# A simple random sample is drawn from a population that is known to be normally distributed. The sample variance, s2, is determined to be 12.6 Construct a 90\% confidence interval for \sigma if the sample size, n, is 20 and construct a 90\% confidence interval for \sigma if the sample size, n, is 30.

A simple random sample is drawn from a population that is known to be normally distributed. The sample variance, s2, is determined to be 12.6
A. Construct a $90\mathrm{%}$ confidence interval for $\sigma$ if the sample size, n, is 20. (Hint: use the result obtained from part (a)) 7.94 (LB) & 23.66 (UB))
B. Construct a $90\mathrm{%}$ confidence interval for $\sigma$ if the sample size, n, is 30. (Hint: use the result obtained from part (b)) 8.59 (LB) &amp. 20.63 (UB)
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Step 1
Here $\sigma 2=12.6$ and $c=90\mathrm{%}$ which is 0.9.
Therefore $\frac{1-c}{2}=\frac{1-0.9}{2}=0.05$ and $1-\left(\frac{1-c}{2}\right)=0.95$
Part A
For $n=20$ the degrees of freedom are:
$df=20-1=19$
So the $\chi$ squared table values are:
${X}_{0.95}=10.117$ and ${X}_{0.05}=30.144$
So the upper and lower bounds are given by:
$UB=\left(\frac{19}{10.117}\right)\left(12.6\right)$
$UB=23.66$
$LB=\left(\frac{19}{30.144}\right)\left(12.6\right)$
$LB=7.94$
Part B
For $n=30$ the degrees of freedom are:
$df=30-1=29$
So the $\chi$ squared table values are:
${X}_{0.95}=17.708$ and ${X}_{0.05}=42.557$
So the upper and lower bounds are given by:
$UB=\left(\frac{29}{17.708}\right)\left(12.6\right)$
$UB=20.63$
$LB=\left(\frac{29}{42.557}\right)\left(12.6\right)$
$LB=8.59$