91Ó°ÊÓ

It is given:

$$\begin{gathered} \stackrel{¯}{x}=332 \\ n=200 \\ s=108 \\ c=0.99 \end{gathered}$$

The degree of freedom will be:

$$\begin{gathered} df=n-1 \\ =200-1 \\ =199 \end{gathered}$$

Now, the value of t will be:

$t_{\mathrm{Î\pm }/2}=2.626$

Now, the confidence interval will be:

$$\begin{gathered} \stackrel{¯}{x}-t_{\mathrm{Î\pm }/2}×\frac{s}{\sqrt{n}} \\ =332-2.626×\frac{108}{\sqrt{200}} \\ =311.8956 \\ =\stackrel{¯}{x}+t_{\mathrm{Î\pm }/2}×\frac{s}{\sqrt{n}} \\ =332+2.626×\frac{108}{\sqrt{200}} \\ =352.1044 \end{gathered}$$

Thus we conclude that we are $99\%$ confident that the true mean increases is between$311.8956and352.1044$.

A completely randomized experiment randomly assigns all subjects to a group.

In this case, the experiment is not completely randomized since we use the subjects before and after the treatment. Thus the experiment is not a completely randomized experiment.

We cannot prove causation if we do not use a completely randomized experiment because it is possible that the causation of the difference is another variable.

For example, inflation could be the cause because everything you buy will be slightly more expensive the next year due to inflation and thus the increase in the amount charged could be due to inflation.