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Consider the hypothesis test of \(H_{0}: \sigma^{2}=10\) against \(H_{1}: \sigma^{2}>10\). Approximate the \(P\) -value for each of the following test statistics. (a) \(x_{0}^{2}=25.2\) and \(n=20\) (b) \(x_{0}^{2}=15.2\) and \(n=12\) (c) \(x_{0}^{2}=4.2\) and \(n=15\)

Short Answer

Expert verified
(a) \(P\)-value is approximately 0.10, (b) \(P\)-value is approximately 0.19, (c) \(P\)-value is approximately 0.99.

Step by step solution

01

Understand the Context

This is a chi-square hypothesis test problem, where we test if the population variance \(\sigma^2\) is equal to a certain value. The null hypothesis \(H_0: \sigma^2 = 10\) means that the population variance is 10, while the alternative hypothesis \(H_1: \sigma^2 > 10\) means that the population variance is greater than 10.
02

Recall the Test Statistic Formula

The test statistic for a chi-square test of variance is given by \( x_0^2 = \frac{(n-1)s^2}{\sigma_0^2} \), where \( n \) is the sample size, \( s^2 \) is the sample variance, and \( \sigma_0^2 \) is the hypothesized population variance which in this case is 10.
03

Identify the Degrees of Freedom

The degrees of freedom for the chi-square distribution in this test is \( n-1 \). For each scenario provided, compute \( n-1 \) to know which chi-square distribution to use for determining the \( P \)-value.
04

Calculate the P-value for Case (a)

For case (a), \( n = 20 \), so the degrees of freedom are \( 19 \). Given \( x_0^2 = 25.2 \), find the right-tail \( P \)-value: this is the probability \( P(\chi_{19}^2 > 25.2) \). Use a chi-square table or statistical software to find this probability.
05

Calculate the P-value for Case (b)

For case (b), \( n = 12 \), so the degrees of freedom are \( 11 \). Given \( x_0^2 = 15.2 \), compute the \( P \)-value \( P(\chi_{11}^2 > 15.2) \). Use chi-square tables or software to determine this probability.
06

Calculate the P-value for Case (c)

For case (c), \( n = 15 \), so the degrees of freedom are \( 14 \). Given \( x_0^2 = 4.2 \), the \( P \)-value is \( P(\chi_{14}^2 > 4.2) \). Again, consult a chi-square table or use statistical software to find this probability.

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Key Concepts

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

Hypothesis Testing
In the realm of statistics, hypothesis testing is a systematic method used to decide on the validity of a hypothesis. We start with two opposite claims: the null hypothesis (H_0) and the alternative hypothesis (H_1). In our exercise, the null hypothesis is H_0: \(\sigma^2 = 10\), meaning the population variance is equal to 10.
The alternative hypothesis, H_1: \(\sigma^2>10\), suggests the population variance is greater than 10. Our task is to determine if the data supports rejecting the null hypothesis or not. A hypothesis test assesses how likely the observed data occurred under the null hypothesis.
This chi-square test of variance is designed especially for testing the variance of a normally distributed population. The decision to accept or reject H_0 hinges on the calculated p-value and a pre-selected significance level, typically denoted by \(\alpha\).
These tests help in decision-making processes across various fields such as quality control, finance, and scientific research.
P-value Calculation
The p-value is a crucial concept in hypothesis testing as it helps to quantify the evidence against the null hypothesis. In simple terms, the p-value indicates the probability of obtaining test results at least as extreme as the ones observed, assuming the null hypothesis is true.
In our chi-square test for variance, calculating the p-value involves determining the right-tail probability of our test statistic exceeding a certain value, given the degrees of freedom. If the p-value is smaller than the significance level \(\alpha\), it indicates that observed data is unlikely under H_0, thus, we consider rejecting H_0.
For each part of the exercise, the process to approximate the p-value involves:
  • Determining the chi-square distribution's degrees of freedom based on the sample size \(n\).
  • Calculating the probability that a chi-square random variable with such degrees of freedom would be greater than the test statistic provided.
Consulting statistical tables or software aids in obtaining these probabilities, facilitating the decision-making process in hypothesis testing.
Degrees of Freedom
One essential step in performing a chi-square test for variance is determining the degrees of freedom. This concept is vital as it directly impacts the shape of the chi-square distribution used in hypothesis testing. Degrees of freedom refer to the number of independent values in a calculation that are free to vary.
In the context of our exercise, the degrees of freedom are calculated as \(n - 1\), where \(n\) is the sample size. Each condition provided in the exercise has different degrees of freedom:
  • For a sample size of 20 in case (a), there are 19 degrees of freedom.
  • With a sample size of 12 in case (b), we have 11 degrees of freedom.
  • Case (c) with a sample size of 15 has 14 degrees of freedom.
The degrees of freedom determine the specific chi-square distribution to be referenced for p-value calculation, ensuring accurate interpretation of the test results.
Population Variance
Population variance (\(\sigma^2\)) is a measure that represents the spread of a set of values in a population. It's an indicator of how much the data points differ from the mean value. In hypothesis testing, the population variance is often the focal point when assessing the variability within a data set.
In this exercise, the hypothesized population variance under the null hypothesis is 10. The goal is to test if the actual population variance exceeds this value based on the sample data provided.
Understanding population variance is pivotal in situations that demand quality assurance or consistency analysis. It provides insights into the stability of a process or product, guiding interventions when necessary. The chi-square test for variance is tailored to investigate these deviations from the hypothesized variance, thus ensuring reliable conclusions are derived from statistical data.

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Most popular questions from this chapter

An article in the British Medical Journal [“Comparison of Treatment of Renal Calculi by Operative Surgery, Percutaneous Nephrolithotomy, and Extra- Corporeal Shock Wave Lithotrips," (1986, Vol. 292, pp. \(879-882\) ) ] found that percutaneous nephrolithotomy (PN) had a success rate in removing kidney stones of 289 out of 350 patients. The traditional method was \(78 \%\) effective. (a) Is there evidence that the success rate for \(\mathrm{PN}\) is greater than the historical success rate? Find the \(P\) -value. (b) Explain how the question in part (a) could be answered with a confidence interval.

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Consider the following frequency table of observations on the random variable \(X\). \(\begin{array}{lllll}\text { Values } & 0 & 1 & 2 & 3 & 4\end{array}\) \(\begin{array}{llllll}\text { Observed Frequency } & 24 & 30 & 31 & 11 & 4\end{array}\) (a) Based on these 100 observations, is a Poisson distribution with a mean of 1.2 an appropriate model? Perform a goodness-of-fit procedure with \(\alpha=0.05 .\) (b) Calculate the \(P\) -value for this test.

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A semiconductor manufacturer collects data from a new tool and conducts a hypothesis test with the null hypothesis that a critical dimension mean width equals \(100 \mathrm{nm}\). The conclusion is to not reject the null hypothesis. Does this result provide strong evidence that the critical dimension mean equals \(100 \mathrm{nm}\) ? Explain.

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