A simple random sample of size n is drawn from a population that is known to be normally distributed. The sample variance, s, is determined to be 12.2. Complete parts (a) through (c). ... (a) Construct a 90% confidence interval for o² if the sample size, n, is 20. The lower bound is. (Round to two decimal places as needed.) 4
A simple random sample of size n is drawn from a population that is known to be normally distributed. The sample variance, s, is determined to be 12.2. Complete parts (a) through (c). ... (a) Construct a 90% confidence interval for o² if the sample size, n, is 20. The lower bound is. (Round to two decimal places as needed.) 4
MATLAB: An Introduction with Applications
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![### Constructing a 90% Confidence Interval for Population Variance
A simple random sample of size \( n \) is drawn from a population that is known to be normally distributed. The sample variance, \( s^2 \), is determined to be 12.2. Here, we are required to complete parts (a) through (c) based on the given information.
**(a) Construct a 90% confidence interval for \( \sigma^2 \) if the sample size, \( n \), is 20.**
To find the confidence interval, we use the Chi-Square distribution. The formula for the confidence interval for the population variance \( \sigma^2 \) is:
\[
\left( \frac{(n-1)s^2}{\chi^2_{\alpha/2}}, \frac{(n-1)s^2}{\chi^2_{1-\alpha/2}} \right)
\]
Given:
- Sample variance \( s^2 = 12.2 \)
- Sample size \( n = 20 \)
- Degrees of freedom \( df = n-1 = 19 \)
- Confidence level = 90% (which implies \( \alpha = 0.10 \) and \( \alpha/2 = 0.05 \))
We need the critical values from the Chi-Square distribution for \( df = 19 \) at \(\alpha/2\) and \(1-\alpha/2\):
\[
\chi^2_{0.05, 19} = 10.117
\]
\[
\chi^2_{0.95, 19} = 30.144
\]
Using these values, we can calculate the lower and upper bounds:
Lower bound:
\[
\frac{(20-1) \cdot 12.2}{30.144} = \frac{19 \cdot 12.2}{30.144} \approx 7.68
\]
Upper bound:
\[
\frac{(20-1) \cdot 12.2}{10.117} = \frac{19 \cdot 12.2}{10.117} \approx 22.91
\]
Therefore, the 90% confidence interval for \( \sigma^2 \) is approximately \((7.68, 22.91)\).
### Graphs or Diagrams](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fb7da3211-2184-47c6-9dc5-31c87a612d2a%2F55e98c58-db5f-47f1-a611-04d8d7bcda0b%2Fhf67e1_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Constructing a 90% Confidence Interval for Population Variance
A simple random sample of size \( n \) is drawn from a population that is known to be normally distributed. The sample variance, \( s^2 \), is determined to be 12.2. Here, we are required to complete parts (a) through (c) based on the given information.
**(a) Construct a 90% confidence interval for \( \sigma^2 \) if the sample size, \( n \), is 20.**
To find the confidence interval, we use the Chi-Square distribution. The formula for the confidence interval for the population variance \( \sigma^2 \) is:
\[
\left( \frac{(n-1)s^2}{\chi^2_{\alpha/2}}, \frac{(n-1)s^2}{\chi^2_{1-\alpha/2}} \right)
\]
Given:
- Sample variance \( s^2 = 12.2 \)
- Sample size \( n = 20 \)
- Degrees of freedom \( df = n-1 = 19 \)
- Confidence level = 90% (which implies \( \alpha = 0.10 \) and \( \alpha/2 = 0.05 \))
We need the critical values from the Chi-Square distribution for \( df = 19 \) at \(\alpha/2\) and \(1-\alpha/2\):
\[
\chi^2_{0.05, 19} = 10.117
\]
\[
\chi^2_{0.95, 19} = 30.144
\]
Using these values, we can calculate the lower and upper bounds:
Lower bound:
\[
\frac{(20-1) \cdot 12.2}{30.144} = \frac{19 \cdot 12.2}{30.144} \approx 7.68
\]
Upper bound:
\[
\frac{(20-1) \cdot 12.2}{10.117} = \frac{19 \cdot 12.2}{10.117} \approx 22.91
\]
Therefore, the 90% confidence interval for \( \sigma^2 \) is approximately \((7.68, 22.91)\).
### Graphs or Diagrams
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