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Chapter 9: Problem 16

Fifty rounds of a new type of ammunition were fired from a test weapon, and the muzzle velocity of the projectile was measured. The sample had a mean muzzle velocity of 863 meters per second and a standard deviation of 2.7 meters per second. Construct and interpret a \(99 \%\) confidence interval for the mean muzzle velocity.

Short Answer

Expert verified
The 99% confidence interval for the mean muzzle velocity is approximately \(861.98\) to \(864.02\) meters per second.

Step by step solution

01

- Identify the sample statistics

Determine the sample mean \(\bar{x}=863\) meters per second and the standard deviation \(s=2.7\) meters per second. The sample size is \(n=50\).
02

- Determine the critical value

For a 99% confidence interval, the critical value \(t^*\) from the t-distribution table based on \(n-1=50-1=49\) degrees of freedom is approximately \(2.68\).
03

- Calculate the standard error

The standard error (SE) is calculated using \(SE = \frac{s}{\sqrt{n}} = \frac{2.7}{\sqrt{50}} \approx 0.3816\).
04

- Compute the margin of error

The margin of error (ME) is obtained by multiplying the critical value by the standard error: \(ME = t^* \times SE = 2.68 \times 0.3816 \approx 1.0227\).
05

- Construct the confidence interval

The confidence interval is \(\bar{x} \pm ME = 863 \pm 1.0227\). Thus, the 99% confidence interval for the mean muzzle velocity is approximately \(861.98\) to \(864.02\) meters per second.
06

- Interpret the confidence interval

With 99% confidence, it can be stated that the true mean muzzle velocity of the projectile lies between approximately \(861.98\) and \(864.02\) meters per second.

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Key Concepts

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

sample statistics
Sample statistics are valuable when analyzing data from experiments or surveys. They provide insights into the sample being studied and help make inferences about the larger population.
An essential sample statistic is the mean (\(\bar{x}\)), calculated as the sum of all sample values divided by the number of values. For the ammunition example, the mean muzzle velocity is \(863\) meters per second.
The standard deviation (\(s\)) indicates the spread of sample values around the mean. In our case, it is \(2.7\) meters per second. Sample size (\(n\)) is also vital, representing the number of observations. Here, we have \(n=50\).

critical value
The critical value is a threshold value that helps determine the margin of error in a confidence interval. It comes from specific statistical distributions, such as the normal or t-distribution, depending on the sample size and whether the population standard deviation is known.

For a 99% confidence interval, we use the t-distribution because our sample size is relatively small (\(n=50\)). The critical value (\(t^*\)) depends on the degrees of freedom (\(df\)), calculated as \(df = n - 1\). Here, \(df = 50 - 1 = 49\).
Using a t-distribution table, the critical value for 49 degrees of freedom at a 99% confidence level is approximately \(2.68\).

standard error
The standard error (SE) measures the accuracy of a sample mean by indicating how much it is likely to vary from the true population mean.
Standard error is calculated by dividing the sample standard deviation (\(s\)) by the square root of the sample size (\(\)), represented as:
\(SE = \frac{s}{\sqrt{n}} \)

In our example, the SE is calculated as:
\(SE = \frac{2.7}{\sqrt{50}} \approx 0.3816 \)
Smaller standard errors imply more precise sample means. Here, \(0.3816\) is the SE, indicating the sample mean's variability around the true population mean.

margin of error
The margin of error (ME) reflects the range within which the true population mean is expected to lie, given the sample data. It is calculated by multiplying the critical value (\(t^*\)) by the standard error (SE).
The formula is:
\(ME = t^* \times SE \)

Using our data, the margin of error is calculated as:
\(ME = 2.68 \times 0.3816 \approx 1.0227 \)
This margin of error indicates that our sample mean muzzle velocity can vary by \(\pm1.0227\) meters per second from the true mean.

t-distribution
The t-distribution is a probability distribution used when the sample size is small and the population standard deviation is unknown. It is similar to the normal distribution but has thicker tails, which account for additional variability in small samples.

The shape of the t-distribution changes with degrees of freedom (\(df\)), calculated as the sample size minus one (\(df = n - 1\)). For our example, \(df = 49\). As the sample size increases, the t-distribution approaches the normal distribution.In the context of confidence intervals, the t-distribution provides the critical value (\(t^*\)) for calculating the margin of error and making more accurate inferences about the population mean.

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Most popular questions from this chapter

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.

In a Gallup poll, \(64 \%\) of the people polled answered yes to the following question: "Are you in favor of the death penalty for a person convicted of murder?" The margin of error in the poll was \(3 \%,\) and the estimate was made with \(95 \%\) confidence. At least how many people were surveyed?

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.

Alan wants to estimate the proportion of adults who walk to work. In a survey of 10 adults, he finds 1 who walk to work. Explain why a \(95 \%\) confidence interval using the normal model yields silly results. Then compute and interpret a \(95 \%\) confidence interval for the proportion of adults who walk to work using Agresti and Coull's method.

A Bernoulli random variable is a variable that is either 0 (a failure) or 1 (a success). The probability of success is denoted \(p\). (a) Use a statistical spreadsheet to generate 1000 Bernoull samples of size \(n=20\) with \(p=0.15\) (b) Estimate the population proportion for each of the 1000 Bernoulli samples. (c) Draw a histogram of the 1000 proportions from part (b). What is the shape of the histogram? (d) Construct a \(95 \%\) confidence interval for each of the 1000 Bernoulli samples using the normal model. (e) What proportion of the intervals do you expect to include the population proportion, \(p ?\) What proportion of the intervals actually captures the population proportion? Explain any differences.
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