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Consider the test of \(H_{0}: \sigma^{2}=7\) against \(H_{1}: \sigma^{2} \neq 7\). 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
(a) Critical values are 6.84 and 36.19. (b) Critical values are 3.82 and 21.92. (c) Critical values are 6.57 and 23.68.

Step by step solution

01

Determine Degrees of Freedom

To find the critical values for a chi-squared test, calculate the degrees of freedom \( \text{df} \) using the formula \( \text{df} = n - 1 \), where \( n \) is the sample size.
02

Locate Critical Chi-squared Values for α=0.01, n=20

Calculate \( \text{df} = 20 - 1 = 19 \). For a two-tailed test at \( \alpha = 0.01 \), the critical values are \( \chi^2_{0.005, 19} \) for the upper tail and \( \chi^2_{0.995, 19} \) for the lower tail. Using chi-squared distribution tables or a calculator, \( \chi^2_{0.005, 19} \approx 36.19 \) and \( \chi^2_{0.995, 19} \approx 6.84 \).
03

Locate Critical Chi-squared Values for α=0.05, n=12

Calculate \( \text{df} = 12 - 1 = 11 \). For a two-tailed test at \( \alpha = 0.05 \), the critical values are \( \chi^2_{0.025, 11} \) for the upper tail and \( \chi^2_{0.975, 11} \) for the lower tail. Use chi-squared distribution tables or a calculator to find \( \chi^2_{0.025, 11} \approx 21.92 \) and \( \chi^2_{0.975, 11} \approx 3.82 \).
04

Locate Critical Chi-squared Values for α=0.10, n=15

Calculate \( \text{df} = 15 - 1 = 14 \). For a two-tailed test at \( \alpha = 0.10 \), the critical values are \( \chi^2_{0.05, 14} \) for the upper tail and \( \chi^2_{0.95, 14} \) for the lower tail. Find these values using chi-squared distribution tables or a calculator: \( \chi^2_{0.05, 14} \approx 23.68 \) and \( \chi^2_{0.95, 14} \approx 6.57 \).

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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 degrees of freedom (df) is a concept that helps us determine the number of values in a calculation that are free to vary. When you are using a chi-squared test to evaluate hypotheses, calculating the degrees of freedom is crucial. To calculate the degrees of freedom for the chi-squared test related to variance, we use the formula:\[ \text{df} = n - 1 \]where \( n \) represents the sample size. For example, if the sample size is 20, the degrees of freedom will be:\[ 20 - 1 = 19 \]The degrees of freedom are essential because they help us locate the correct critical values for our test. The more degrees of freedom we have, the more our chi-squared distribution looks like a normal distribution.- **Key point:** Degrees of freedom influence the shape of the chi-squared distribution.- It determines how many values can independently vary, which affects the critical values we use.
Critical Values
Critical values in the chi-squared test are thresholds that determine whether we reject the null hypothesis (\( H_0 \)) or not. These values define the boundaries of our acceptance region. If our test statistic falls beyond these critical values, the result suggests a significant difference from what we would expect under the null hypothesis.To find the critical values, you need the degrees of freedom and the significance level (\( \alpha \)) for your test. Chi-squared distribution tables or calculators can aid in finding these values. The critical values are twofold in a two-tailed test: one for the lower end and one for the upper end of the chi-squared distribution.- For example, in a two-tailed test with \( \alpha = 0.05 \) and \( \text{df} = 11 \), your critical values might be \( \chi^2_{0.975, 11} \approx 3.82 \) and \( \chi^2_{0.025, 11} \approx 21.92 \).- **Key point:** Critical values help us decide if our observations are significantly different from our initial assumptions.
Significance Level
The significance level, denoted by \( \alpha \), is a probability threshold set by the researcher. It reflects the risk one is willing to take of rejecting a true null hypothesis (Type I error). This value is chosen before the test and is usually set at 0.05, 0.01, or 0.10 depending on the rigor required in your analysis.- **Common values:** - **0.05 (5%):** Often used in many scientific studies, offering a moderate level of confidence. - **0.01 (1%):** Offers a high confidence level, reducing risk but requiring stronger evidence. - **0.10 (10%):** Provides a lower confidence threshold, allowing more room for Type I error.The selected significance level affects the critical values. A lower \( \alpha \) value leads to more stringent critical values, meaning that more compelling evidence is needed to reject the null hypothesis.- **Key point:** Significance level determines how conservative the test is in making a Type I error.
Two-tailed Test
A two-tailed test is a statistical method where the area for rejection of the null hypothesis is in both tails of the distribution. In simpler terms, this means we are examining whether a parameter is either too high or too low compared to what we would expect. This type of test is useful when you are checking for any significant difference, whether the value falls excessively above or below a certain point under the null hypothesis.- **Why use a two-tailed test?** - When the direction of the deviation from the expected value is unknown or of no particular interest. - It allows us to identify deviations in both directions, offering a more comprehensive analysis.In the chi-squared test, this involves calculating critical values for both high and low ends of the chi-squared distribution. For example, with a two-tailed test at \( \alpha = 0.01 \) and \( \text{df} = 19 \), you'd consider critical values like \( \chi^2_{0.005, 19} \approx 36.19 \) and \( \chi^2_{0.995, 19} \approx 6.84 \) to decide on rejecting the null hypothesis.- **Key point:** A two-tailed test examines for deviations on both ends, accepting variability in either direction.

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Most popular questions from this chapter

For the hypothesis test \(H_{0}: \mu=7\) against \(H_{1}: \mu \neq 7\) and variance known, calculate the \(P\) -value for each of the following test statistics. (a) \(z_{0}=2.05\) (b) \(z_{0}=-1.84\) (c) \(z_{0}=0.4\)

The life in hours of a battery is known to be approximately normally distributed with standard deviation \(\sigma=1.25\) hours. A random sample of 10 batteries has a mean life of \(\bar{x}=40.5\) hours. (a) Is there evidence to support the claim that battery life exceeds 40 hours? Use \(\alpha=0.05 .\) (b) What is the \(P\) -value for the test in part (a)? (c) What is the \(\beta\) -error for the test in part (a) if the true mean life is 42 hours? (d) What sample size would be required to ensure that \(\beta\) does not exceed 0.10 if the true mean life is 44 hours? (e) Explain how you could answer the question in part (a) by calculating an appropriate confidence bound on life.

Supercavitation is a propulsion technology for undersea vehicles that can greatly increase their speed. It occurs above approximately 50 meters per second when pressure drops sufficiently to allow the water to dissociate into water vapor, forming a gas bubble behind the vehicle. When the gas bubble completely encloses the vehicle, supercavitation is said to occur. Eight tests were conducted on a scale model of an undersea vehicle in a towing basin with the average observed speed \(\bar{x}=102.2\) meters per second. Assume that speed is normally distributed with known standard deviation \(\sigma\) \(=4\) meters per second. (a) Test the hypothesis \(H_{0}: \mu=100\) versus \(H_{1}: \mu<100\) using \(\alpha=0.05\) (b) What is the \(P\) -value for the test in part (a)? (c) Compute the power of the test if the true mean speed is as low as 95 meters per second. (d) What sample size would be required to detect a true mean speed as low as 95 meters per second if you wanted the power of the test to be at least \(0.85 ?\) (e) Explain how the question in part (a) could be answered by constructing a one-sided confidence bound on the mean speed.

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A melting point test of \(n=10\) samples of a binder used in manufacturing a rocket propellant resulted in \(\bar{x}=154.2^{\circ} \mathrm{F}\). Assume that the melting point is normally distributed with \(\sigma=1.5^{\circ} \mathrm{F}\). (a) Test \(H_{0}: \mu=155\) versus \(H_{1}: \mu \neq 155\) using \(\alpha=0.01\). (b) What is the \(P\) -value for this test? (c) What is the \(\beta\) -error if the true mean is \(\mu=150 ?\) (d) What value of \(n\) would be required if we want \(\beta<0.1\) when \(\mu=150 ?\) Assume that \(\alpha=0.01\)

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