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A $95\%$confidence interval for the population mean weight of the heads of lettuce.

If $\stackrel{¯}{x}$is the sample mean of a random sample of size $n$from a normal population with unknown variance ${\mathrm{σ}}^2$,

$\stackrel{¯}{x}-z_{\frac{\mathrm{Î\pm }}{2}}\frac{\mathrm{σ}}{\sqrt{n}}≤\mathrm{μ}≤\stackrel{¯}{x}+z_{\frac{\mathrm{Î\pm }}{2}}\frac{\mathrm{σ}}{\sqrt{n}}$

where $z_{\frac{\mathrm{Î\pm }}{2}}$is the upper $100\frac{\mathrm{Î\pm }}{2}$percentage point of the standard normal distribution.

The population standard deviation is known to be $\mathrm{σ}=0.2$pounds. A random sample size is$n=20$and a sample mean $\stackrel{¯}{x}=2.2$pounds.

We need find a $95\%$confidence interval estimate for the population mean weight of the heads of lettuce. Therefore,

$\frac{\mathrm{Î\pm }}{2}=\frac{1-0.95}{2}=0.025⇒z_{\frac{\mathrm{Î\pm }}{2}}=z_{0.025}=1.96$

The previous implication was obtained on a probability table for the standard normal distribution.

From (1) and (2) we get

$2.2-1.96\frac{0.2}{\sqrt{20}}≤\mathrm{μ}≤2.2+1.96\frac{0.2}{\sqrt{20}}$$2.2-0.0876≤\mathrm{μ}≤2.2+0.0876$

Therefore,

$EBM=0.0876$

and $95\%\mathrm{Cl}$for $\mathrm{μ}$is

$2.112≤\mathrm{μ}≤2.288$

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