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Similarly, for a 95 % confidence level, find where
\begin{equation*} F(z_{ \frac{0.05}{2}}) = P( z \lt z_{ \frac{0.05}{2}}) = 0.975 = 1 - \frac{0.05}{2} . \end{equation*}
The calculators InvNorm(0.975) gives
\(z_{ 0.025} \approx 1.960\text{.}\)
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