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The ages of a random sample of five university professors are \(39,54,61,72,\) and \(59 .\) Using this information, find a \(99 \%\) confidence interval for the population standard deviation of the ages of all professors at the university, assuming that the ages of university professors are normally distributed.

Short Answer

Expert verified
99% CI for the population standard deviation is approximately \((5.96, 34.56)\).

Step by step solution

01

Calculate the Sample Size (n) and Degrees of Freedom (df)

The sample size \( n \) is the number of data points in our sample. Here, we have 5 ages: \( 39, 54, 61, 72, 59 \). Therefore, \( n = 5 \).The degrees of freedom \( df \) for estimating the standard deviation in a sample is calculated as \( df = n - 1 \). In this case, \( df = 5 - 1 = 4 \).
02

Compute the Sample Mean

The sample mean \( \bar{x} \) is calculated by summing all the sample values and dividing by the sample size \( n \).\[ \bar{x} = \frac{39 + 54 + 61 + 72 + 59}{5} = \frac{285}{5} = 57 \]
03

Calculate the Sample Variance

The sample variance \( s^2 \) is given by the formula:\[ s^2 = \frac{\sum_{i=1}^{n}(x_i - \bar{x})^2}{n-1} \]Calculate the deviations and their squares:- \( (39 - 57)^2 = 324 \)- \( (54 - 57)^2 = 9 \)- \( (61 - 57)^2 = 16 \)- \( (72 - 57)^2 = 225 \)- \( (59 - 57)^2 = 4 \)Sum of squared deviations: \( 324 + 9 + 16 + 225 + 4 = 578 \)Now calculate the sample variance:\[ s^2 = \frac{578}{4} = 144.5 \]
04

Determine the Chi-Square Values

For a 99% confidence interval and \( df = 4 \), find the chi-square critical values using a chi-square distribution table:- \( \chi^2_{0.005,4} = 0.484 \)- \( \chi^2_{0.995,4} = 16.264 \)
05

Compute the Confidence Interval for Standard Deviation

The confidence interval for the population standard deviation \( \sigma \) is given by:\[ \left( \sqrt{\frac{(n-1) \cdot s^2}{\chi^2_{\text{upper}}}}, \sqrt{\frac{(n-1) \cdot s^2}{\chi^2_{\text{lower}}}} \right) \]Substituting the known values:- \( \chi^2_{\text{lower}} = 0.484 \)- \( \chi^2_{\text{upper}} = 16.264 \)- \( n-1 = 4 \)- \( s^2 = 144.5 \)Calculate the interval:\[ \left( \sqrt{\frac{4 \times 144.5}{16.264}}, \sqrt{\frac{4 \times 144.5}{0.484}} \right) = \left( \sqrt{35.542}, \sqrt{1194.214} \right) \approx (5.96, 34.56) \]
06

Conclusion: State the Confidence Interval

The 99% confidence interval for the population standard deviation of the ages of all professors at the university is approximately \((5.96, 34.56)\).

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Key Concepts

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

Population Standard Deviation
The population standard deviation is a crucial statistic that measures the spread of individual data points in a population around the mean. It provides insight into how much variation exists within a set of data. To understand this concept, imagine you have a large group of data points, like ages or test scores. If the data points are close to the mean, the standard deviation is small. Conversely, if they are spread out, the standard deviation is larger.

In the context of our exercise, the population standard deviation is what we are attempting to estimate through the sample of university professors' ages. Unlike the sample standard deviation, which applies to just a subset of the population, the population standard deviation is a parameter of the entire dataset we want to understand better. Using a confidence interval allows us to infer an estimate for this population standard deviation, even when we cannot measure it directly from the entire population.
Degrees of Freedom
Degrees of freedom refer to the number of values in a statistical calculation that are free to vary. In simpler terms, it is the number of independent values or quantities which can be assigned to a statistical distribution. When it comes to estimating parameters like the variance of a sample, degrees of freedom play a vital role.

In our exercise, since we’re dealing with samplings from the population, the degrees of freedom help adjust our estimate to become more reliable. They are calculated by taking the sample size and subtracting one:
  • In the exercise: Sample size, \( n = 5 \)
  • Degrees of freedom, \( df = n - 1 = 5 - 1 = 4 \)
The concept of degrees of freedom ensures that statistical estimates account for estimating one parameter, the mean, which requires us to use this adjusted calculation.
Chi-Square Distribution
The Chi-Square distribution is a fundamental concept when it comes to statistical inference, especially related to variances. It is a family of distributions that are defined by the degrees of freedom. Its importance arises because it helps in deciding whether or not a specific variance estimate is consistent with a distribution of sample means under the assumption of a normally distributed dataset.

In our exercise, we utilize the Chi-Square distribution to determine the critical values necessary for constructing a confidence interval for the population standard deviation. These values provide insight into how variance within our sample compares and helps identify the range we expect the population variance to be within.

The Chi-Square critical values we use are derived from statistical tables based on our calculated degrees of freedom. Thus, understanding this sor t of distribution aids in accurately making predictions and estimations on the larger population parameters with confidence.
Sample Variance
Sample variance is pivotal because it evaluates the variability within a sample. By calculating the sample variance, we can estimate how much the ages of the university professors in our sample differ from their mean age. Numerically, the variance is represented as the square of the sample standard deviation, and it plays an essential role in the computation of confidence intervals for the population parameter.

Here's how we derive it from a sample:
  • Subtract each value in the sample from the mean, resulting in the deviations
  • Square these deviations to eliminate negative values
  • Sum these squared deviations
  • Finally, divide by the degrees of freedom \( (n-1) \)
The computation in our exercise shows that the sample variance is 144.5, demonstrating the spread of ages in our sample of professors. Knowing the sample variance allows us to proceed with confidence interval computations on the population variance and standard deviation, further aided by statistical distributions like Chi-Square.

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