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Chapter 12: Problem 35

The formula used to calculate a confidence interval for the mean of a normal population when \(n\) is small is $$ \bar{x} \pm(t \text { critical value }) \frac{s}{\sqrt{n}} $$ What is the appropriate \(t\) critical value for each of the following confidence levels and sample sizes? a. \(95 \%\) confidence, \(n=15\) b. \(99 \%\) confidence, \(n=20\) c. \(90 \%\) confidence, \(n=26\)

Short Answer

Expert verified
For the given confidence levels and sample sizes, the appropriate t critical values are: a. \(95\%\) confidence, \(n=15\): \(t=2.145\) b. \(99\%\) confidence, \(n=20\): \(t=2.861\) c. \(90\%\) confidence, \(n=26\): \(t=1.708\)

Step by step solution

01

Understanding the formula

The formula given calculates the confidence interval for the mean of a normal population when the sample size \(n\) is small: \[ \bar{x} \pm (t \text { critical value }) \frac{s}{\sqrt{n}} \] Here, \(\bar{x}\) is the sample mean, \(s\) is the sample standard deviation, and \(n\) is the sample size. The t critical value depends on the desired confidence level and the degrees of freedom, which is given by \(n-1\).
02

Using a t-distribution table

To find the appropriate t critical values for each of the given confidence levels and sample sizes, we will use a t-distribution table. We need to look up the t critical value based on the desired confidence level (in the column headings) and the degrees of freedom (in the row headings). a. \(95\%\) confidence, \(n=15\) Degrees of freedom: \(n - 1 = 15 - 1 = 14\) Using a t-distribution table, the corresponding t critical value is: \(t_{14, 0.975} = 2.145\) b. \(99\%\) confidence, \(n=20\) Degrees of freedom: \(n - 1 = 20 - 1 = 19\) Using a t-distribution table, the corresponding t critical value is: \(t_{19, 0.995} = 2.861\) c. \(90\%\) confidence, \(n=26\) Degrees of freedom: \(n - 1 = 26 - 1 = 25\) Using a t-distribution table, the corresponding t critical value is: \(t_{25, 0.95} = 1.708\)
03

Writing the final answers

a. For a \(95\%\) confidence level and \(n=15\), the appropriate t critical value is \(t=2.145\). b. For a \(99\%\) confidence level and \(n=20\), the appropriate t critical value is \(t=2.861\). c. For a \(90\%\) confidence level and \(n=26\), the appropriate t critical value is \(t=1.708\).

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Key Concepts

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

Understanding the T-Distribution
The t-distribution is an important concept when dealing with small sample sizes in statistics. Unlike the normal distribution, the t-distribution adjusts for the added uncertainty associated with estimating the population standard deviation from a small sample.
This distribution is symmetric and bell-shaped, like the normal distribution, but has heavier tails. Those heavier tails are what allow for better estimates when dealing with smaller samples.
  • The t-distribution is primarily used in hypothesis testing and in calculating confidence intervals.
  • It is especially useful in cases where the sample size is less than 30, or when the population standard deviation is unknown.
  • As the sample size increases, the t-distribution begins to resemble the normal distribution more closely.
To choose the correct t-distribution, we rely on degrees of freedom, which makes it flexible for different sample sizes and confidence levels.
Determining Appropriate Sample Sizes
Sample size, noted as \(n\), plays a critical role in statistical analysis since it affects the reliability of our estimates. Smaller samples have less representation of the population, therefore increasing variability.
When calculating confidence intervals, sample size affects the width of the interval; smaller samples typically result in wider confidence intervals, reflecting the higher uncertainty in our estimations.
  • In the t-distribution formula \(\bar{x} \pm (t \text{ critical value}) \frac{s}{\sqrt{n}}\), \(n\) is crucial in determining the margin of error \((ME)\), which influences the interval width.
  • The larger the sample size, the more the t-distribution approaches a normal distribution and the less impact the sample size has on the margin of error.
  • In practical terms, selecting an appropriate sample size is often a balance between resource limitations and desired confidence in results.
Understanding the impact of sample size helps in interpreting statistical results and making informed decisions based on statistical analysis.
Role of Degrees of Freedom in a Sample
Degrees of freedom (df) are a concept used in statistics that refer to the number of independent values or quantities that can vary in an analysis, while still adhering to certain constraints. When creating confidence intervals with the t-distribution, degrees of freedom are calculated as \(n-1\), where \(n\) is the sample size.
Degrees of freedom impact the shape of the t-distribution:
  • As degrees of freedom increase, the t-distribution curve becomes closer to a standard normal distribution curve.
  • A lower number of degrees of freedom results in a wider and flatter curve, indicating more variability and uncertainty.
  • The degrees of freedom adjust the t critical value, thereby influencing the resulting confidence interval's precision.
This adjustment ensures that the confidence intervals used are suitable for the specific data involved, maintaining their accuracy across different sample sizes.

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Most popular questions from this chapter

A sign in the elevator of a college library indicates a limit of 16 persons. In addition, there is a weight limit of 2500 pounds. Assume that the average weight of students, faculty, and staff at this college is 150 pounds, that the standard deviation is 27 pounds, and that the distribution of weights of individuals on campus is approximately normal. A random sample of 16 persons from the campus will be selected. a. What is the mean of the sampling distribution of \(\bar{x} ?\) b. What is the standard deviation of the sampling distribution of \(\bar{x} ?\) c. What average weights for a sample of 16 people will result in the total weight exceeding the weight limit of 2500 pounds? d. What is the probability that a random sample of 16 people will exceed the weight limit?

The dodo was a species of flightless bird that lived on the island of Mauritius in the Indian Ocean. The first record of human interaction with the dodo occurred in 1598 , and within 100 years the dodo was extinct due to hunting by humans and other newly introduced invasive species. After the extinction, the word "dodo" became synonymous with stupidity, implying that the birds lacked the intelligence to avoid or escape extinction. The closest existing relatives of the dodo are pigeons and doves. Researchers at the American Museum of Natural History used computed tomography (CT) scans to measure the brain size ("endocranial capacity") of one of the few existing preserved dodo birds, and to measure the brain sizes in samples of eight birds that are close relatives of dodos ("The First Endocast of the Extinct Dodo and an Anatomical Comparison Amongst Close Relatives," Zoological Journal of the Linnean Society [2016]: 950-953) The brain size for the dodo was \(4.17 \log \mathrm{mm}^{3}\). The following table contains the brain sizes for the sample of birds from related species (approximate values from a graph in the paper). a. Use the output at the bottom of the page from the Shiny App "Randomization Test for One Mean" to help you to carry out a randomization test of the hypothesis that the population mean brain size for birds that are relatives of the dodo differs from the established dodo brain size of 4.17 . b. What does the result of your test indicate about the brain size of the dodo?

An automobile manufacturer decides to carry out a fuel efficiency test to determine if it can advertise that one of its models achieves \(30 \mathrm{mpg}\) (miles per gallon). Six people each drive a car from Phoenix to Los Angeles. The resulting fuel efficiencies (in miles per gallon) are: \(\begin{array}{ll}30.3 & 29.6\end{array}\) \(\begin{array}{llll}27.2 & 29.3 & 31.2 & 28.4\end{array}\) Assuming that fuel efficiency is normally distributed, do these data provide evidence against the claim that actual mean fuel efficiency for this model is (at least) \(30 \mathrm{mpg}\) ?

Suppose that a random sample of size 100 is to be drawn from a population with standard deviation 10 . a. What is the probability that the sample mean will be within 20 of the value of \(\mu\) ? b. For this example \((n=100, \sigma=10)\), complete each of the following statements by calculating the appropriate value: i. Approximately \(95 \%\) of the time, \(\bar{x}\) will be within of \(\mu\). ii. Approximately \(0.3 \%\) of the time, \(\bar{x}\) will be farther than from \(\mu\).

The time that people have to wait for an elevator in an office building has a uniform distribution over the interval from 0 to 1 minute. For this distribution, \(\mu=0.5\) and \(\sigma=0.289\) a. If \(\bar{x}\) is the average waiting time for a random sample of \(n=16\) waiting times, what are the values of the mean and standard deviation of the sampling distribution of \(\bar{x} ?\) b. Answer Part (a) for a random sample of 50 waiting times. Draw a picture of the approximate sampling distribution of \(\bar{x}\) when \(n=50\).
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