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.

Step 1
${\displaystyle A100(1-\alpha )\mathrm{\%}}$ confidence interval on the difference is
${\displaystyle {\bar{x}}_1-{\bar{x}}_2-t_{\frac{\alpha }{2},n_1+n_2-1}{s}_p\sqrt{\frac{1}{n_1}+\frac{1}{n_2}}\le {\mu }_1-{\mu }_2\le {\bar{x}}_1-{\bar{x}}_2,+t_{\frac{\alpha }{2},n_1+n_2-1}{s}_p\sqrt{\frac{1}{n_1}+\frac{1}{n_2}}}$
where ${\displaystyle s_p=\sqrt{\frac{(n_1-1)s_1^2+(n_2-1)s_2^2}{n_1+n_2-1}}}$
Step 2
Considering 95% confidence interval,
${\displaystyle \alpha =0.05}$
${\displaystyle {\bar{x}}_1=4.8}$
${\displaystyle {\bar{x}}_2=5.6}$
${\displaystyle s_1^2=8.64}$
${\displaystyle s_2^2=7.88}$
${\displaystyle n_1=10}$
${\displaystyle n_2=10}$
Therefore
${\displaystyle t_{\frac{\alpha }{2},n_1+n_2-1}=t_{\frac{0.05}{2},10+10-1}}$
${\displaystyle =t_{0.025,19}}$
=2.093
Step 3
To calculate ${\displaystyle s_p}$
${\displaystyle s_p=\sqrt{\frac{(n_1-1)s_1^2+(n_2-1)s_2^2}{n_1+n_2-1}}}$
${\displaystyle s_p=(\frac{\sqrt{(10-1)8.64+(10-1)7.88}}{19})}$
${\displaystyle s_p=2.797}$
Step 4
Now, by substituting all the values in the equation, we get
${\displaystyle {\bar{x}}_1-{\bar{x}}_2-t_{\frac{\alpha }{2},n_1+n_2-1}{s}_p\sqrt{\frac{1}{n_1}+\frac{1}{n_2}}\le {\mu }_1-{\mu }_2\le {\bar{x}}_1-{\bar{x}}_2}$

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