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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.

Super Rhinocerus
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  • 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 06:50
  • Do you know that the estimated standard deviation is $$\large{s=\sqrt{\frac{1}{n-1}\sum_{i=1}^{n} (x_i-\overline x )^2}}$$ ? – callculus42 May 24 '15 at 08:29
  • @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. – callculus42 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 00:54

1 Answers1

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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.

BruceET
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  • 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 07:22
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    @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. – callculus42 May 24 '15 at 08:48
  • 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