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Chapter 9: Problem 78

Consider the test of \(H_{0}: \sigma^{2}=10\) against \(H_{1}: \sigma^{2}>10 .\) What are the critical values for the test statistic \(\chi_{0}^{2}\) for the following significance levels and sample sizes? (a) \(\alpha=0.01\) and \(n=20\) (b) \(\alpha=0.05\) and \(n=12\) (c) \(\alpha=0.10\) and \(n=15\)

Short Answer

Expert verified
Critical values: (a) 36.19, (b) 19.68, (c) 21.06.

Step by step solution

01

Identify the Test Statistic

The test statistic for a chi-squared test of variance is given by: \[ \chi^2 = \frac{(n-1)s^2}{\sigma_0^2} \] where \( n \) is the sample size, \( s^2 \) is the sample variance, and \( \sigma_0^2 \) is the variance under the null hypothesis.
02

Determine the Degrees of Freedom

For a chi-squared test, the degrees of freedom are given by \( n - 1 \). You apply this for each of the cases provided: (a) For \( n = 20 \), the degrees of freedom are \( 19 \). (b) For \( n = 12 \), the degrees of freedom are \( 11 \). (c) For \( n = 15 \), the degrees of freedom are \( 14 \).
03

Locate the Critical Value in Chi-Square Distribution Table

Using a chi-square distribution table, locate the critical value using the specified significance level \( \alpha \) and the calculated degrees of freedom:* (a) With \( \alpha = 0.01 \) and \( 19 \) degrees of freedom, the critical value is approximately \( 36.19 \).* (b) With \( \alpha = 0.05 \) and \( 11 \) degrees of freedom, the critical value is approximately \( 19.68 \).* (c) With \( \alpha = 0.10 \) and \( 14 \) degrees of freedom, the critical value is approximately \( 21.06 \).

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Key Concepts

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

Critical Values
In hypothesis testing, critical values are thresholds that define regions where we either reject or fail to reject a null hypothesis. For a Chi-Squared Test of Variance, these values indicate the limits beyond which observed data deviates too much from what the null hypothesis predicts.
  • These values are directly connected to the significance level (\( \alpha \)) and degrees of freedom (df).
  • Critical values from the chi-square distribution help us determine if the sample variance significantly differs from the hypothesized variance.
To find critical values, you consult a chi-square distribution table. The table matches degrees of freedom against the significance level.
For instance:
  • For a significance level of 0.01 and 19 degrees of freedom, the critical value is 36.19.
  • For \( \alpha = 0.05 \) and 11 degrees of freedom, it is 19.68.
  • At \( \alpha = 0.10 \) and 14 degrees of freedom, it is 21.06.
Understanding how to find and interpret critical values is essential for accurately conducting statistical tests.
Degrees of Freedom
Degrees of Freedom (df) play a vital role in statistical tests like the Chi-Squared Test of Variance. Essentially, degrees of freedom refer to the number of values in the final calculation of a statistic that are free to vary. In simpler terms, they are like the number of independent pieces of information you have:
  • For a sample of size \( n \), the degrees of freedom are calculated as \( n - 1 \).
  • Each degree of freedom adds complexity and precision to calculations.
In the context of the original exercise,
  • With \( n = 20 \), df = 19.
  • For \( n = 12 \), df = 11.
  • When \( n = 15 \), df = 14.
Knowing the degrees of freedom is necessary for looking up critical values in chi-square tables. More degrees of freedom typically mean a more accurate estimation of the population variance.
Significance Levels
The significance level, denoted by \( \alpha \), measures the probability of rejecting a true null hypothesis (Type I error). This critical concept defines how strongly you want to test the hypotheses:
  • A lower significance level means you require stronger evidence to reject the null hypothesis.
  • Commonly used levels include 0.01, 0.05, and 0.10, but the choice depends on the field of study or nature of the test.
In statistical testing, the significance level sets the threshold for deciding on the critical values:
  • If \( \alpha = 0.01 \), the test is very stringent, accepting a low probability of Type I error.
  • At \( \alpha = 0.05 \), it's a moderate level, often used in social sciences.
  • With \( \alpha = 0.10 \), you're slightly more lenient with potential errors.
Choosing the right significance level affects the test's sensitivity and ensures appropriate decisions are made based on data observations.

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Most popular questions from this chapter

The titanium content in an aircraft-grade alloy is an important determinant of strength. A sample of 20 test coupons reveals the following titanium content (in percent): \(8.32,8.05,8.93,8.65,8.25,8.46,8.52,8.35,8.36,8.41,8.42,\) \(8.30,8.71,8.75,8.60,8.83,8.50,8.38,8.29,8.46\) The median titanium content should be \(8.5 \%\). (a) Use the sign test with \(\alpha=0.05\) to investigate this hypothesis. Find the \(P\) -value for this test. (b) Use the normal approximation for the sign test to test \(H_{0}: \tilde{\mu}=8.5\) versus \(H_{1}: \tilde{\mu} \neq 8.5\) with \(\alpha=0.05 .\) What is the \(P\) -value for this test?

Did survival rate for passengers on the Titanic really depend on the type of ticket they had? Following are the data for the 2201 people on board listed by whether they survived and what type of ticket they had. Does survival appear to be independent of ticket class? (Test the hypothesis at \(\alpha=0.05 .)\) What is the \(P\) -value of the test statistic? $$\begin{array}{lccccc} & \text { Crew } & \text { First } & \text { Second } & \text { Third } & \text { Total } \\\\\hline \text { Alive } & 212 & 202 & 118 & 178 & 710 \\\\\text { Dead } & 673 & 123 & 167 & 528 & 1491 \\\\\hline \text { Total } & 885 & 325 & 285 & 706 & 2201\end{array}$$

Output from a software package follows: Test of \(m u=35\) vs not \(=35\) The assumed standard deviation \(=1.8\) $$\begin{array}{ccccccc}\text { Variable } & \mathrm{N} & \text { Mean } & \text { StDev } & \text { SE Mean } & \mathrm{Z} & \mathrm{P} \\\x & 25 & 35.710 & 1.475 & ? & ? & ? \\\\\hline\end{array}$$ (a) Fill in the missing items. What conclusions would you draw? (b) Is this a one-sided or a two-sided test? (c) Use the normal table and the preceding data to construct a \(95 \%\) two- sided \(\mathrm{CI}\) on the mean. (d) What would the \(P\) -value be if the alternative hypothesis is \(H_{1}: \mu>35 ?\)

A primer paint can be used on aluminum panels. The primer's drying time is an important consideration in the manufacturing process. Twenty panels are selected, and the drying times are as follows: \(1.6,1.3,1.5,1.6,1.7,1.9,1.8,1.6,1.4,\) \(1.8,1.9,1.8,1.7,1.5,1.6,1.4,1.3,1.6,1.5,\) and \(1.8 .\) Is there evidence that the mean drying time of the primer exceeds \(1.5 \mathrm{hr}\) ?

Consider the computer output below Test of \(p=0.25\) vs \(p<0.25\) $$\begin{array}{cccccc}\mathrm{X} & \mathrm{N} & \text { Sample } p & \text { Bound } & Z \text { -Value } & P \text { -Value } \\\53 & 225 & 0.235556 & 0.282088 & ? & ?\end{array}$$ Using the normal approximation: (a) Fill in the missing information. (b) What are your conclusions if \(\alpha=0.05 ?\) (c) The normal approximation to the binomial was used here. Was that appropriate? (d) Find a \(95 \%\) upper-confidence bound on the true proportion. (e) What are the \(P\) -value and your conclusions if the alternative hypothesis is \(H_{1}: p \neq 0.25 ?\)
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