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Chapter 7: Problem 8

Finding Critical Values and Confidence Intervals. In Exercises \(5-8,\) use thegiven information to find the number of degrees of freedom, the criticalvalues \(\mathcal{X}_{L}^{2}\) and \(\mathcal{X}_{R}^{2},\) and the confidenceinterval estimate of \(\boldsymbol{\sigma} .\) The samples are from Appendix\(\boldsymbol{B}\) and it is reasonable to assume that a simple random samplehas been selected from a population with a normal distribution. Heights of Men \(99 \%\) confidence; \(n=153, s=7.10 \mathrm{cm}\)

### Short Answer

Expert verified

Degrees of Freedom: 152. Critical values: \(\mathcal{X}_{L}^{2} = 118.528\) and \(\mathcal{X}_{R}^{2} = 191.036\). Confidence interval for \(\sigma\): \(6.32\) to \(8.02\).

## Step by step solution

01

## Determine Degrees of Freedom

The degrees of freedom (df) can be calculated using the formula: \[ df = n - 1 \] Given: \( n = 153 \) Therefore, \[ df = 153 - 1 = 152 \]

02

## Determine Critical Values

For a 99% confidence interval, the significance level \( \alpha \) is 0.01. The left and right critical values are found from the chi-square distribution table corresponding to \( df = 152 \) and \( \alpha/2 \) and \( 1 - \alpha/2 \) respectively. Thus: \( \alpha/2 = 0.005 \) and \( 1 - \alpha/2 = 0.995 \).From the chi-square distribution table, for \( df = 152 \): \[ \mathcal{X}_{L}^{2} = 118.528 \space (value \space at \space 0.995) \] \[ \mathcal{X}_{R}^{2} = 191.036 \space (value \space at \space 0.005) \]

03

## Calculate Confidence Interval for \(\sigma^2\)

The confidence interval for \(\sigma^2\) can be calculated using the formula: \[ \left( \frac{(n-1)s^2}{\mathcal{X}_{R}^2}, \frac{(n-1)s^2}{\mathcal{X}_{L}^2} \right) \] Given: \( n = 153 \), \( s = 7.10 \mathrm{cm} \), and previously calculated \( df = 152 \) \[ \text{CI}(\sigma^2) = \left( \frac{152 \cdot (7.10)^2}{191.036}, \frac{152 \cdot (7.10)^2}{118.528} \right) \] \[ = \left( \frac{7630.84}{191.036}, \frac{7630.84}{118.528} \right) \] \[ = (39.93, 64.39) \]

04

## Calculate Confidence Interval for \(\sigma\)

Take the square root of the confidence interval limits for \(\sigma^2\) to obtain the confidence interval for \(\sigma\). \[ \text{CI}(\sigma) = \left( \sqrt{39.93}, \sqrt{64.39} \right) \] \[ \approx (6.32, 8.02) \]

## Key Concepts

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

###### degrees of freedom

In statistics, the concept of 'degrees of freedom' is essential in various analyses. It signifies the number of values in a calculation that are free to vary. For example, if you have 153 observations (sample size), the sum of deviations from the mean must be zero. This imposes one constraint, thus reducing the degrees of freedom by one. Therefore, with 153 observations, your degrees of freedom (often denoted as df) is 152. The formula to compute it is: \[ df = n - 1 \] It is a fundamental concept when determining how much independent information you have in your data set. In the given problem, df is calculated as follows:

- Given: n = 153
- df = 153 - 1 = 152

###### chi-square distribution

The chi-square distribution is a key probability distribution in statistics, primarily used for hypothesis testing and constructing confidence intervals for variances and standard deviations. This distribution is defined by the degrees of freedom and varies with the sample size. A chi-square ( \(\chi^2\) ) distribution is not symmetric, and its shape depends on the degrees of freedom. In the provided exercise, two critical values from the chi-square distribution table are required: one for the 0.005 percentile and the other for the 0.995 percentile, specifically for 152 degrees of freedom. These values represent the tails where a very small amount of the data (0.5%) lies. Here are the values:

- Left critical value ( \(\mathcal{X}_{L}^{2}\) ): 118.528
- Right critical value ( \(\mathcal{X}_{R}^{2}\) ): 191.036

These values serve as critical points in our confidence interval estimation.

###### confidence interval

The confidence interval provides a range of values within which we can expect a population parameter to lie with a certain level of confidence. To estimate a confidence interval for the standard deviation \(\sigma\) , the problem utilizes the sample size, standard deviation, degrees of freedom, and chi-square critical values. The steps to compute the confidence interval for \(\sigma\) are:

- Calculate the confidence interval for the variance \(\sigma^2\) using the formula: \(\left( \frac{(n-1)s^2}{\mathcal{X}_{R}^2}, \frac{(n-1)s^2}{\mathcal{X}_{L}^2} \right)\)
- For \( n = 153, s = 7.10 \mathrm{cm}, \text{df} = 152 \)
- \(\text{CI}(\sigma^2) = \left( \frac{152 \cdot (7.10)^2}{191.036}, \frac{152 \cdot (7.10)^2}{118.528} \right) \)
- \( = \left( \frac{7630.84}{191.036}, \frac{7630.84}{118.528} \right) \)
- \( = (39.93, 64.39) \)

Finally, take the square root of these bounds to get the confidence interval for \(\sigma\):

- \( \text{CI}(\sigma) = \left( \sqrt{39.93}, \sqrt{64.39} \right) \)
- \( \approx (6.32, 8.02) \)

Thus, with 99% confidence, the population standard deviation \(\sigma \) is estimated to lie between 6.32 cm and 8.02 cm.

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