In constructing a confidence interval for $\sigma$ or $\sigma ^ { 2 }$, a table is used to find the critical values $\chi _ { L } ^ { 2 }$ and $\chi \underset { R } { 2 }$ for values of $n \leq 101$. For larger values of $n , \chi _ { L } ^ { 2 }$ and $\chi { } _ { \mathrm { R } } ^ { 2 }$ can be approximated by using the following formula: $\chi ^ { 2 } = \frac { 1 } { 2 } [ \pm \mathrm { z } \alpha / 2 + \sqrt { 2 \mathrm { k } - 1 } ] ^ { 2 }$ where $\mathrm { k }$ is the number of degrees of freedom and $\mathrm { z } \alpha / 2$ is the critical z score. Construct the $90 \%$ confidence interval for $\sigma$ using the following sample data: a sample of size $\mathrm { n } = 234$ yields a mean weight of $155 \mathrm { lb }$ and a standard deviation of $26.0 \mathrm { lb }$. Round the confidence interval limits to the nearest hundredth.
A) $24.13 \mathrm { lb }

Question

In constructing a confidence interval for $\sigma$ or $\sigma ^ { 2 }$, a table is used to find the critical values $\chi _ { L } ^ { 2 }$ and $\chi \underset { R } { 2 }$ for values of $n \leq 101$. For larger values of $n , \chi _ { L } ^ { 2 }$ and $\chi { } _ { \mathrm { R } } ^ { 2 }$ can be approximated by using the following formula: $\chi ^ { 2 } = \frac { 1 } { 2 } [ \pm \mathrm { z } \alpha / 2 + \sqrt { 2 \mathrm { k } - 1 } ] ^ { 2 }$ where $\mathrm { k }$ is the number of degrees of freedom and $\mathrm { z } \alpha / 2$ is the critical z score. Construct the $90 \%$ confidence interval for $\sigma$ using the following sample data: a sample of size $\mathrm { n } = 234$ yields a mean weight of $155 \mathrm { lb }$ and a standard deviation of $26.0 \mathrm { lb }$. Round the confidence interval limits to the nearest hundredth.
A) $24.13 \mathrm { lb } < \sigma < 28.11 \mathrm { lb }$
B) $24.18 \mathrm { lb } < \sigma < 28.18 \mathrm { lb }$
C) $24.57 \mathrm { lb } < \sigma < 27.67 \mathrm { lb }$
D) $23.86 \mathrm { lb } < \sigma < 28.63 \mathrm { lb }$

Quizplus · Accepted Answer

The formula for the confidence interval for $\sigma$ uses the chi-squared distribution. For large $n$, the approximation formula is used. With $n = 234$, the degrees of freedom $k = 233$. Using the approximation formula and the critical z-score for a 90% confidence interval, the calculated interval is closest to option B.

In Constructing a Confidence Interval For σ\sigmaσ Or σ2\sigma ^ { 2 }σ2

In Constructing a Confidence Interval For $\sigma$ Or $\sigma ^ { 2 }$