Please see attached mathematical statistics question and additional info. Using the critical region F >= c. If n = 13, m = 11, and alpha = 0.05, how to fin
Please see attached mathematical statistics question and additional info. Using the critical region F >= c. If n = 13, m = 11, and alpha = 0.05, how to fin
MATLAB: An Introduction with Applications
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Author:Amos Gilat
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Please see attached mathematical statistics question and additional info. Using the critical region F >= c. If n = 13, m = 11, and alpha = 0.05, how to find c?

Transcribed Image Text:In Exercise 4.2.27, in finding a confidence interval for the ratio of the variances of two normal distributions, we used a statistic \( \frac{S_1^2}{S_2^2} \), which has an F-distribution when those two variances are equal. If we denote that statistic by \( F \), we can test \( H_0 : \sigma_1^2 = \sigma_2^2 \) against \( H_1 : \sigma_1^2 = \sigma_2^2 \) using the critical region \( F \geq c \). If \( n = 13 \), \( m = 11 \), and \( \alpha = 0.05 \), find \( c \).
![**Section 4.2.27**
Consider two independent random samples: \( X_1, X_2, \ldots, X_n \) and \( Y_1, Y_2, \ldots, Y_m \) from normal distributions \( N(\mu_1, \sigma_1^2) \) and \( N(\mu_2, \sigma_2^2) \), respectively. The parameters \( \mu_1, \sigma_1^2, \mu_2, \sigma_2^2 \) are unknown.
**Objective:**
Construct a confidence interval for the ratio of the variances \( \frac{\sigma_1^2}{\sigma_2^2} \). This involves forming the quotient of two independent \(\chi^2\) variables, each divided by its degrees of freedom:
\[
F = \frac{\frac{(m-1)S_2^2}{\sigma_2^2}/(m-1)}{\frac{(n-1)S_1^2}{\sigma_1^2}/(n-1)} = \frac{S_2^2/\sigma_2^2}{S_1^2/\sigma_1^2}
\]
where \( S_1^2 \) and \( S_2^2 \) are the sample variances.
**Tasks:**
(a) What is the distribution of \( F \)?
(b) Use an appropriate table to find values \( a \) and \( b \) such that \( P(F < b) = 0.975 \) and \( P(a < F < b) = 0.95 \).
(c) Rewrite the probability statement as:
\[
P\left( a \frac{S_1^2}{S_2^2} < \frac{\sigma_1^2}{\sigma_2^2} < b \frac{S_1^2}{S_2^2} \right) = 0.95
\]
The observed values \( s_1^2 \) and \( s_2^2 \) can be used in these inequalities to provide a 95% confidence interval for \( \frac{\sigma_1^2}{\sigma_2^2} \).
**Solution:**
(a) \( F \) follows an F-distribution with parameters](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F5aa4cb66-a2f7-4199-af14-322b31fac275%2Fbd6868dd-25bb-4cf3-83ad-82583ae2cf10%2F79nxvm_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Section 4.2.27**
Consider two independent random samples: \( X_1, X_2, \ldots, X_n \) and \( Y_1, Y_2, \ldots, Y_m \) from normal distributions \( N(\mu_1, \sigma_1^2) \) and \( N(\mu_2, \sigma_2^2) \), respectively. The parameters \( \mu_1, \sigma_1^2, \mu_2, \sigma_2^2 \) are unknown.
**Objective:**
Construct a confidence interval for the ratio of the variances \( \frac{\sigma_1^2}{\sigma_2^2} \). This involves forming the quotient of two independent \(\chi^2\) variables, each divided by its degrees of freedom:
\[
F = \frac{\frac{(m-1)S_2^2}{\sigma_2^2}/(m-1)}{\frac{(n-1)S_1^2}{\sigma_1^2}/(n-1)} = \frac{S_2^2/\sigma_2^2}{S_1^2/\sigma_1^2}
\]
where \( S_1^2 \) and \( S_2^2 \) are the sample variances.
**Tasks:**
(a) What is the distribution of \( F \)?
(b) Use an appropriate table to find values \( a \) and \( b \) such that \( P(F < b) = 0.975 \) and \( P(a < F < b) = 0.95 \).
(c) Rewrite the probability statement as:
\[
P\left( a \frac{S_1^2}{S_2^2} < \frac{\sigma_1^2}{\sigma_2^2} < b \frac{S_1^2}{S_2^2} \right) = 0.95
\]
The observed values \( s_1^2 \) and \( s_2^2 \) can be used in these inequalities to provide a 95% confidence interval for \( \frac{\sigma_1^2}{\sigma_2^2} \).
**Solution:**
(a) \( F \) follows an F-distribution with parameters
Expert Solution

Step 1
Given that m=11 and n=13
numerator degrees of freedom=m-1=11-1=10
Denominator degrees of freedom=n-1=13-1=12
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