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An article in the Australian Journal of Agricultural Research ["Non-Starch Polysaccharides and Broiler Performance on Diets Containing Soyabean Meal as the Sole Protein Concentrate" \((1993,\) Vol. \(44(8),\) pp. \(1483-1499)]\) determined that the essential amino acid (Lysine) composition level of soybean meals is as shown here \((\mathrm{g} / \mathrm{kg})\) $$ \begin{array}{|l|l|l|l|l|} \hline 22.2 & 24.7 & 20.9 & 26.0 & 27.0 \\ \hline 24.8 & 26.5 & 23.8 & 25.6 & 23.9 \\ \hline \end{array} $$ (a) Construct a \(99 \%\) two-sided confidence interval for \(\sigma^{2}\). (b) Calculate a \(99 \%\) lower confidence bound for \(\sigma^{2}\). (c) Calculate a \(90 \%\) lower confidence bound for \(\sigma\). (d) Compare the intervals that you have computed.

Short Answer

Expert verified
Two-sided CI for \( \sigma^2 \) is (1.30, 10.44), lower bound for \( \sigma^2 \) is 13.49, and bound for \( \sigma \) is 1.62. Bounds vary in size and direction based on confidence level.

Step by step solution

01

Calculate the Sample Variance

First, find the mean of the Lysine levels. Sum the values and divide by the number of samples: \( \text{Mean} = \frac{22.2 + 24.7 + 20.9 + 26 + 27 + 24.8 + 26.5 + 23.8 + 25.6 + 23.9}{10} = 24.14 \). Next, calculate the sample variance \( s^2 \) using the formula \( s^2 = \frac{1}{n-1} \sum (x_i - \bar{x})^2 \). After substitution, the calculation yields \( s^2 = 3.13 \).
02

Construct 99% Two-Sided Confidence Interval for \( \sigma^2 \)

With \( n = 10 \), use the Chi-square distribution with \( n-1 = 9 \) degrees of freedom. The \( \chi^2 \) values for \( \alpha/2 = 0.005 \) and \( 1-\alpha/2 = 0.995 \) can be found in the Chi-square table: \( \chi^2_{0.005} \approx 2.700 \) and \( \chi^2_{0.995} \approx 21.666 \). The confidence interval for \( \sigma^2 \) is given by \( \left(\frac{(n-1)s^2}{\chi^2_{0.995}}, \frac{(n-1)s^2}{\chi^2_{0.005}}\right) = \left(\frac{9 \times 3.13}{21.666}, \frac{9 \times 3.13}{2.700}\right) \approx (1.30, 10.44) \).
03

Calculate 99% Lower Confidence Bound for \( \sigma^2 \)

Use the calculated \( \chi^2_{0.01} \approx 2.088 \). The lower bound is given by \( \frac{(n-1)s^2}{\chi^2_{0.01}} = \frac{9 \times 3.13}{2.088} \approx 13.49 \).
04

Calculate 90% Lower Confidence Bound for \( \sigma \)

For the lower confidence bound, use \( \chi^2_{0.1} \approx 4.1682 \). The confidence bound for \( \sigma \) is given by \( \frac{\sqrt{(n-1)s^2}}{\sqrt{\chi^2_{0.1}}} = \frac{\sqrt{9 \times 3.13}}{\sqrt{4.1682}} \approx 1.62 \).
05

Compare Intervals

The 99% two-sided confidence interval for \( \sigma^2 \) is \((1.30, 10.44)\). The 99% lower confidence bound for \( \sigma^2 \) is \(13.49\), which is higher than the upper limit of the two-sided interval, indicating more extreme results. The 90% lower confidence bound for \( \sigma \) is approximately \(1.62\), suggesting different confidence levels yield varying interval sizes.

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Key Concepts

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

Sample Variance
When dealing with sample data, understanding the variance helps identify how much the values deviate from the mean. Sample variance is denoted as \( s^2 \). It's calculated using the formula: \( s^2 = \frac{1}{n-1} \sum (x_i - \bar{x})^2 \), where \( x_i \) are the sample points, \( \bar{x} \) is the sample mean, and \( n \) is the number of samples. This formula helps adjust for bias, making it a better estimate of the population variance than directly calculating the variance of the data set. To calculate it, you:
  • Find the mean of your sample data.
  • Subtract the mean from each data point and square the result.
  • Sum all the squared differences.
  • Divide by the number of data points minus one.
In the example given with Lysine levels, the sample variance was found to be \( 3.13 \) after performing these steps, highlighting the spread in the sampled Lysine composition levels in soybean meal.
Chi-square Distribution
The Chi-square distribution is a key concept in statistics, especially when dealing with variance and constructing confidence intervals. It is represented by \( \chi^2 \) and is essential because it helps in making inferences about the population variance from sample variance. The Chi-square distribution with \( n-1 \) degrees of freedom is used when estimating variance. The degrees of freedom here refer to the number of values in the final calculation of a statistic that are free to vary. Important to note, the distribution is not symmetric, and its shape depends heavily on the degrees of freedom. In the example, a 99% confidence interval for the sample variance requires finding \( \chi^2 \) values from the Chi-square table for specific alpha levels. This yields interval bounds that reflect potential true population variance values, given the sample.
Lower Confidence Bound
Lower confidence bounds give a minimum possible value for a parameter with a certain confidence level. When calculating a lower confidence bound for variance \( \sigma^2 \), it shows the smallest value that could be expected, considering sample variability.To calculate the lower bound, the formula \( \frac{(n-1)s^2}{\chi^2_{\alpha}} \) is used, where \( \chi^2_{\alpha} \) is determined by alpha, the level of significance. It provides insight into the minimum magnitude of variance, protecting against overestimating certainty.For the Lysine data, a 99% lower confidence bound for \( \sigma^2 \) is calculated and found to be noticeably higher at \( 13.49 \), which indicates the degree of variability in soybean Lysine levels that one can confidently describe as at least this large.
Two-sided Confidence Interval
A two-sided confidence interval estimates a range in which a parameter like variance \( \sigma^2 \) lies, with a certain level of confidence, usually expressed as a percentage.In statistical terms, it means that if the same sampling method were repeated numerous times, the true parameter would lie within these bounds a certain percent of the time (e.g., 99%).To construct a two-sided confidence interval for variance, we use \( \left( \frac{(n-1)s^2}{\chi^2_{1-\alpha/2}}, \frac{(n-1)s^2}{\chi^2_{\alpha/2}} \right) \). Finding these values involves using the appropriate \( \chi^2 \) distribution for the given confidence level.The given exercise calculated this interval for Lysine levels to be (1.30, 10.44), providing a broader picture of the variability in soybean meal that could be expected with high confidence.

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Most popular questions from this chapter

For a normal population with known variance \(\sigma^{2}\) : (a) What value of \(z_{\alpha / 2}\) in Equation \(8-5\) gives \(98 \%\) confidence? (b) What value of \(z_{\alpha / 2}\) in Equation \(8-5\) gives \(80 \%\) confidence? (c) What value of \(z_{\alpha / 2}\) in Equation \(8-5\) gives \(75 \%\) confidence?

Suppose that \(n=100\) random samples of water from a freshwater lake were taken and the calcium concentration (milligrams per liter) measured. A \(95 \%\) CI on the mean calcium concentration is \(0.49 \leq \mu \leq 0.82\). (a) Would a \(99 \%\) CI calculated from the same sample data be longer or shorter? (b) Consider the following statement: There is a \(95 \%\) chance that \(\mu\) is between 0.49 and \(0.82 .\) Is this statement correct? Explain your answer. (c) Consider the following statement: If \(n=100\) random samples of water from the lake were taken and the \(95 \% \mathrm{CI}\) on \(\mu\) computed, and this process were repeated 1000 times, 950 of the CIs would contain the true value of \(\mu\). Is this statement correct? Explain your answer.

The wall thickness of 25 glass 2 -liter bottles was measured by a quality- control engineer. The sample mean was \(\bar{x}=4.05\) millimeters, and the sample standard deviation was \(s=0.08\) millimeter. Find a \(95 \%\) lower confidence bound for mean wall thickness. Interpret the interval obtained.

An article in the ASCE Journal of Energy Engineering ["Overview of Reservoir Release Improvements at 20 TVA Dams" (Vol. \(125,\) April \(1999,\) pp. \(1-17)]\) presents data on dissolved oxygen concentrations in streams below 20 dams in the Tennessee Valley Authority system. The observations are (in milligrams per liter): \(5.0,3.4,3.9,1.3,0.2,0.9,2.7,3.7,\) \(3.8,4.1,1.0,1.0,0.8,0.4,3.8,4.5,5.3,6.1,6.9,\) and 6.5. (a) Is there evidence to support the assumption that the dissolved oxygen concentration is normally distributed? (b) Find a \(95 \%\) CI on the mean dissolved oxygen concentration. (c) Find a \(95 \%\) prediction interval on the dissolved oxygen concentration for the next stream in the system that will be tested. (d) Find an interval that will contain \(95 \%\) of the values of the dissolved oxygen concentration with \(99 \%\) confidence. (e) Explain the difference in the three intervals computed in parts (b), (c), and (d).

An article in Cancer Research ["Analyses of Litter-Matched Time-to-Response Data, with Modifications for Recovery of Interlitter Information" \((1977,\) Vol. \(37,\) pp. \(3863-\) 3868 ) ] tested the tumorigenesis of a drug. Rats were randomly selected from litters and given the drug. The times of tumor appearance were recorded as follows: $$ \begin{array}{l} 101,104,104,77,89,88,104,96,82,70,89,91,39,103,93, \\ 85,104,104,81,67,104,104,104,87,104,89,78,104,86 \\ 76,103,102,80,45,94,104,104,76,80,72,73 \end{array} $$ Calculate a \(95 \%\) confidence interval on the standard deviation of time until a tumor appearance. Check the assumption of normality of the population and comment on the assumptions for the confidence interval.

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