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Consider the test of \(H_{0}: \sigma^{2}=5\) against \(H_{1}: \sigma^{2}<5 .\) 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
Calculate each \(P\)-value using chi-square distribution tables or software given the provided test statistics and sample sizes.

Step by step solution

01

Understand the Hypothesis Test

We are conducting a chi-square test for variance.The null hypothesis, \(H_0\), is that the variance \(\sigma^2 = 5\), and the alternative hypothesis, \(H_1\), is that \(\sigma^2 < 5\). We will use the critical value of the chi-square distribution to approximate the \(P\)-value.
02

Define the Chi-Square Statistic Formula

The chi-square statistic is computed as: \(\chi^2 = \frac{(n-1)s^2}{\sigma_0^2}\), where \(s^2\) is the sample variance and \(\sigma_0^2\) is the hypothesized population variance (5, in this case). Here, we have been provided test statistics \(x_0^2\) as 25.2, 15.2, and 4.2, for different sample sizes.
03

Identify the Degrees of Freedom

The degrees of freedom (df) are calculated as \(n-1\). We will compute df for each part of the problem:- For \(n=20\), df = 19.- For \(n=12\), df = 11.- For \(n=15\), df = 14.
04

Calculate P-Value for (a)

With \(x_0^2 = 25.2\), \(n = 20\), and df = 19:To find the \(P\)-value, we look at the right tail of the chi-square distribution. We look for the probability that a chi-square random variable with 19 degrees of freedom is greater than 25.2 using chi-square tables or software. This provides the \(P\)-value.
05

Calculate P-Value for (b)

With \(x_0^2 = 15.2\), \(n = 12\), and df = 11:Again, check the right tail of the chi-square distribution with df = 11 for a value of 15.2. This provides the \(P\)-value for part (b).
06

Calculate P-Value for (c)

With \(x_0^2 = 4.2\), \(n = 15\), and df = 14:Look for the probability in the left tail of the chi-square distribution with 14 degrees of freedom that corresponds to 4.2. This provides the \(P\)-value for part (c).
07

Interpret the P-Values

\(P\)-values help determine if we reject \(H_0\). A small \(P\)-value (usually \(< 0.05\)) indicates strong evidence against \(H_0\), so we reject it in favor of \(H_1\). Calculate \(P\)-values for each scenario using chi-square distribution tables or software, comparing them each to a significance level (e.g., 0.05).

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Key Concepts

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

Variance Hypothesis Testing
Variance hypothesis testing is a statistical method used to determine if there is a significant difference between a sample variance and a hypothesized population variance. It helps in assessing whether the variability observed in the sample data is due to chance or some significant effect.

In the given exercise, we are testing a hypothesis about the variance using a chi-square test. The null hypothesis (\( H_0 \) ) states that the population variance (\( \sigma^2 \) ) is equal to a specified value (5 in this case), while the alternative hypothesis (\( H_1 \) ) proposes the variance is less than this specified value. This is known as a one-sided test, aiming to detect a decrease in variance. Hypothesis testing for variance can reveal insights into the stability or consistency of a process or quality characteristic.
  • Null Hypothesis (\( H_0 \) ): States what is assumed to be true for the population variance.
  • Alternative Hypothesis (\( H_1 \) ): Contradicts the null, suggesting a different variance.
  • Statistical Inference: We use test statistics to decide whether the observed variance aligns with \( H_0 \) or supports \( H_1 \) .
Degrees of Freedom
Degrees of freedom (df) are an important concept in hypothesis testing that reflect the number of values in the final calculation of a statistic that are free to vary. In many statistical tests, including the chi-square test for variance, degrees of freedom are crucial for determining the distribution that applies to the test statistic.

For the chi-square test, the degrees of freedom are calculated as \( n-1 \) , where \( n \) is the sample size. This accounts for the fact that one parameter was estimated (the sample variance, \( s^2 \) , for example), and so we adjust for that by reducing the freedom typically by one.
  • In part (a) of the exercise: \( df = 20 - 1 = 19 \)
  • In part (b): \( df = 12 - 1 = 11 \)
  • In part (c): \( df = 15 - 1 = 14 \)
Degrees of freedom help map the test statistic to the appropriate chi-square distribution, which is essential for computing other test metrics like P-values.
Chi-Square Distribution
The chi-square distribution is a continuous distribution that is particularly useful in hypothesis testing concerning variances. It is used in the context of the chi-square test to analyze the goodness-of-fit of the observed data with the assumed distribution under the null hypothesis. This approach is primarily useful for testing hypotheses about the variance of a single population.

In this exercise, the chi-square distribution helps calculate test statistics, which, in combination with degrees of freedom, are used to find the P-values.
  • Right-tailed Test: As variance increases, the chi-square statistic moves towards the right tail of the distribution.
  • Left-tailed Test: If the test is to determine if variance decreases, as in the given problem, interest lies in the left tail.
  • Shape: Varies with df; more degrees of freedom result in a wider and flatter distribution.
Interpreting the position of the test statistic within the chi-square distribution helps us determine the probability of observing a given result if the null hypothesis is true.
P-value Calculation
The P-value is a critical part of hypothesis testing, representing the probability of obtaining a test statistic equal to or more extreme than the one observed, under the assumption that the null hypothesis is true. A low P-value indicates that such an extreme result is unlikely under the null hypothesis, suggesting that the null may not hold.

In this exercise, we calculate the P-value for the given test statistics using the chi-square distribution. It's essential to know the degrees of freedom to select the correct chi-square distribution curve. For instance, with a test statistic of 25.2 and df of 19, we look in chi-square tables or algorithms to find this probability.
  • Interpretation: A P-value less than \( 0.05 \) commonly indicates significant results, leading to rejecting the null hypothesis.
  • Exact Calculation: Typically done using statistical software or chi-square distribution tables.
  • Critical Value: Comparison point that helps decide between retention or rejection of \( H_0 \) .
Understanding the P-value aids in assessing whether the observed variance is statistically significant or falls within expected random variation under the null hypothesis.

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Most popular questions from this chapter

A random sample of students is asked their opinions on a proposed core curriculum change. The results are as follows. $$\begin{array}{lcc} & {\text { Opinion }} \\\\\text { Class } & \text { Favoring } & \text { Opposing } \\\\\text { Freshman } & 120 & 80 \\\\\text { Sophomore } & 70 & 130 \\\\\text { Junior } & 60 & 70 \\\\\text { Senior } & 40 & 60\end{array}$$ Test the hypothesis that opinion on the change is independent of class standing. Use \(\alpha=0.05 .\) What is the \(P\) -value for this test?

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An article in Experimental Brain Research ["Synapses in the Granule Cell Layer of the Rat Dentate Gyrus: SerialSectionin Study" (1996, Vol. \(112(2),\) pp. \(237-243\) ) ] showed the ratio between the numbers of symmetrical and total synapses on somata and azon initial segments of reconstructed granule cells in the dentate gyrus of a 12 -week-old rat: $$\begin{array}{llllll}0.65 & 0.90 & 0.78 & 0.94 & 0.40 & 0.94 \\\0.91 & 0.86 & 0.53 & 0.84 & 0.42 & 0.50 \\\0.50 & 0.68 & 1.00 & 0.57 & 1.00 & 1.00 \\\0.84 & 0.9 & 0.91 & 0.92 & 0.96 & \\\0.96 & 0.56 & 0.67 & 0.96 & 0.52 & \\\0.89 & 0.60 & 0.54 & & &\end{array}$$ (a) Use the data to test \(H_{0}: \sigma^{2}=0.02\) versus \(H_{1}: \sigma^{2} \neq 0.02\) using \(\alpha=0.05 .\) (b) Find the \(P\) -value for the test.

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