# The data, recorded in days, represents the recovery time, for patients who are randomly treated with one of two medications.Find the 99\% confidence interval for \mu1-\mu2, the difference in mean drug recovery times, and INTERPRET it to get a helpful conclusion about the drugs.

The following data, recorded in days, represents the recovery time, for patients who are randomly treated with one of two medications to cure servere bladder infections:

Find the $99\mathrm{%}$ confidence interval for $\mu 1-\mu 2$, the difference in mean drug recovery times, and INTERPRET it to get a helpful conclusion about the drugs.
Assume normal populations, with equal variances.

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Step 1
Given data:
${n}_{1}=13$,
${\stackrel{―}{x}}_{1}=20$
${\sigma }_{1}^{2}=1$,
${\sigma }_{1}=1$
${n}_{2}=16$,
${\stackrel{―}{x}}_{2}=15$,
${\sigma }_{2}^{2}=1.8$
${\sigma }_{2}=1.34$
Confidence level $=99\mathrm{%}$
The formula for confidence interval is:
$C.I.=\stackrel{―}{X}±Z×\frac{\sigma }{\sqrt{n}}$
For the sample 1:
Put the values for sample 1:
$C.I.=\stackrel{―}{X}±Z×\frac{\sigma }{\sqrt{n}}$
$C.I.=20±2.5758×\frac{1}{\sqrt{13}}$
$C.I.=20±0.714$
$C.I.=\left[19.28-2.71\right]$
Step 2
For sample 2 the CI is:
$C.I.=\stackrel{―}{X}±Z×\frac{\sigma }{\sqrt{n}}$
$C.I.=15±2.5758×\frac{1.34}{\sqrt{16}}$
$C.I.=15±0.863$
$C.I.=\left[14.14-15.86\right]$
For ${\mu }_{1}-{\mu }_{2}$:
${\mu }_{1}-{\mu }_{2}={\stackrel{―}{X}}_{1}-{\stackrel{―}{X}}_{2}±Z\left(\sqrt{\frac{{\sigma }_{1}^{2}}{{n}_{1}}-\frac{{\sigma }_{2}^{2}}{{n}_{2}}}\right)$
${\mu }_{1}-{\mu }_{2}=20-15±2.77\left(0.45\right)$
${\mu }_{1}-{\mu }_{2}=5±1.24$
${\mu }_{1}-{\mu }_{2}=\left[3.76-6.24\right]$
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