91Ó°ÊÓ

The $95\%$ confidence interval for the mean age of all millionaires in the United States is $54.3$ to $62.8$

The formula used:$\left(\stackrel{¯}{x}-z_{\frac{\mathrm{Î\pm }}{2}}\frac{\mathrm{σ}}{\sqrt{n}},\stackrel{¯}{x}+z_{\frac{\mathrm{Î\pm }}{2}}\frac{\mathrm{σ}}{\sqrt{n}}\right)$

Part (C) is the correct interpretation of the given statement, because a $95\%$confidence interval of $\mathrm{μ}$means that if a large number of samples of the same size are obtained from a population with a mean of $\mathrm{μ}$, and if the interval $\left(\stackrel{¯}{x}-z_{\frac{\mathrm{Î\pm }}{2}}\frac{\mathrm{σ}}{\sqrt{n}},\stackrel{¯}{x}+z_{\frac{\mathrm{Î\pm }}{2}}\frac{\mathrm{σ}}{\sqrt{n}}\right)$is used for each sample or $\left(\stackrel{¯}{x}-t_{\frac{\mathrm{Î\pm }}{2}}\frac{s}{\sqrt{n}},\stackrel{¯}{x}+t_{\frac{\mathrm{Î\pm }}{2}}\frac{s}{\sqrt{n}}\right)$When is found, the interval will encompass $\mathrm{μ}$in around $95\%$ of the situations. That is, we are $95\%$ certain that the obtained (confidence) interval will include $\mathrm{μ}$ for a given sample.

Because a $95\%$ confidence interval of $\mathrm{μ}$ does not indicate that $95\%$ of the population units are inside that interval, portion (a) is incorrect.

Both parts (b) and (d) are incorrect interpretations. Because the likelihood (or chance) that the random interval $\left(\stackrel{¯}{x}-z_{\frac{\mathrm{Î\pm }}{2}}\frac{\mathrm{σ}}{\sqrt{n}},\stackrel{¯}{x}+z_{\frac{\mathrm{Î\pm }}{2}}\frac{\mathrm{σ}}{\sqrt{n}}\right)$will include $\mathrm{μ}$is $0.95(95\%)$An obtained confidence interval is a specific observation of the aforesaid random interval for a given sample. That is, the supplied $(54.3,62.8)$confidence interval of $\mathrm{μ}$is an observed (i.e. constant) interval. In addition, $\mathrm{μ}$is a constant population parameter. As a result, claiming that the chance of a constant interval containing a fixed parameter $\mathrm{μ}$is $0.95(or95\%)$is ridiculous, because probability can only be connected with random occurrences.