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Use the information given in Exercises 9-15 to find the necessary confidence interval for the population mean \(\mu .\) Interpret the interval that you have constructed. \(\alpha=.01, n=38, \bar{x}=34, s^{2}=12\)

Short Answer

Expert verified
Answer: The 99% confidence interval for the population mean is (32.47, 35.53).

Step by step solution

01

Calculate the sample standard deviation (\(s\))

To find the sample standard deviation (\(s\)), take the square root of the sample variance \(s^2\): \(s = \sqrt{12} \approx 3.46\)
02

Determine the degrees of freedom

The degrees of freedom for a t-distribution are given by \(n-1\): df = \(38 - 1 = 37\)
03

Find the critical t-value

Using a t-table or calculator, look up the critical t-value for the given alpha (\(\alpha\)) and degrees of freedom (df). We have \(\alpha = .01\), which corresponds to a \(99\%\) confidence level, and df = \(37\). The critical t-value is approximately \(2.704\).
04

Compute the margin of error

The margin of error is the product of the critical t-value, the sample standard deviation, and the square root of the inverse of the sample size: Margin of error = \(2.704 \times 3.46 \times \sqrt{\frac{1}{38}} \approx 1.53\)
05

Construct the confidence interval

The confidence interval is found by adding and subtracting the margin of error from the sample mean: Lower bound = \(\bar{x} - \text{Margin of error} = 34 - 1.53 \approx 32.47\) Upper bound = \(\bar{x} + \text{Margin of error} = 34 + 1.53 \approx 35.53\) Thus, the confidence interval for the population mean μ is \((32.47, 35.53)\).
06

Interpretation

The confidence interval \((32.47, 35.53)\) means that we are 99% confident that the true population mean \(\mu\) lies between \(32.47\) and \(35.53\).

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Key Concepts

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

Population Mean
Understanding the concept of population mean is crucial in statistics as it represents the average value in a population. A population is the entire collection of elements or outcomes that you're interested in studying. In many situations, it is impractical or impossible to measure every individual in the population, so we use a sample - a smaller, manageable number of observations - to make inferences about the population mean, denoted by \(\mu\).

The sample mean, denoted by \(\bar{x}\), is used as an estimate of the true population mean. However, because it's derived from a sample, it's subject to sampling error, and it may not exactly match the true mean. That's where confidence intervals come in, as they provide a range of values within which the true population mean is likely to fall with a certain level of confidence.
t-Distribution
The t-distribution, also known as Student's t-distribution, is a type of probability distribution that is symmetric and bell-shaped, like the normal distribution, but has heavier tails. These heavier tails mean there is a greater probability for extreme values, which compensates for the additional variability and uncertainty when estimating the population mean with a small sample size.

The t-distribution is used instead of the normal distribution when the sample size is small (commonly n<30) and/or when the population standard deviation is unknown. As the sample size increases, the t-distribution approaches the normal distribution. This distribution is pivotal when it comes to calculating confidence intervals for the population mean when using small samples. The t-value, which varies based on the sample size through degrees of freedom (df), allows us to determine how far our sample statistic lies from the population parameter, adjusting for sample size.
Sample Standard Deviation
The sample standard deviation (denoted as \(s\)) is a measure of the amount of variation or dispersion of a set of values. A low \(s\) indicates that the data points tend to be close to the sample mean (\bar{x}) and hence to the population mean (\(\mu\)). Conversely, a high sample standard deviation signifies that the data points are spread out over a wider range of values.

It is important to note that when calculating the confidence interval for the population mean, we use the sample standard deviation as an estimate of the true population standard deviation. This is because, in many cases, the actual population standard deviation is unknown. The accuracy and reliability of the confidence interval heavily depend on the accuracy of the sample standard deviation. When applying this measure in the steps to find a confidence interval, it is squared to find the variance (\(s^2\)) before taking the square root to return to standard deviation, ensuring the units match those of the mean.

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Most popular questions from this chapter

A survey is designed to estimate the proportion of sports utility vehicles (SUVs) being driven in the state of California. A random sample of 500 registrations are selected from a Department of Motor Vehicles database, and 68 are classified as SUVs. a. Use a \(95 \%\) confidence interval to estimate the proportion of SUVs in California. b. How can you estimate the proportion of SUVs in California with a higher degree of accuracy? (HINT: There are two answers.)

What \(i s\) normal, when it comes to people's body temperatures? A random sample of 130 human body temperatures, provided by Allen Shoemaker \(^{11}\) in the Journal of Statistical Education, had a mean of \(98.25^{\circ}\) Fahrenheit and a standard deviation of \(0.73^{\circ}\) Fahrenheit. a. Construct a \(99 \%\) confidence interval for the average body temperature of healthy people. b. Does the confidence interval constructed in part a contain the value \(98.6^{\circ}\) Fahrenheit, the usual average temperature cited by physicians and others? If not, what conclusions can you draw?

Auto Accidents A recent year's records of auto accidents occurring on a given section of highway were classified according to whether the resulting damage was \(\$ 3000\) or more and to whether a physical injury resulted from the accident. The data follows: $$\begin{array}{lcc}\hline & \text { Under } \$ 3000 & \$ 3000 \text { or More } \\\\\hline \text { Number of Accidents } & 32 & 41 \\\\\text { Number Involving Injuries } & 10 & 23\end{array}$$ a. Estimate the true proportion of accidents involving injuries when the damage was \(\$ 3000\) or more for similar sections of highway and find the margin of error. b. Estimate the true difference in the proportion of accidents involving injuries for accidents with damage under \(\$ 3000\) and those with damage of \(\$ 3000\) or more. Use a \(95 \%\) confidence interval.

Independent random samples were selected from two binomial populations, with sample sizes and the number of successes given in Exercises \(1-2 .\) Construct a \(95 \%\) and a \(99 \%\) confidence interval for the difference in the population proportions. What does the phrase "95\% confident" or "99\% confident" mean? $$\begin{array}{lcc}\hline & \multicolumn{2}{c} {\text { Population }} \\\\\cline { 2 - 3 } & 1 & 2 \\\\\hline \text { Sample Size } & 500 & 500 \\\\\text { Number of Successes } & 120 & 147 \\\\\hline\end{array}$$

A pediatrician randomly selected 50 six-month-old boys from her practice's database and recorded an average weight of 8.0 kilograms with a standard deviation of 0.30 kilogram. She also recorded an average length of 67.3 centimeters with a standard deviation of 0.64 centimeter. a. Find a \(95 \%\) confidence interval for the average weight of all six-month- old boys. b. Find a \(99 \%\) confidence interval for the average length of all six-month- old boys. c. What do you have to assume about the pediatrician's database in order to make inferences about all sixmonth-old boys?

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