# Confidence Interval Probabilty

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.

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 $b$?

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You want to find $b$ such that $P(−b \leq N(0, 1)\leq b) = 0.9$. From the symmetry of the normal density function, this means that $$P(b < N(0,1) < \infty) = P(-\infty < N(0,1) < b) = 0.05$$ since $10\%$ of the area under the normal density function is outside the interval $[-b,b]$. You have a table of values of the cumulative standard normal distribution function, which tells you the value of $$P(-\infty < N(0,1) \leq b) = P(-\infty < N(0,1) < b) + P(−b \leq N(0, 1)\leq b) = 0.05 + 0.9 = 0.95$$ So you look in the table and find that $b = 1.645$ gives a value of $0.9495$, just a little less than the desired $0.9500$, and so the exact value of $b$ is a tad more that $1.645$.

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$0.95$ comes from the fact that $\lambda(b) = 1 - \lambda(-b)$ and that you want $b$ to satisfy $\lambda(b) - \lambda(-b) = 0.9$. The number $b$ is found by looking up in the table of values of $\lambda$.

(I assume that $\lambda$ is the cumulative distribution function of the normal distribution with zero mean and variance $1$.)

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Yes, you assumed right. But how do you know what b is? It says in the answer that b = 1.645. But in the table it shows thats 45% and I still don't know how they came up with 1.645. –  Q.matin Oct 10 '12 at 1:17
I am not sure which table you use, but I found this which shows $b = 1.64$ corresponds to $0.9495$. –  Tunococ Oct 10 '12 at 1:46
No wonder, thank you! sorry for my ignorance, but i still dont know how they knew to look up 95 % . You said 0.95 comes from the fact that λ(b)=1−λ(−b). But how is that what will make λ(b)=1−λ(−b) equal to 95%? –  Q.matin Oct 10 '12 at 2:20
$\lambda(b) = 1 - \lambda(-b)$ is a property of $\lambda$. The confidence interval you want satisfies $\lambda(b) - \lambda(-b) = 0.9$. Solving these two equations for $\lambda(b)$ and $\lambda(-b)$ will give $\lambda(b) = 0.95$ and $\lambda(-b) = 0.05$. –  Tunococ Oct 12 '12 at 18:22

I'm not fully certain what the question is, but for I'll guess that it's something like this: The probability that a normally distributed random variable is less than its expected value plus $1.645$ standard deviations is about $95\%$, so how is that related to the $90\%$ figure mentioned after that?

$$\Pr(Z<1.645) = 0.95.$$ So $$\Pr(Z>1.645) = 0.05.$$ By symmetry, then, $$\Pr(Z<-1.645) = 0.05.$$ So you have $5\%$ of the probability in the upper tail and $5\%$ in the lower tail. That makes $10\%$. So $90\%$ remains in the middle.

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