# Calculate the confidence intervals for the ratio of the two population variances and the ratio of standard deviations. Suppose the samples are simple

Calculate the confidence intervals for the ratio of the two population variances and the ratio of standard deviations. Suppose the samples are simple random samples taken from normal populations.
a. $\alpha =0.05,{n}_{1}=30,{s}_{1}=16.37,{n}_{2}=39,{s}_{2}=9.88,$
b. $\alpha =0.01,{n}_{1}=25,{s}_{1}=5.2,{n}_{2}=20,{s}_{2}=6.8$
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Step 1
a)
Given:
Sample size, ${n}_{1}=30$
Sample size, ${n}_{2}=39$
Sample standard deviation 1, ${s}_{1}=16.37$
Sample standard deviation 2, ${s}_{2}=9.88$
Let's calculate $95\mathrm{%}$ confidence interval for the ratio of two population variances.
$CI=\left(\frac{{s}_{1}^{2}}{{s}_{2}^{2}}{F}_{1-\alpha \text{/}2,{n}_{2}-1,{n}_{1}-1},\frac{{s}_{1}^{2}}{{s}_{2}^{2}}{F}_{\alpha \text{/}2,{n}_{2}-1,{n}_{1}-1}\right)$
$CI=\left(\frac{{16.37}^{2}}{{9.88}^{2}}{F}_{1-0.05\text{/}2,39-1,30-1},\frac{{16.37}^{2}}{{9.88}^{2}}{F}_{0.05\text{/}2,39-1,30-1}\right)$
$CI=\left(\frac{{16.37}^{2}}{{9.88}^{2}}{F}_{0.95\text{/}2,38,29},\frac{{16.37}^{2}}{{9.88}^{2}}{F}_{0.05\text{/}2,38,29}\right)$
Using the critical value table,
$CI=\left(\frac{{16.37}^{2}}{{9.88}^{2}}×0.507,\frac{{16.37}^{2}}{{9.88}^{2}}2.038\right)$
$CI=\left(1.3919,5.5949\right)$
Thus, it is $95\mathrm{%}$ confidence that the true ratio of population variances lies in the interval (1.3919, 5.5949).
Step 2
b) Given:
Sample size, ${n}_{1}=25$
Sample size, ${n}_{2}=20$
Sample standard deviation 1, ${s}_{1}=5.2$
Sample standard deviation 2, ${s}_{2}=6.8$
Let's calculate $99\mathrm{%}$ confidence interval for the ratio of two population variances.
$CI=\left(\frac{{s}_{1}^{2}}{{s}_{2}^{2}}{F}_{1-\alpha \text{/}2,{n}_{2}-1,{n}_{1}-1},\frac{{s}_{1}^{2}}{{s}_{2}^{2}}{F}_{\alpha \text{/}2,{n}_{2}-1,{n}_{1}-1}\right)$
$CI=\left(\frac{{5.2}^{2}}{{6.8}^{2}}{F}_{1-0.01\text{/}2,39-1,30-1},\frac{{5.2}^{2}}{{6.8}^{2}}{F}_{0.01\text{/}2,39-1,30-1}\right)$