Question

In an experiment using 30 mice, the sample proportion of the mice that gained weight after a drug injection is 0.65. What is the 99.7% confidence interval for the actual proportion in the population?

Answer 1

The confidence interval for the actual proportion will have the form
\[\hat{p}\pm Z_{\alpha/2}\sqrt{\frac{\hat{p}(1-\hat{p})}{n}}\]
where \(\hat{p}\) is the sample proportion, 0.65, and \(Z_{\alpha/2}\) is the cutoff values, so the speak, for the \(\alpha\) significance level, for which \(1-\alpha=99.7\%\)

The 99.7% CI will thus be
\[\left(0.65-Z_{.0015}\sqrt{\frac{0.65\times0.35}{30}},0.65+Z_{.0015}\sqrt{\frac{0.65\times0.35}{30}}\right)\]
According to this chart ( http://4.bp.blogspot.com/_5u1UHojRiJk/TEh9BHxxPUI/AAAAAAAAAIQ/DafeQNMYFoE/s1600/ztable.gif), \(Z_{.0015}=2.96\), so the CI is
\[\left(0.65-0.2578,0.65+0.2578\right)~~\Rightarrow~~\left(0.3922,0.9078\right)\]

Answer 2

i still don't understand what is the Z for?

Answer 3

I think it's best understood if you have a drawing of a curve:

The shaded area is contained by the confidence interval I found.