Construct a $95\%$ confidence interval for the population mean weight of the heads of lettuce. State the confidence interval, sketch the graph and calculate the error bound.

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

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

$\overline{x}-z_{\frac{\alpha }{2}}\frac{\sigma }{\sqrt{n}}\le \mu \le \overline{x}+z_{\frac{\alpha }{2}}\frac{\sigma }{\sqrt{n}}$

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

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

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

$\frac{\alpha }{2}=\frac{1-0.95}{2}=0.025\Rightarrow z_{\frac{\alpha }{2}}=z_{0.025}=1.96$

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}}\le \mu \le 2.2+1.96\frac{0.2}{\sqrt{20}}$$2.2-0.0876\le \mu \le 2.2+0.0876$

Therefore,

$EBM=0.0876$

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

$2.112\le \mu \le 2.288$

The graph for this problem: