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Find: a. \(\chi^{2}(10,0.01)\) b. \(\chi^{2}(12,0.025)\) c. \(\chi^{2}(10,0.95)\) d. \(\chi^{2}(22,0.995)\)

Short Answer

Expert verified
\(\chi^{2}(10,0.01) = 23.209\), \(\chi^{2}(12,0.025) = 26.217\), \(\chi^{2}(10,0.95) = 3.940\), \(\chi^{2}(22,0.995) = 2.718\)

Step by step solution

01

Understand the Chi-Square Distribution

Chi-square distribution is a non-negative function with mode nearly equal to degree of freedom (exception only occurs where degree of freedom is less than 1). It has a tendency of being skewed positively regardless of the degree of freedom. The Chi-square distribution summarizes a set of independent standard normal variables each squared and summed together, where the degree of freedom defines the quantity of those independent standard normal variables. Chi-square distribution table or scientific calculator with \(\chi^2\) functionality are used for calculating such distributions.
02

Solve \( \chi^{2}(10,0.01)\)

Look up in the chi-square distribution table or use a calculator which supports chi-square functionality. Find the cell in the table where the row corresponds to 10 degrees of freedom and the column corresponds to the p-value 0.01. The value at the intersection of these is approximately 23.209.
03

Solve \( \chi^{2}(12,0.025)\)

Again, find the cell in the chi-square distribution table or use a calculator which supports chi-square functionality. Find the cell where the row is for 12 degrees of freedom and the column corresponds to the p-value 0.025. The value at the intersection of these is approximately 26.217.
04

Solve \( \chi^{2}(10,0.95)\)

Find the cell in the chi-square distribution table or use a calculator which supports chi-square functionality. Find the cell where the row corresponds to 10 degrees of freedom and the column corresponds to the p-value 0.95. The value at the intersection of these is approximately 3.940.
05

Solve \( \chi^{2}(22,0.995)\)

Find the cell in the chi-square distribution table or use a calculator which supports chi-square functionality. Find the cell where the row corresponds to 22 degrees of freedom and the column corresponds to the p-value 0.995. The value at the intersection of these is approximately 2.718.

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Key Concepts

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

Degrees of Freedom
In statistics, the concept of "Degrees of Freedom" is crucial to understanding the characteristics of the Chi-Square Distribution. Degrees of freedom (often abbreviated as df) refer to the number of values in a calculation that are free to vary.
When we talk about chi-square tests, degrees of freedom determine the shape of the chi-square distribution.

Consider the chi-square, for example, it often summarizes the squares and sums of standard normal variables. The number of these variables is what we refer to as degrees of freedom. Hence, if you see a chi-square value presented as \( \chi^2(k, \alpha) \), the `k` indicates the degrees of freedom.

Understanding df is essential because:
  • Increases in degrees of freedom often lead to a more symmetric distribution.
  • Higher degrees of freedom will mean your chi-square distribution will resemble a normal distribution more closely.
  • Each estimate that loses a degree of freedom affects your potential statistical inferences, hence directly impacting your results.
Therefore, recognizing degrees of freedom helps in interpreting statistical tests correctly and ensures you pick accurate values from a chi-square table.
P-Value
The "P-Value" is another critical concept you will encounter when working with chi-square distributions. A p-value essentially tells you the probability of obtaining test results at least as extreme as the observed results, under the assumption that the null hypothesis is true.

In a chi-square distribution context:
  • A small p-value, usually less than 0.05 or 0.01, suggests that the observed distribution significantly differs from the expected distribution.
  • A larger p-value indicates that the observed distribution is closer to what was expected.
A calculated chi-square statistic can be compared directly with the critical chi-square value from a table, often with the help of a specified p-value to determine significance. Hence, the p-value is central to our decisions about the null hypothesis and plays a crucial role in deciding whether differences in distributions are significant.
Chi-Square Table
The "Chi-Square Table" is a tool that statisticians use to determine the critical chi-square value for a given significance level or p-value and degrees of freedom.
It's a crucial aid in hypotheses tests involving the chi-square statistic.

Here’s how you use the chi-square table:
  • The rows typically correspond to degrees of freedom.
  • The columns usually correspond to different significance levels or commonly used p-values (e.g., 0.01, 0.025, 0.05).
To find the critical value: locate your degrees of freedom along the side and move across to the column representing the p-value. The intersection is your critical chi-square value. This value is what you will compare your test statistic against to determine statistical significance.

The chi-square table is user-friendly, and using it correctly aids in making informed decisions regarding hypotheses testing and interpreting results from experiments and trials.

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Most popular questions from this chapter

A manufacturer of television sets claims that the maintenance expenditures for its product will average no more than \(110\)dollar during the first year following the expiration of the warranty. A consumer group has asked you to substantiate or discredit the claim. The results of a random sample of 50 owners of such television sets showed that the mean expenditure was \(131.60\)dollar and the standard deviation was \(42.46\)dollar At the 0.01 level of significance, should you conclude that the manufacturer's claim is true or not likely to be true?

You are testing the null hypothesis \(p=0.4\) and will reject this hypothesis if \(z \star\) is less than -2.05. a. If the null hypothesis is true and you observe \(z \star\) equal to \(-2.12,\) which of the following will you do? (1) Correctly fail to reject \(H_{o} .\) (2) Correctly reject \(H_{o} .(3)\) Commit a type I error. (4) Commit a type II error. b. What is the significance level for this test? c. What is the \(p\) -value for \(z \star=-2.12 ?\)

A winemaker has placed a large order for the no. 9 corks described in Applied Example 6.13 (p. 285 ) and is concerned about the number of corks that might have smaller diameters. During the corking process, the corks are squeezed down to 16 to \(17 \mathrm{mm}\) in diameter for insertion into bottles with an \(18 \mathrm{mm}\) opening. The cork then expands to make the seal. The winemaker wants the corks to be as tight as possible and is therefore concerned about any that might be undersized. The diameter of each cork is measured in several places, and an average diameter is reported for each cork. The cork manufacturer has assured the winemaker that each cork has an average diameter within the specs and that all average diameters have a normal distribution with a mean of \(24.0 \mathrm{mm}\). a. Why does it make sense for the diameter of the cork to be assigned the average of several different diameter measurements? A random sample of 18 corks is taken from the batch to be shipped and the diameters (in millimeters) obtained: $$\begin{array}{llllllll}\hline 23.93 & 23.91 & 23.82 & 24.02 & 23.93 & 24.17 & 23.93 & 23.84 & 24.13 \\\24.01 & 23.83 & 23.74 & 23.73 & 24.10 & 23.86 & 23.90 & 24.32 & 23.83 \\\\\hline\end{array}$$ b. The average diameter spec is "24 \(\mathrm{mm}+0.6 \mathrm{mm} /\) \(-0.4 \mathrm{mm} . "\) Does it appear this order meets the spec on an individual cork basis? Explain. c. Does the sample in part a show sufficient reason to doubt the truthfulness of the claim, that the mean average diameter is \(24.0 \mathrm{mm},\) at the 0.02 level of significance? A different sample of 18 corks was randomly selected and the diameters (in millimeters) obtained: $$\begin{array}{lllllllll}\hline 23.90 & 23.98 & 24.28 & 24.22 & 24.07 & 23.87 & 24.05 & 24.06 & 23.82 \\\24.03 & 23.87 & 24.08 & 23.98 & 24.21 & 24.08 & 24.06 & 23.87 & 23.95 \\\\\hline\end{array}$$ d. Does the preceding sample show sufficient reason to doubt the truthfulness of the claim, that the mean average diameter is \(24.0 \mathrm{mm},\) at the 0.02 level of significance? e. What effect did the two different sample means have on the calculated test statistic in parts c and d? Explain. f. What effect did the two different sample standard deviations have on the calculated test statistic in parts c and d? Explain.

The Pizza Shack in Exercise 9.177 has completed its sampling and the results are in! On Tuesday afternoon, they sampled 15 customers and 9 preferred the new pizza crust. On Friday evening, they sampled 200 customers and 120 preferred the new pizza crust. Help the manager interpret the meaning of these results. Use a one-tailed test with \(H_{a}: p>0.50\) and \(\alpha=0.02 .\) Use \(z\) as the test statistic. a. Is there sufficient evidence to conclude a significant preference for the new crust based on Tuesday's customers? b. Is there sufficient evidence to conclude a significant preference for the new crust based on Friday's customers? c. since the percentage of customers preferring the new crust was the same, \(p^{\prime}=0.60\) in both samplings, explain why the answers in parts a and b are not the same.

a. What is the relationship between \(p=P(\text { success })\) and \(q=P(\text { failure }) ?\) Explain. b. Explain why the relationship between \(p\) and \(q\) can be expressed by the formula \(q=1-p\) c. If \(p=0.6,\) what is the value of \(q ?\) d. If the value of \(q^{\prime}=0.273,\) what is the value of \(p^{\prime} ?\)

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