1) We have discovered a new method to produce a liquid containing Zinc. We measure the concentration μ four times independently (from the same solution) and find the values
0.3, 0.33, 0.3, 0.27
a) Assuming that the standard deviation of the measurement error is 0.02 (when we make one measurement) estimate μ and give a 90%-confidence interval.
Answer:
We assume the measurement errors to be normal with zero expectation. The estimate for μ is
$μ_X = (X_1 + X_2 + X_3 + X_4)/4 = (0.3 + 0.33 + 0.3 + 0.27)/4 = 0.3.$
The same procedure as usual leads to a confidence interval
$[0.3 - b_\text{standard deviation}/\sqrt{n},\, 0.3 + b_\text{standard deviation}/\sqrt{n}]$
where the standard deviation equals $0.02$ and $\sqrt{n} = 2$, whilst $b$ is defined by the equation
$P(−b \leq N(0, 1)\leq b) = 90\%.$
The last equation above is equivalent to $\lambda(b) = 0.95$ which we can solve with the help of a table and find $b = 1.645$. Hence the $90\%$ confidence interval is equal to $[0.3 − 0.0164, 0.3 + 0.0164] = [0.2836, 0.3164]$. The true concentration is thus between $0.2836$ and $0.3164$ with at least $95\%$ probability.
My question is how were they able to get $95\%$ probability and $b$?