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Determine the values of the following percentiles: \(\chi_{0.05,10}^{2}, \chi_{0.025,15}^{2}, \chi_{0.01,12}^{2}, \chi_{0.95,20}^{2}, \chi_{0.99,18}^{2}, \chi_{0.995,16}^{2},\) and \(\chi_{0.005,25}^{2}\).

Short Answer

Expert verified
18.31, 27.488, 26.217, 10.851, 6.256, 5.812, and 44.314 for each respective percentile.

Step by step solution

01

Understanding Chi-Square Percentiles

The exercise involves finding specific percentiles of the chi-square distribution, denoted as \( \chi^2_{\alpha, n} \), where \( \alpha \) represents the upper-tail probability, and \( n \) is the degrees of freedom. Each value represents the point at which the area to the right under the chi-square distribution curve equals \( \alpha \).
02

Using Chi-Square Distribution Tables

To determine the specific percentiles, we reference a chi-square distribution table or use statistical software. These tables relate \( \alpha \) and \( n \) to the chi-square value. For example, to find \( \chi^2_{0.05, 10} \), locate the row for 10 degrees of freedom and find the column for the 0.05 upper-tail probability.
03

Calculate \( \chi^2_{0.05, 10} \)

Using a chi-square table, we find \( \chi^2_{0.05, 10} \) correlates to a value of approximately 18.31.
04

Calculate \( \chi^2_{0.025, 15} \)

For \( \chi^2_{0.025, 15} \), look up 15 degrees of freedom and the 0.025 tail probability. This value is approximately 27.488.
05

Calculate \( \chi^2_{0.01, 12} \)

Find \( \chi^2_{0.01, 12} \) by checking the 12 degrees of freedom against the 0.01 upper-tail probability, resulting in approximately 26.217.
06

Calculate \( \chi^2_{0.95, 20} \)

To find \( \chi^2_{0.95, 20} \), use 20 degrees of freedom with a 0.95 tail probability. This is approximately 10.851.
07

Calculate \( \chi^2_{0.99, 18} \)

For \( \chi^2_{0.99, 18} \), use the table or software for 18 degrees of freedom and 0.99 probability. This yields approximately 6.256.
08

Calculate \( \chi^2_{0.995, 16} \)

Find \( \chi^2_{0.995, 16} \) with 16 degrees of freedom at a 0.995 tail probability. This is approximately 5.812.
09

Calculate \( \chi^2_{0.005, 25} \)

Finally, to get \( \chi^2_{0.005, 25} \), look at 25 degrees of freedom and a 0.005 tail probability giving a value of approximately 44.314.

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Key Concepts

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

upper-tail probability
In statistics, when we talk about upper-tail probability, we are referring to the area under a probability distribution curve that lies to the right of a specified point. This is essentially the probability that a random variable from a given distribution is greater than a certain value.

The chi-square distribution often uses this concept, especially when determining chi-square percentiles. For example, in the expression \( \chi^2_{0.05, 10} \), the 0.05 represents an upper-tail probability, indicating that the probability of the chi-square statistic being larger than this particular value is 0.05, or 5%.

This upper-tail approach is useful in hypothesis testing and confidence interval estimation, where it helps establish whether observed data significantly deviate from the expected outcomes.
degrees of freedom
Degrees of freedom (often abbreviated as \( df \)) are important in statistics as they refer to the number of independent values or quantities which can be assigned to a statistical distribution.

In the context of the chi-square distribution, degrees of freedom are crucial because they shape the distribution's appearance. A larger number of degrees of freedom skews the distribution to be more symmetric and normal-like.

Within the chi-square distribution, degrees of freedom often relate to the sample size or the number of categories minus one in a test. For instance, \( \chi^2_{0.025, 15} \) uses 15 degrees of freedom, indicating a specific table row or parameter group in statistical software.

Knowing the correct degrees of freedom is essential for accurate analysis as it impacts the critical values and test results.
chi-square distribution table
A chi-square distribution table is a valuable tool that provides the critical values of the chi-square distribution for various upper-tail probabilities and degrees of freedom.

This table is instrumental when you need to find the percentile values specified by expressions like \( \chi^2_{0.05, 10} \) or \( \chi^2_{0.01, 12} \).

To use this table, you locate the row corresponding to the number of degrees of freedom and find the column that represents the desired upper-tail probability. The intersection is the critical value of the chi-square distribution for those conditions.

This tool aids students and statisticians in determining test statistics and is an efficient method for manual calculations without needing computational software.
probability percentiles
Probability percentiles in the chi-square distribution context refer to specific points on the distribution curve that correspond to given cumulative probabilities.

Percentiles, such as those in \( \chi^2_{0.99, 18} \) or \( \chi^2_{0.005, 25} \), mark where the proportion of the distribution falls below or above certain values.

These are particularly useful for hypothesis testing where test results are compared against these percentiles to determine statistical significance. For instance, if a calculated chi-square statistic is higher than the critical value at \( \chi^2_{0.05, 10} \), it suggests that the observed variation might not just be due to random chance.

Understanding and using probability percentiles can improve one's ability to interpret and evaluate statistical models and data.

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Most popular questions from this chapter

For a normal population with known variance \(\sigma^{2}\), answer the following questions: (a) What is the confidence level for the interval \(\bar{x}-2.14 \sigma / \sqrt{n}\) \(\leq \mu \leq \bar{x}+2.14 \sigma / \sqrt{n} ?\) (b) What is the confidence level for the interval \(\bar{x}-2.49 \sigma / \sqrt{n}\) \(\bar{x}-2.49 \sigma / \sqrt{n} \leq \mu \leq \bar{x}+2.49 \sigma / \sqrt{n} ?\) (c) What is the confidence level for the interval \(\bar{x}-1.85 \sigma / \sqrt{n}\) \(\leq \mu \leq \bar{x}+1.85 \sigma / \sqrt{n} ?\) (d) What is the confidence level for the interval \(\mu \leq \bar{x}+\) \(2.00 \sigma / \sqrt{n} ?\) (e) What is the confidence level for the interval \(\bar{x}-1.96 \sigma / \sqrt{n} \leq \mu ?\)

Consider the one-sided confidence interval expressions for a mean of a normal population. (a) What value of \(z_{\alpha}\) would result in a \(90 \%\) CI? (b) What value of \(z_{\alpha}\) would result in a \(95 \%\) CI? (c) What value of \(z_{\alpha}\) would result in a \(99 \%\) CI?

A random sample of 50 suspension helmets used by motorcycle riders and automobile race-car drivers was subjected to an impact test, and some damage was observed on 18 of these helmets. (a) Find a \(95 \%\) two-sided confidence interval on the true proportion of helmets that would show damage from this test. (b) Using the point estimate of \(p\) from the 50 helmets, how many helmets must be tested to be \(95 \%\) confident that the error in estimating \(p\) is less than \(0.02 ?\) (c) How large must the sample be if we wish to be at least \(95 \%\) confident that the error in estimating \(p\) is less than 0.02 regardless of the true value of \(p ?\)

An operating system for a personal computer has been studied extensively, and it is known that the standard deviation of the response time following a particular command is \(\sigma=8\) milliseconds. A new version of the operating system is installed, and you wish to estimate the mean response time for the new system to ensure that a \(95 \%\) confidence interval for \(\mu\) has a length of at most 5 milliseconds. (a) If you can assume that response time is normally distributed and that \(\sigma=8\) for the new system, what sample size would you recommend? (b) Suppose that the vendor tells you that the standard deviation of the response time of the new system is smaller, say, \(\sigma=6\); give the sample size that you recommend and comment on the effect the smaller standard deviation has on this calculation.

Following are two confidence interval estimates of the mean \(\mathrm{m}\) of the cycles to failure of an automotive door latch mechanism (the test was conducted at an elevated stress level to accelerate the failure). $$ 3124.9 \leq \mu \leq 3215.7 \quad 3110.5 \leq \mu \leq 3230.1 $$ (a) What is the value of the sample mean cycles to failure? (b) The confidence level for one of these CIs is \(95 \%\) and for the other is \(99 \%\). Both CIs are calculated from the same sample data. Which is the \(95 \%\) CI? Explain why.

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