91Ó°ÊÓ

The formula used: The confidence interval$\left(\stackrel{¯}{x}-t_{\frac{a}{2}}×\frac{s}{\sqrt{n}},\stackrel{¯}{x}+t_{\frac{a}{2}}×\frac{s}{\sqrt{n}}\right)$

Let $\mathrm{μ}$population s.d. be the population mean additional sleep hours, and $\mathrm{σ}$be unknown.

Here the sample size$n=10$

We will utilize the $t-$interval approach to find $95\%\mathrm{Cl}$of $\mathrm{μ}$as population s.d., $\mathrm{σ}$is unknown.

According to one mean t-interval technique, the percent $\mathrm{Cl}$of $\mathrm{μ}$is $100(1-\mathrm{Î\pm })\%$

$\left(\stackrel{¯}{x}-t_{\frac{a}{2}}×\frac{s}{\sqrt{n}},\stackrel{¯}{x}+t_{\frac{a}{2}}×\frac{s}{\sqrt{n}}\right)$

Where $t_{\frac{\mathrm{Î\pm }}{2}}$is the $t-$value with area $\frac{\mathrm{Î\pm }}{2}$to its right, $\frac{\mathrm{Î\pm }}{2}$is the sample size, and$n$ is the sample size.

Here confidence level $=95\%$

$$\begin{gathered} =100×0.95\% \\ ∴1-\mathrm{Î\pm }=0.95 \\ ⇒\mathrm{Î\pm }=1-0.95 \\ ⇒\mathrm{Î\pm }=0.05 \\ ⇒\frac{\mathrm{Î\pm }}{2}=0.025 \\ df=n-1 \\ =10-1=9 \\ ∴\text{For}df=9,t_{\mathrm{Î\pm }}=t_{0025}=2.262 \end{gathered}$$

Given that sample mean $\stackrel{¯}{x}=2.33$

Sample S.D $s=2.002$

$$\begin{gathered} ∴95\%\mathrm{Cl}\text{of}\mathrm{μ} \\ =\left(\stackrel{¯}{x}-t_{0.025}\frac{s}{\sqrt{n}},\stackrel{¯}{x}+t_{0.025}\frac{s}{\sqrt{n}}\right) \\ =\left(2.33-2.262×\frac{2.002}{\sqrt{10}},2.33+2.262×\frac{2.002}{\sqrt{10}}\right) \\ =(2.33-1.43,2.33+1.43) \\ =(0.8980,3.7620) \end{gathered}$$

i.e., we are $95\%$certain that the mean additional hours of sleep for all patients using laevohysocymine hydrobromide is between $0.90$ and $3.76$ hours.

We discovered that the mean hour of increased sleep for all patients using laevohysocymine bromide lies between two positive numbers $0.90$ hrs and $3.76$ hrs based on the $95\%$ confidence interval. As a result, we are at least $95\%$ confident that an increase in mean hours of sleep is beneficial. As a result, we can conclude that the medicine was successful in improving sleep.