Plastic sheets produced by a machine are periodically monitored for possible fluctuations in thickness. Typically, a variance in thickness around 1.95 square millimeters is regarded as acceptable. As part of quality control, a random sample of 30 plastic sheets are taken and the sample variance is calculated to be 3.17. Assume that the population is normally distributed. Let a = 0.05. (a) Construct a confidence interval for population variance at (1 − a) confidence level. (b) Test against an upper-tail hypothesis that the true population variance is larger than 1.95.
Plastic sheets produced by a machine are periodically monitored for possible fluctuations in thickness. Typically, a variance in thickness around 1.95 square millimeters is regarded as acceptable. As part of quality control, a random sample of 30 plastic sheets are taken and the sample variance is calculated to be 3.17. Assume that the population is normally distributed. Let a = 0.05. (a) Construct a confidence interval for population variance at (1 − a) confidence level. (b) Test against an upper-tail hypothesis that the true population variance is larger than 1.95.
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![### Quality Control in Manufacturing: Variance Analysis of Plastic Sheets
Plastic sheets produced by a machine are periodically monitored for possible fluctuations in thickness. Typically, a variance in thickness around 1.95 square millimeters is regarded as acceptable. As part of the quality control process, a random sample of 30 plastic sheets is taken and the sample variance is calculated to be 3.17 square millimeters. Assume that the population is normally distributed. Let \(\alpha = 0.05\).
#### (a) Constructing a Confidence Interval for Population Variance
To construct a confidence interval for the population variance at \((1 - \alpha)\) confidence level, we need to utilize the chi-square distribution because the sample variance follows a scaled chi-square distribution if the population is normally distributed.
Given:
- Sample size (\(n\)) = 30
- Sample variance (\(s^2\)) = 3.17 square millimeters
- \(\alpha\) = 0.05
The confidence interval for the population variance \(\sigma^2\) is given by:
\[
\left( \frac{(n-1)s^2}{\chi^2_{\alpha/2, n-1}}, \frac{(n-1)s^2}{\chi^2_{1-\alpha/2, n-1}} \right)
\]
Where:
- \(\chi^2_{\alpha/2, n-1}\) and \(\chi^2_{1-\alpha/2, n-1}\) are the critical values from the chi-square distribution with \(n-1\) degrees of freedom.
#### (b) Hypothesis Testing: Upper-Tail Test on Population Variance
To test the hypothesis that the true population variance is larger than 1.95, we perform an upper-tail hypothesis test.
The hypotheses are:
- Null hypothesis (\(H_0\)): \(\sigma^2 \leq 1.95\)
- Alternative hypothesis (\(H_1\)): \(\sigma^2 > 1.95\)
Test statistic:
\[
\chi^2 = \frac{(n-1)s^2}{\sigma_0^2}
\]
Where \(\sigma_0^2\) is the variance under the null hypothesis (1.95 square millimeters).
Given:
- Sample size (\(n\)) = 30](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fa420ac02-9353-4fae-9e0a-5b221122383e%2F0540f304-d03b-4b43-accd-120bf43c06cf%2F7jkiyf9_processed.png&w=3840&q=75)
Transcribed Image Text:### Quality Control in Manufacturing: Variance Analysis of Plastic Sheets
Plastic sheets produced by a machine are periodically monitored for possible fluctuations in thickness. Typically, a variance in thickness around 1.95 square millimeters is regarded as acceptable. As part of the quality control process, a random sample of 30 plastic sheets is taken and the sample variance is calculated to be 3.17 square millimeters. Assume that the population is normally distributed. Let \(\alpha = 0.05\).
#### (a) Constructing a Confidence Interval for Population Variance
To construct a confidence interval for the population variance at \((1 - \alpha)\) confidence level, we need to utilize the chi-square distribution because the sample variance follows a scaled chi-square distribution if the population is normally distributed.
Given:
- Sample size (\(n\)) = 30
- Sample variance (\(s^2\)) = 3.17 square millimeters
- \(\alpha\) = 0.05
The confidence interval for the population variance \(\sigma^2\) is given by:
\[
\left( \frac{(n-1)s^2}{\chi^2_{\alpha/2, n-1}}, \frac{(n-1)s^2}{\chi^2_{1-\alpha/2, n-1}} \right)
\]
Where:
- \(\chi^2_{\alpha/2, n-1}\) and \(\chi^2_{1-\alpha/2, n-1}\) are the critical values from the chi-square distribution with \(n-1\) degrees of freedom.
#### (b) Hypothesis Testing: Upper-Tail Test on Population Variance
To test the hypothesis that the true population variance is larger than 1.95, we perform an upper-tail hypothesis test.
The hypotheses are:
- Null hypothesis (\(H_0\)): \(\sigma^2 \leq 1.95\)
- Alternative hypothesis (\(H_1\)): \(\sigma^2 > 1.95\)
Test statistic:
\[
\chi^2 = \frac{(n-1)s^2}{\sigma_0^2}
\]
Where \(\sigma_0^2\) is the variance under the null hypothesis (1.95 square millimeters).
Given:
- Sample size (\(n\)) = 30
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