# Confidence Interval Question - Confused on Approaching the Problem

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.

• I did a very minimal edit on your question so that it could be read without scrolling from side to side. Still not sure what you mean by "Heading." – BruceET May 24 '15 at 6:50
• Do you know that the estimated standard deviation is $$\large{s=\sqrt{\frac{1}{n-1}\sum_{i=1}^{n} (x_i-\overline x )^2}}$$ ? – callculus May 24 '15 at 8:29
• @calculus: Correct, but not relevant here. No need to estimate $\sigma$ when it is already given exactly. – BruceET May 24 '15 at 13:38
• @Brucetrumbo You are right. I missunderstood the exercise. But it seems, that $E(\overline x)\neq \mu$. This fact is confusing me. – callculus May 24 '15 at 16:04
• @calculus: I assume $E(\bar X) = \mu$ and that the variation represented by $\sigma$ is unbiased. BTW, see OP's note of appreciation to you following my Answer. – BruceET May 25 '15 at 0:54

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.

• Sorry for the late reply, but could you further explain 1.96? Is it just obtained using that Z = ... formula (i.e., we get that value and plug it into a table)? – Super Rhinocerus May 24 '15 at 7:22
• @SuperRhinocerus If you use a one-sided table you have to look at which value (z or x) you get the corresponding value $1-\frac{\alpha}{2}=1-\frac{1-0.95}{2}=1-\frac{0.05}{2}=0.975$ It is just the mean of 0.95 and 1. – callculus May 24 '15 at 8:48
• Yes. To have 95% in the middle you must cut half of 5% from each tail. – BruceET May 24 '15 at 13:36
• Yes, I found the value in the table, calculus' explanation helped a lot and I was able to solve the problem. Thanks to both you! – Super Rhinocerus May 24 '15 at 17:19