Question

a statistics professor was analyzing the grades on the final exam for statistics students on his campus. the professor randomly selects 50 statistics students across campus and calculates the standard deviation for their final exam grades to be 10.68 points. identify the correct formula and/or calculator function that would be used to construct a confidence interval for the standard deviation of final exam grades for statistics students across his campus.
formula: \\( \hat { p } \pm z _ { \frac { a } { 2 } } \sqrt { \frac { \hat { p } ( 1 - \hat { p } ) } { n } } \\)
calculator function: 1-propzint
formula: \\( \overline { x } \pm t _ { \frac { a } { 2 } } \frac { s } { \sqrt { n } } \\)
calculator function: t-interval
formula: \\( l b = \sqrt { \frac { ( n - 1 ) s ^ { 2 } } { x ^ { 2 } _ { \frac { a } { 2 } } } } \\)
and
\\( u b = \sqrt { \frac { ( n - 1 ) s ^ { 2 } } { x ^ { 2 } _ { 1 - \frac { a } { 2 } } } } \\)
formula:
\\( l b = \frac { ( n - 1 ) s ^ { 2 } } { x ^ { 2 } _ { \frac { a } { 2 } } } \\)
and
\\( u b = \frac { ( n - 1 ) s ^ { 2 } } { x ^ { 2 } _ { 1 - \frac { a } { 2 } } } \\)

Explanation:

1. First option analysis:

• The formula \(\hat{p}\pm z_{\frac{\alpha}{2}}\sqrt{\frac{\hat{p}(1 - \hat{p})}{n}}\) and the calculator function \(1 - \text{PropZInt}\) are used for confidence intervals of proportions. Since we are dealing with standard deviation (not a proportion), this option is incorrect.

1. Second option analysis:

• The formula \(\bar{x}\pm t_{\frac{\alpha}{2}}\frac{s}{\sqrt{n}}\) and the calculator function \(T - \text{Interval}\) are used for confidence intervals of the population mean (when the population standard deviation is unknown). We are interested in the standard deviation, so this option is wrong.

1. Third option analysis:

• The formula for the lower bound \(LB=\sqrt{\frac{(n - 1)s^{2}}{\chi_{\frac{\alpha}{2}}^{2}}}\) and upper - bound \(UB=\sqrt{\frac{(n - 1)s^{2}}{\chi_{1-\frac{\alpha}{2}}^{2}}}\) is the correct formula for a confidence interval of the population standard deviation \(\sigma\). The reason is that when constructing a confidence interval for the population standard deviation \(\sigma\) (assuming the population is normally distributed), we use the chi - square distribution \(\chi^{2}\). The formula is derived from the fact that \(\frac{(n - 1)s^{2}}{\sigma^{2}}\sim\chi^{2}(n - 1)\) (chi - square distribution with \(n - 1\) degrees of freedom).

1. Fourth option analysis:

• The formula \(LB=\frac{(n - 1)s^{2}}{\chi_{\frac{\alpha}{2}}^{2}}\) and \(UB=\frac{(n - 1)s^{2}}{\chi_{1-\frac{\alpha}{2}}^{2}}\) gives a confidence interval for the population variance \(\sigma^{2}\), not the standard deviation \(\sigma\).

Answer:

The third option (Formula: \(LB=\sqrt{\frac{(n - 1)s^{2}}{\chi_{\frac{\alpha}{2}}^{2}}}\) and \(UB=\sqrt{\frac{(n - 1)s^{2}}{\chi_{1-\frac{\alpha}{2}}^{2}}}\)) is the correct one.