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Determine the \(t\) -percentile that is required to construct each of the following two-sided confidence intervals: (a) Confidence level \(=95 \%,\) degrees of freedom \(=12\) (b) Confidence level \(=95 \%,\) degrees of freedom \(=24\) (c) Confidence level \(=99 \%,\) degrees of freedom \(=13\) (d) Confidence level \(=99.9 \%,\) degrees of freedom \(=15\)

Short Answer

Expert verified
(a) 2.179, (b) 2.064, (c) 3.012, (d) 4.073

Step by step solution

01

Understanding the Percentile Requirement

When constructing a two-sided confidence interval with a given confidence level and degrees of freedom, find the critical value \( t \) from the \( t \)-distribution table. This value corresponds to the \( \frac{\alpha}{2} \) percentile where \( \alpha = 1 - \text{confidence level} \).
02

Calculate \( \alpha \)

For the given confidence levels, calculate \( \alpha = 1 - \text{confidence level}\). For example:(a) 95% confidence, \( \alpha = 0.05 \)(b) 95% confidence, \( \alpha = 0.05 \)(c) 99% confidence, \( \alpha = 0.01 \)(d) 99.9% confidence, \( \alpha = 0.001 \)
03

Calculate \( \frac{\alpha}{2} \)

Split \( \alpha \) into two tails since it is two-sided:(a) \( \frac{0.05}{2} = 0.025 \)(b) \( \frac{0.05}{2} = 0.025 \)(c) \( \frac{0.01}{2} = 0.005 \)(d) \( \frac{0.001}{2} = 0.0005 \)
04

Find \( t \)-Critical Values

Use the \( t \)-distribution table to find the critical \( t \)-value for each case:(a) With 12 degrees of freedom, \( t_{0.025,12} \approx 2.179 \)(b) With 24 degrees of freedom, \( t_{0.025,24} \approx 2.064 \)(c) With 13 degrees of freedom, \( t_{0.005,13} \approx 3.012 \)(d) With 15 degrees of freedom, \( t_{0.0005,15} \approx 4.073 \)

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Key Concepts

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

Confidence Intervals
Confidence intervals are a fundamental concept in statistics, providing a range within which we are confident a parameter lies. For instance, if we say we have a 95% confidence interval for the mean of a dataset, we mean there's a 95% chance that the true mean is within this range. It's like saying we're 95% sure this specific range has our answer. We calculate confidence intervals to communicate the uncertainty around our estimates due to sampling errors.
A higher confidence level (like 99% instead of 95%) will create a wider interval because we want to be more sure that the parameter is captured within the range. As we want to be more confident, the span of possible values increases. Therefore, understanding confidence intervals helps us make more informed decisions in research by acknowledging the uncertainty involved.
Degrees of Freedom
Degrees of freedom (df) are an essential aspect of statistical calculations, particularly when working with t-distributions. They can be thought of as the number of values in the final calculation of a statistic that are free to vary. In essence, it tells you how many values in your statistical calculation are allowed to be independent.
For example, if you're calculating the variance or standard deviation of a sample, the degrees of freedom equal the number of observations in your data set minus the number of parameters you're estimating. In many practical scenarios, it's usually the sample size minus one (n-1). Understanding degrees of freedom helps in ensuring accurate estimations and test results by appropriately scaling the sample variability to reflect the data's true distribution.
Percentile
The percentile in a statistical distribution refers to the value below which a given percentage of observations falls. If you're looking at a t-distribution, and you need to find a 95th percentile, you're interested in the value below which 95% of the distribution’s observations lie.
When constructing two-sided confidence intervals, the concept of percentiles comes into play to define the critical values that determine the interval's boundaries. These percentiles help in understanding where your data point lies in comparison to the rest of the dataset, putting it in a more comprehensive context.
Critical Value
Critical values are pivotal in hypothesis testing and the creation of confidence intervals. In the context of a t-distribution, the critical value is a cutoff point that demarcates the threshold beyond which we consider data results to be statistically significant or extreme. For constructing confidence intervals, we locate the critical value in a t-distribution table using both the confidence level and the degrees of freedom in our data.
It's calculated based on the tail area (\( \frac{\alpha}{2} \)) of the distribution, where we find the exact point that marks off the extreme 5% or 1% of the data in a two-tailed test. A crucial part of analysis, critical values help us understand under what conditions we can consider the observed data to deviate significantly from assumed distributions or parameters.

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Most popular questions from this chapter

A confidence interval estimate is desired for the gain in a circuit on a semiconductor device. Assume that gain is normally distributed with standard deviation \(\mathrm{s}=20 .\) (a) Find a \(95 \% \mathrm{CI}\) for \(\mathrm{m}\) when \(n=10\) and \(\bar{x}=1000\). (b) Find a \(95 \%\) CI for \(m\) when \(n=25\) and \(\bar{x}=1000\). (c) Find a \(99 \% \mathrm{CI}\) for \(\mathrm{m}\) when \(n=10\) and \(\bar{x}=1000\). (d) Find a \(99 \%\) CI for \(m\) when \(n=25\) and \(\bar{x}=1000\). (e) How does the length of the CIs computed change with the changes in sample size and confidence level?

The confidence interval for a population proportion depends on the central limit theorem. A common rule of thumb is that to use the normal approximation for the sampling distribution for \(\hat{p},\) you should have at least 10 "successes" and 10 "failures." However, Agresti and Coull developed a method that can be used for smaller samples and increases the accuracy of all confidence intervals for proportions. The idea is simply to add 4 "pseudo observations" to the data set- 2 successes and 2 failures. That is, if you have \(X\) successes from of \(n\) trials, use \(\tilde{p}=\frac{X+2}{n+4}\) instead of the usual \(\hat{p}=\frac{X}{n}\) in the formulas for the confidence interval. A quality control engineer is inspecting defects on a newly designed printed circuit board. She inspects 50 boards and finds no defects. The usual estimate would be \(\hat{p}=0,\) but she does not believe that there will ever be a nodefects situation for this product. Use this Agresi-Coull estimate to come up with a \(95 \%\) confidence interval for the true proportion of defects. [Hint: The lower bound may be negative, which is not feasible, so simply use the upper limit and report that she is \(95 \%\) confident that the true defect proportion is no higher than this number.

A rivet is to be inserted into a hole. A random sample of \(n=15\) parts is selected, and the hole diameter is measured. The sample standard deviation of the hole diameter measurements is \(s=0.008\) millimeters. Construct a \(99 \%\) lower confidence bound for \(\sigma^{2}\).

A civil engineer is analyzing the compressive strength of concrete. Compressive strength is normally distributed with \(\sigma^{2}=1000(\mathrm{psi})^{2} .\) A random sample of 12 specimens has a mean compressive strength of \(\bar{x}=3250\) psi. (a) Construct a \(95 \%\) two-sided confidence interval on mean compressive strength. (b) Construct a \(99 \%\) two-sided confidence interval on mean compressive strength. Compare the width of this confidence interval with the width of the one found in part (a).

The 2004 presidential election exit polls from the critical state of Ohio provided the following results. The exit polls had 2020 respondents, 768 of whom were college graduates. Of the college graduates, 412 voted for George Bush. (a) Calculate a \(95 \%\) confidence interval for the proportion of college graduates in Ohio who voted for George Bush. (b) Calculate a \(95 \%\) lower confidence bound for the proportion of college graduates in Ohio who voted for George Bush.

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