I'm working on a few confidence interval questions, each with slight differences that make it very hard to determine what to use.
One question asks:
An electric scale gives a reading equal to the true weight plus a random
error that is normally distributed with mean 0 mg and standard deviation = 0.1 mg. Suppose
that the results of five successive weighings (in mg) of the same object are as follows:Heading 3.142, 3.163, 3.155, 3.150, 3.141 .
a) Compute a 95 percent confidence interval estimate of the true weight.
b) Compute a 99 percent confidence interval estimate of the true weight.
So here, the true mean is known. The standard deviation is known and all stats associated with the sample is known. What formula is used in this situation? The student-t? I tried but it's long and I think there's a faster way.
The other question asks:
Each of 16 science students independently measured the melting point of
lead. The sample mean of these measurements was 330.2 degrees centigrade.a) If the standard deviation of such measurements is known to be 14, find a 99 percent two sided confidence interval estimate of the true melting point of lead.
b) Suppose that the population variance is not known in advance. If the sample standard
deviation is 15.4 degrees centigrade, compute a 99 percent two-sided confidence interval
of the true melting point of lead.
In one situation we have the Standard Deviation, in the other we only have the sample standard deviation. What is used here?
Best Answer
First question: Imagine doing the weighing $5$ times. Let random variables $X_1,\dots, X_5$ be the results. Now look at the random variable $Y=(X_1+\cdots+X_5)/5$.
This random variable is normal, with mean the true mean, and standard deviation equal to $\dfrac{1}{\sqrt{5}}$ times the standard deviation of any $X_i$, in this case $0.1$.
Now compute the sample mean, that is, add up the observations, divide by $5$. Suppose the result is $a$. Then our symmetric $95\%$ confidence interval is the interval that goes from $a-1.96\dfrac{0.1}{\sqrt{5}}$ to $a+1.96\dfrac{0.1}{\sqrt{5}}$. The $99\%$ confidence interval is obtained similarly, with $2.57$ replacing $1.96$. I assume you know where these numbers come from.
For the situation where the population standard deviation is not known, one uses Student's $t$-distribution with $n-1$ degrees of freedom instead of the normal. Information can be found here. You will need to locate appropriate tables. (For largish $n$, there are no tables, but in that case one can safely assume that the sample standard deviation is the true standard deviation.)