My question is almost exactly the same as another one here on math stack exchange, but it isn't as explanatory as I'd like it to be with some parts. I was unsure of whether or not it'd be "okay" to repost the question, especially since the other one was asked about three years ago.
Question is:
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:
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.
Based on the answer that was given, we first find the sample mean, which is just:
$(3.142+3.163+3.155+3.150+3.141)/5.$
Then, we find the standard deviation, which is where I get some confusion. We're supposed to use the "other" standard deviation, but I don't know what the differences would be. The apparent "new" standard deviation would be:
std deviation = $(1/\sqrt5) * .1$
Then, we multiply that by plus or minus 1.96. However, I am completely confused as to how 1.96 is obtained. There is a table we can consult, but it seems that there are different tables depending on whether or not the confidence interval is two-sided – in which case, how would we know if the CI is two-sided?
I hope I'm being clear enough here, a slow explanation would be very helpful.
Best Answer
I believe you are supposed to take $\sigma = 0.1$ and compute a z-interval $\bar X \pm 1.96\sigma/\sqrt{n}.$
This formula for a 95% confidence interval arises from the normality of the data, whence $$Z = \frac{\bar X - \mu}{\sigma/\sqrt{n}} \sim Norm(0,1).$$ Then from tables of the standard normal distribution, you have $$P(-1.96 \le Z \le 1.96) = 0.95.$$
The number 1.96 cuts 2.5% of the area from the upper tail of the standard normal density curve, and -1.96 cuts 2.5% from its lower tail, leaving 95% of the area between $\pm$ 1.96.
Notes:
(1) I think your data are intended to be results of five successive weighings of the same object on a scale that does not give exactly the same result every time.
(2) A two sided confidence interval (CI) gives an upper and a lower bound. It can be written as $\bar X \pm 1.96\sigma/\sqrt{n}$ or as $$(\bar X - 1.96\sigma/\sqrt{n},\;\bar X + 1.96\sigma/\sqrt{n}).$$
(3) For a 99% CI use 2.576 instead of 1.96. Hope you can find these numbers in your normal table.
(4) If you were not given the value of the population standard deviation $\sigma$ you would need to compute the sample deviation $S = 0.0092$ and find a CI using Student's t distribution.