Problem 63 The error $X$ in a measurement... [FREE SOLUTION]

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Modern Mathematical Statistics with Applications

Devore, Jay L., Berk, Kenneth N.

$Math Studyset 91影视 Explanations$ Math

2 Edition

Chapter 9: Problem 63

The error $X$ in a measurement has a normal distribution with mean value 0 and variance $\sigma^{2}$. Consider testing $H_{0}: \sigma^{2}=2$ versus $H_{\mathrm{a}}: \sigma^{2}=3$ based on a random sample $X_{1}, \ldots, X_{n}$ of errors. a. Show that a most powerful test rejects $H_{0}$ when $\sum x_{i}^{2} \geq c$. b. For $n=10$, find the value of $c$ for the test in (a) that results in $\alpha=.05$. c. Is the test of (a) UMP for $H_{0}: \sigma^{2}=2$ versus $H_{\mathrm{a}}: \sigma^{2}>2 ?$ Justify your assertion.

Short Answer

Expert verified

a) Reject $H_0$ when $\sum x_i^2 \geq c$. b) $c = 36.62$. c) No, it's not UMP for $\sigma^2 > 2$.

Step by step solution

Model the problem

The problem is structured around hypothesis testing for the variance of a normal distribution. The null hypothesis, $H_0: \sigma^2 = 2$, is tested against the alternative, $H_a: \sigma^2 = 3$. The test involves comparing the variance calculated from the sample, $\sum x_i^2$, to a critical value $c$. For a sample size $n$, the sum of squares follows a chi-squared distribution.

Determine critical region for most powerful test

According to the Neyman-Pearson Lemma, the most powerful test at a significance level $\alpha$ rejects the null hypothesis $H_0$ of $\sigma^2 = 2$ when $\sum x_i^2 \geq c$. Here, $\sum x_i^2$ is the test statistic that measures how far the estimated variance in a sample differs from the theoretical value $\sigma^2 = 2$.

Calculate the threshold for $\alpha = 0.05$

To find the value of $c$ for $n = 10$, we use the fact that $\sum x_i^2 / 2$ follows a chi-squared distribution with $n$ degrees of freedom when $H_0$ is true. Let's look at the chi-squared distribution table for $\chi^2_{0.95, 10}$.

Use chi-square table or tool

From chi-squared distribution tables or calculators, we find $\chi^2_{0.95, 10} \approx 18.31$. Since $\sum x_i^2 / 2$ follows a chi-squared distribution under $H_0$: $\sigma^2=2$, it implies $\sum x_i^2 \geq 36.62$ (because $18.31 \times 2 = 36.62$). Therefore, the rejection criterion is $\sum x_i^2 \geq 36.62$.

Determine if the test is UMP

A uniformly most powerful test (UMP) test represents the most powerful test for the entire range of alternative hypotheses. The test is not UMP when the alternative hypothesis is two-tailed (e.g., $\sigma^2 > 2$) because the rejection criteria $\sum x_i^2 \geq c$ only apply in one direction, assuming the mean doesn't change. To test UMP for all $\sigma^2 > 2$, Neyman-Pearson is applicable primarily for simple hypotheses, not composite.

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Key Concepts

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

Neyman-Pearson Lemma

The Neyman-Pearson Lemma is a fundamental concept in statistical hypothesis testing that helps determine the most powerful test for a given size of test (type I error rate, denoted as $\alpha$). This lemma gives a statistical test rule to best differentiate between two competing hypotheses.

Null Hypothesis $H_0$: This is a statement that there is no effect or difference, and in our exercise, it's $\sigma^2 = 2$.
Alternative Hypothesis $H_a$: This is what we believe to be true if $H_0$ is false, here $\sigma^2 = 3$.
Powerful Test: This is the test that has the highest probability of rejecting $H_0$ if $H_a$ is true, ensuring that we make the correct decision more often.

The lemma tells us that for tests with a fixed $\alpha$, the most powerful test rule involves comparing a test statistic (here, the sum of squares $\sum x_i^2$) to some critical value $c$. If $\sum x_i^2 \geq c$, we reject $H_0$. This ensures the test maximizes power under the constraint of a fixed significance level.

Chi-Squared Distribution

The chi-squared distribution is crucial when dealing with tests involving variances, especially in normal distributions. In our context, we're testing variance $\sigma^2$ of normally distributed errors.

Definition: The chi-squared distribution is a sum of the squares of independent standard normal random variables, and it's parametrized by degrees of freedom ($df$).
Degrees of Freedom: For a sample size $n$, $df = n$.
Usage: Under $H_0$, $\sum x_i^2 / \sigma^2$ follows a chi-squared distribution. This is used to derive the critical value $c$ for hypothesis testing.

In the example exercise, $\sum x_i^2 / 2$ follows a chi-squared distribution with $n=10$ degrees of freedom. By referencing a chi-squared table or calculator, one finds critical values which establish the rejection region for $H_0$. This helps in making statistical decisions regarding variance.

Most Powerful Test

In the context of hypothesis testing, a most powerful test is one that maximizes the power, i.e., the probability of correctly rejecting the null hypothesis $H_0$ when the alternative $H_a$ is true.

Power of a Test: This is the probability that the test will reject $H_0$ when $H_a$ is actually true. It quantifies the test's efficacy.
Why Most Powerful: Among various statistical tests, the most powerful ensures a higher rate of true positives (correctly identifying the situation under $H_a$).

For the variance example, the test is said to be the most powerful for $H_0$ against $H_a$ because it uses the Neyman-Pearson criteria ensuring maximum power at a set $\alpha$. This is achieved by appropriately setting the critical region, often relying on the chi-squared distribution.

Variance Testing

Variance testing is a statistical method used to infer the variability within a dataset. It's critical in determining whether the variance of a population is equal to a specific value, different from it, or whether two populations have the same variance.

Objective: To test if the variance equals a hypothesized value (e.g., testing $\sigma^2 = 2$ against $\sigma^2 = 3$).
Test Statistic: For samples from a normal distribution, you often use $\sum x_i^2$, distributed as chi-squared when divided by $\sigma^2$.
Example in Exercise: The test for $\sigma^2$ is constructed by comparing the calculated sample variance against a critical value derived from the chi-squared distribution.

In the exercise, because $\sigma^2$ is being tested against specific values, it illustrates variance testing's role in making determinations about a population's spread, leveraging the chi-squared distribution's properties.

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Short Answer

Step by step solution

Model the problem

Determine critical region for most powerful test

Calculate the threshold for \(\alpha = 0.05\)

Use chi-square table or tool

Determine if the test is UMP

Key Concepts

Neyman-Pearson Lemma

Chi-Squared Distribution

Most Powerful Test

Variance Testing

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