Use a sample proportion of \(0.40\) throughout. What happens to interval width as sample size increases? Why? $$ \begin{aligned} P_{s} &=0.40 \\ \text { Sample A: } N &=100 \\ \text { Sample B: } N &=1000 \\ \text { Sample C: } N &=10,000 \end{aligned} $$

A random sample of 260 workers in a high-rise office building revealed that \(30 \%\) were very satisfied with the quality of elevator service. At the \(99 \%\) level, what is your estimate of the population value?

Two individuals are running for mayor of your town. You conduct an election survey a week before the election and find that \(51 \%\) of the respondents prefer candidate A. Can you predict a winner? Use the \(99 \%\) level. (HINT: In a two-candidate race, what percentage of the vote would the winner need? Does the confidence interval indicate that candidate \(A\) has a sure margin of victory? Remember that while the population parameter is probably \([\alpha=0.01]\) in the confidence interval, it may be anywhere in the interval.) $$ \begin{aligned} &P_{s}=0.51 \\ &N=578 \end{aligned} $$

You have developed a series of questions to measure job satisfaction of bus drivers in New York City. A random sample of 100 drivers has an average score of \(10.6\), with a standard deviation of \(2.8\). What is your estimate of the average job satisfaction score for the population of all NYC bus drivers? Use the \(95 \%\) confidence level.
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