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Chapter 8: Problem 47

Determine the \(\chi^{2}\) percentile that is required to construct each of the following CIs: (a) Confidence level \(=95 \%,\) degrees of freedom \(=24,\) onesided (upper) (b) Confidence level \(=99 \%,\) degrees of freedom \(=9,\) one-sided (lower) (c) Confidence level \(=90 \%,\) degrees of freedom \(=19,\) two-sided.

Short Answer

Expert verified
(a) 36.4150, (b) 1.735, (c) 10.117 and 30.144.

Step by step solution

01

Understanding Chi-Square Distribution

Chi-square distribution depends on two parameters: the degrees of freedom (df) and the percentile we wish to find according to the confidence level. For confidence intervals (CIs), specific percentiles of the chi-square distribution are needed.
02

One-Sided Upper Confidence Interval (CI) for (a)

For a one-sided upper CI with confidence level of 95% and df = 24, we seek the value of \(\chi^2_{0.95, 24}\). This represents the point such that 95% of the distribution is below this point, and 5% is above it. Use a chi-square table or software to find \(\chi^2_{0.95, 24} = 36.4150\).
03

One-Sided Lower Confidence Interval (CI) for (b)

For a one-sided lower CI with confidence level of 99% and df = 9, we need to find \(\chi^2_{0.01, 9}\). This represents the point such that 1% of the distribution is below this point, and 99% is above it. Use a chi-square table or software to find \(\chi^2_{0.01, 9} = 1.735\).
04

Two-Sided Confidence Interval (CI) for (c)

For a two-sided CI with confidence level of 90% and df = 19, we need to find the percentiles that cut off the lower and upper 5% of the distribution. These are \(\chi^2_{0.05/2, 19}\) and \(\chi^2_{1-(0.05/2), 19}\). Use a chi-square table or software to find \(\chi^2_{0.025, 19} = 10.117\) and \(\chi^2_{0.975, 19} = 30.144\).

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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 vital concept in statistics. They provide a range around a sample estimate which is likely to contain the true population parameter. This range is calculated based on the data from your sample and gives a level of certainty or confidence about where the true parameter lies.

Key characteristics of confidence intervals:
  • The interval is typically given with a specified confidence level, often 95% or 99%.
  • A 95% confidence interval means that if you were to take 100 different samples and calculate a confidence interval for each, approximately 95 out of the 100 confidence intervals will contain the true population parameter.
  • The width of a confidence interval gives a sense of the precision of the sample estimate; narrower intervals suggest more precision.
To calculate a confidence interval in the context of a chi-square distribution, specific percentile points on the chi-square distribution are used to define the bounds of the interval. This is crucial when dealing with parameters like variance or standard deviation.
Degrees of Freedom
Degrees of freedom, often abbreviated as df, are a critical component in statistics, particularly when using chi-square distributions. They determine the shape of the chi-square distribution and impact the calculation of other statistical measures.

Understanding degrees of freedom:
  • Degrees of freedom typically relate to the number of independent values that can vary in an analysis while estimating a parameter.
  • For many tests, such as the chi-square test, the degrees of freedom can be calculated as the number of data points minus the number of calculated parameters.
  • In the context of a chi-square distribution, as the degrees of freedom increase, the distribution becomes more symmetrical and approaches a normal distribution.
The chi-square distribution requires the degrees of freedom to define its probability distribution, making it essential for constructing confidence intervals within this context.
Percentiles
Percentiles play a pivotal role when working with chi-square distributions, especially in confidence interval construction. They help determine which point on the distribution corresponds to a particular level of probability.

Why percentiles matter:
  • A percentile indicates the relative standing of a value within a distribution.
  • For instance, if you are looking for the 95th percentile in a chi-square distribution, it refers to the value below which you would expect to find 95% of the data.
  • In constructing confidence intervals, percentiles are used to mark the bounds of where data is expected to reside with a certain confidence level.
When dealing with chi-square distributions, calculating the desired percentile allows you to create accurate confidence intervals, whether they are one-sided or two-sided.
Statistical Tables
Statistical tables are essential tools in the field of statistics, serving as quick references for retrieving important percentile values in various distributions, such as the chi-square distribution.

Using statistical tables:
  • Chi-square tables list critical values of the chi-square distribution for different degrees of freedom and probability levels.
  • These tables provide a simple way to determine the percentile needed for constructing confidence intervals.
  • By knowing the degrees of freedom and the desired confidence level, you can find the relevant chi-square statistic using the table, or supplemented by statistical software for accuracy.
Whether working on paper or with digital tools, statistical tables help streamline the process of finding critical values needed for precise statistical calculations.

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Most popular questions from this chapter

Information on a packet of seeds claims that \(93 \%\) of them will germinate. Of the 200 seeds that I planted, only 180 germinated. (a) Find a \(95 \%\) confidence interval for the true proportion of seeds that germinate based on this sample. (b) Does this seem to provide evidence that the claim is wrong?

A manufacturer of electronic calculators takes a random sample of 1200 calculators and finds 8 defective units. (a) Construct a \(95 \%\) confidence interval on the population proportion. (b) Is there evidence to support a claim that the fraction of defective units produced is \(1 \%\) or less?

A normal population has a known mean of 50 and unknown variance. (a) A random sample of \(n=16\) is selected from this population, and the sample results are \(\bar{x}=52\) and \(s=8 .\) How unusual are these results? That is, what is the probability of observing a sample average as large as 52 (or larger) if the known, underlying mean is actually \(50 ?\) (b) A random sample of \(n=30\) is selected from this population, and the sample results are \(\bar{x}=52\) and \(s=8\). How unusual are these results? (c) A random sample of \(n=100\) is selected from this population, and the sample results are \(\bar{x}=52\) and \(s=8\). How unusual are these results? (d) Compare your answers to parts (a)-(c) and explain why they are the same or different.

Dairy cows at large commercial farms often receive injections of bST (Bovine Somatotropin), a hormone used to spur milk production. Bauman et al. (Journal of Dairy Science, 1989) reported that 12 cows given bST produced an average of \(28.0 \mathrm{~kg} / \mathrm{d}\) of milk. Assume that the standard deviation of milk production is \(2.25 \mathrm{~kg} / \mathrm{d}\) (a) Find a \(99 \%\) confidence interval for the true mean milk production. (b) If the farms want the confidence interval to be no wider than \(\pm 1.25 \mathrm{~kg} / \mathrm{d},\) what level of confidence would they need to use?

A normal population has known mean \(\mu=50\) and variance \(\sigma^{2}=5\). What is the approximate probability that the sample variance is greater than or equal to \(7.44 ?\) less than or equal to \(2.56 ?\) For a random sample of size (a) \(n=16\) (b) \(n=30\) (c) \(n=71\) (d) Compare your answers to parts (a)-(c) for the approximate probability that the sample variance is greater than or equal to \(7.44 .\) Explain why this tail probability is increasing or decreasing with increased sample size. (e) Compare your answers to parts (a)-(c) for the approximate probability that the sample variance is less than or equal to \(2.56 .\) Explain why this tail probability is increasing or decreasing with increased sample size.
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