Let two independent random samples, each of size 10, from two normal distributions N(mu_1, sigma_2) and N(mu_2, sigma_2) yield x = 4.8, s_1^2 = 8.64, y = 5.6, s_2^2 = 7.88. Find a 95% confidence interval for mu_1 − mu_2.

Let two independent random samples, each of size 10, from two normal distributions $N\left({\mu }_{1},{\sigma }_{2}\right)\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}N\left({\mu }_{2},{\sigma }_{2}\right)$ yield $x=4.8,{s}_{1}^{2}$
$=8.64,y=5.6,{s}_{2}^{2}$
= 7.88.
Find a 95% confidence interval for ${\mu }_{1}-{\mu }_{2}$.
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Step 1
$A100\left(1-\alpha \right)\mathrm{%}$ confidence interval on the difference is
${\stackrel{―}{x}}_{1}-{\stackrel{―}{x}}_{2}-{t}_{\frac{\alpha }{2},{n}_{1}+{n}_{2}-1}{s}_{p}\sqrt{\frac{1}{{n}_{1}}+\frac{1}{{n}_{2}}}\le {\mu }_{1}-{\mu }_{2}\le {\stackrel{―}{x}}_{1}-{\stackrel{―}{x}}_{2},+{t}_{\frac{\alpha }{2},{n}_{1}+{n}_{2}-1}{s}_{p}\sqrt{\frac{1}{{n}_{1}}+\frac{1}{{n}_{2}}}$
where ${s}_{p}=\sqrt{\frac{\left({n}_{1}-1\right){s}_{1}^{2}+\left({n}_{2}-1\right){s}_{2}^{2}}{{n}_{1}+{n}_{2}-1}}$
Step 2
Considering 95% confidence interval,
$\alpha =0.05$
${\stackrel{―}{x}}_{1}=4.8$
${\stackrel{―}{x}}_{2}=5.6$
${s}_{1}^{2}=8.64$
${s}_{2}^{2}=7.88$
${n}_{1}=10$
${n}_{2}=10$
Therefore
${t}_{\frac{\alpha }{2},{n}_{1}+{n}_{2}-1}={t}_{\frac{0.05}{2},10+10-1}$
$={t}_{0.025,19}$
=2.093
Step 3
To calculate ${s}_{p}$
${s}_{p}=\sqrt{\frac{\left({n}_{1}-1\right){s}_{1}^{2}+\left({n}_{2}-1\right){s}_{2}^{2}}{{n}_{1}+{n}_{2}-1}}$
${s}_{p}=\left(\frac{\sqrt{\left(10-1\right)8.64+\left(10-1\right)7.88}}{19}\right)$
${s}_{p}=2.797$
Step 4
Now, by substituting all the values in the equation, we get
${\stackrel{―}{x}}_{1}-{\stackrel{―}{x}}_{2}-{t}_{\frac{\alpha }{2},{n}_{1}+{n}_{2}-1}{s}_{p}\sqrt{\frac{1}{{n}_{1}}+\frac{1}{{n}_{2}}}\le {\mu }_{1}-{\mu }_{2}\le {\stackrel{―}{x}}_{1}-{\stackrel{―}{x}}_{2}$