Problem 8 Finding Critical Values and Conf... [FREE SOLUTION] (2024)

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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

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.

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.