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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

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.

The fraternities and sororities at St. Algebra College have been plagued by declining membership over the past several years and want to know whether the incoming freshman class will be a fertile recruiting ground. Not having enough money to survey all 1600 freshmen, they commission you to survey the interests of a random sample. You find that 35 of your 150 respondents are "extremely" interested in social clubs. At the \(95 \%\) level, what is your estimate of the number of freshmen who would be extremely interested? (HINT: The high and low values of your final confidence interval are proportions. How can proportions be expressed as numbers?)

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