91Ó°ÊÓ

For any normally distributed variable $x$with mean ${\mathrm{μ}}_x$ and standard deviation ${\mathrm{σ}}_x$ probability that the value of the variable will be within the interval.

$\left({\mathrm{μ}}_x-Z_{\frac{\mathrm{Î\pm }}{2}}×{\mathrm{σ}}_x,{\mathrm{μ}}_x+Z_{\frac{\mathrm{Î\pm }}{2}}×{\mathrm{σ}}_x\right)$, is exactly equal to $1-\mathrm{Î\pm }$i.e.,

$P\left[{\mathrm{μ}}_x-Z_{\frac{a}{2}}×{\mathrm{σ}}_x≤x≤{\mathrm{μ}}_x+Z_{\frac{a}{2}}×{\mathrm{σ}}_x\right]=1-\mathrm{Î\pm }$

Where $Z_{\frac{a}{2}}=$upper $\frac{\mathrm{Î\pm }}{2}$point of standard normal distribution

Now we know that sample mean follows normal distribution with mean $\mathrm{μ}$and S.D ${\mathrm{σ}}_{\stackrel{¯}{x}}=\frac{\mathrm{σ}}{\sqrt{n}}$for a normally distributed population variable.

Where population mean and $S.D$ are $\mathrm{μ}$ and $\mathrm{σ}$ respectively

$∴$$(*)$ Holds for normally distributed sample mean $\stackrel{¯}{x}$

i.e., $P\left[\mathrm{μ}-Z_{\frac{\mathrm{Î\pm }}{2}}×{\mathrm{σ}}_{\stackrel{¯}{x}}≤\stackrel{¯}{x}≤\mathrm{μ}+Z_{\frac{\mathrm{Î\pm }}{2}}×{\mathrm{σ}}_{\stackrel{¯}{x}}\right]=1-\mathrm{Î\pm }$

Now, $\mathrm{μ}-Z_{\frac{a}{2}}×{\mathrm{σ}}_{\stackrel{¯}{x}}≤\stackrel{¯}{x}≤\mathrm{μ}+Z_{\frac{a}{2}}×{\mathrm{σ}}_{\stackrel{¯}{x}}$

$⇒-\left(\mathrm{μ}+Z_{\frac{\mathrm{Î\pm }}{2}}×{\mathrm{σ}}_{\stackrel{¯}{x}}\right)≤-\stackrel{¯}{x}≤-\left(\mathrm{μ}-Z_{\frac{\mathrm{Î\pm }}{2}}×{\mathrm{σ}}_{\stackrel{¯}{x}}\right)$

[Multiplying by $-1$]

$$\begin{gathered} ⇒-\mathrm{μ}-Z_{\frac{a}{2}}×{\mathrm{σ}}_{\stackrel{¯}{x}}≤-\stackrel{¯}{x}≤-\mathrm{μ}+Z_{\frac{a}{2}}×{\mathrm{σ}}_{\stackrel{¯}{x}} \\ ⇒-\mathrm{μ}-Z_{\frac{a}{2}}×{\mathrm{σ}}_{\stackrel{¯}{x}}+\mathrm{μ}≤-\stackrel{¯}{x}+\mathrm{μ}≤-\mathrm{μ}+Z_{\frac{a}{2}}×{\mathrm{σ}}_{\stackrel{¯}{x}}+\mathrm{μ} \end{gathered}$$

[Adding $\mathrm{μ}$with each term in the inequality]

$$\begin{gathered} ⇒-Z_{\frac{a}{2}}×{\mathrm{σ}}_{\stackrel{¯}{x}}≤-\stackrel{¯}{x}+\mathrm{μ}≤Z_{\frac{\mathrm{Î\pm }}{2}}×{\mathrm{σ}}_{\stackrel{¯}{x}} \\ ⇒\stackrel{¯}{x}-Z_{\frac{a}{2}}·{\mathrm{σ}}_{\stackrel{¯}{x}}≤-\stackrel{¯}{x}+\mathrm{μ}+\stackrel{¯}{x}≤\stackrel{¯}{x}+Z_{\frac{a}{2}}·{\mathrm{σ}}_{\stackrel{¯}{x}} \end{gathered}$$

[Adding $\stackrel{¯}{x}$with each term]

$$\begin{gathered} ⇒\stackrel{¯}{x}-Z_{\frac{a}{2}}·{\mathrm{σ}}_{\stackrel{¯}{x}}≤\mathrm{μ}≤\stackrel{¯}{x}+Z_{\frac{a}{2}}·{\mathrm{σ}}_{\stackrel{¯}{x}} \\ ⇒\stackrel{¯}{x}-Z_{\frac{a}{2}}·\frac{\mathrm{σ}}{\sqrt{n}}≤\mathrm{μ}≤\stackrel{¯}{x}+Z_{\frac{a}{2}}·\frac{\mathrm{σ}}{\sqrt{n}}\left[âˆ\mu {\mathrm{σ}}_{\stackrel{¯}{x}}=\frac{\mathrm{σ}}{\sqrt{n}}\right] \\ ∴P\left[\mathrm{μ}-Z_{\frac{a}{2}}·{\mathrm{σ}}_{\stackrel{¯}{x}}≤\stackrel{¯}{x}≤\mathrm{μ}+Z_{\frac{a}{2}}·{\mathrm{σ}}_{\stackrel{¯}{x}}\right]=1-\mathrm{Î\pm } \\ ⇒P\left[\stackrel{¯}{x}-Z_{\frac{a}{2}}·{\mathrm{σ}}_{\stackrel{¯}{x}}≤\mathrm{μ}≤\stackrel{¯}{x}+Z_{\frac{a}{2}}·{\mathrm{σ}}_{\stackrel{¯}{x}}\right]=1-\mathrm{Î\pm } \\ ⇒P\left[\stackrel{¯}{x}-Z_{\frac{a}{2}}·\frac{\mathrm{σ}}{\sqrt{n}}≤\mathrm{μ}≤\stackrel{¯}{x}+Z_{\frac{a}{2}}·\frac{\mathrm{σ}}{\sqrt{n}}\right]=1-\mathrm{Î\pm } \end{gathered}$$

i.e., the likelihood that the population means will fall inside the range $\left(\stackrel{¯}{x}-Z_{\frac{a}{2}}·\frac{\mathrm{σ}}{\sqrt{n}},\stackrel{¯}{x}+Z_{\frac{\mathrm{Î\pm }}{2}}·\frac{\mathrm{σ}}{\sqrt{n}}\right)$represents the $100(1-\mathrm{Î\pm })\%$confidence interval of the population mean $\mathrm{μ}$and it is exactly equal to $1-\mathrm{Î\pm }$We assume the normality of the variable under consideration to get the confidence level of the confidence interval exactly equal to $1-\mathrm{Î\pm }$

We know that the sample means for any non-normal population variable follows an essentially normal distribution if and only if the samples are big.

Therefore, probability that sample mean will be within the interval $\left(\mathrm{μ}-Z_{\frac{a}{2}}×\frac{\mathrm{σ}}{\sqrt{n}},\mathrm{μ}+Z_{\frac{a}{2}}×\frac{\mathrm{σ}}{\sqrt{n}}\right)$is approximately equal to $1-\mathrm{Î\pm }$

As a result, the confidence level for the confidence interval $\left(\stackrel{¯}{x}-Z_{\frac{\mathrm{Î\pm }}{2}}×\frac{\mathrm{σ}}{\sqrt{n}},\stackrel{¯}{x}+Z_{\frac{\mathrm{Î\pm }}{2}}×\frac{\mathrm{σ}}{\sqrt{n}}\right)$of the population mean $\mathrm{μ}$will be about equal to $1-\mathrm{Î\pm }$

As a result, regardless of the distribution of the variable under investigation, the assumption of large samples produces approximately correct confidence intervals of $\mathrm{μ}$