Question

# 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.

Confidence intervals
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 $$\displaystyle{90}\%$$ confidence interval for $$\displaystyle\sigma$$ if the sample size, n, is 20. (Hint: use the result obtained from part (a)) 7.94 (LB) & 23.66 (UB))
B. Construct a $$\displaystyle{90}\%$$ confidence interval for $$\displaystyle\sigma$$ if the sample size, n, is 30. (Hint: use the result obtained from part (b)) 8.59 (LB) &amp. 20.63 (UB)

2021-08-17

Step 1
Here $$\displaystyle\sigma{2}={12.6}$$ and $$\displaystyle{c}={90}\%$$ which is 0.9.
Therefore $$\displaystyle{\frac{{{1}-{c}}}{{{2}}}}={\frac{{{1}-{0.9}}}{{{2}}}}={0.05}$$ and $$\displaystyle{1}-{\left({\frac{{{1}-{c}}}{{{2}}}}\right)}={0.95}$$
Part A
For $$\displaystyle{n}={20}$$ the degrees of freedom are:
$$\displaystyle{d}{f}={20}-{1}={19}$$
So the $$\displaystyle\chi$$ squared table values are:
$$\displaystyle{X}_{{{0.95}}}={10.117}$$ and $$\displaystyle{X}_{{{0.05}}}={30.144}$$
So the upper and lower bounds are given by:
$$\displaystyle{U}{B}={\left({\frac{{{19}}}{{{10.117}}}}\right)}{\left({12.6}\right)}$$
$$\displaystyle{U}{B}={23.66}$$
$$\displaystyle{L}{B}={\left({\frac{{{19}}}{{{30.144}}}}\right)}{\left({12.6}\right)}$$
$$\displaystyle{L}{B}={7.94}$$
Part B
For $$\displaystyle{n}={30}$$ the degrees of freedom are:
$$\displaystyle{d}{f}={30}-{1}={29}$$
So the $$\displaystyle\chi$$ squared table values are:
$$\displaystyle{X}_{{{0.95}}}={17.708}$$ and $$\displaystyle{X}_{{{0.05}}}={42.557}$$
So the upper and lower bounds are given by:
$$\displaystyle{U}{B}={\left({\frac{{{29}}}{{{17.708}}}}\right)}{\left({12.6}\right)}$$
$$\displaystyle{U}{B}={20.63}$$
$$\displaystyle{L}{B}={\left({\frac{{{29}}}{{{42.557}}}}\right)}{\left({12.6}\right)}$$
$$\displaystyle{L}{B}={8.59}$$