My Math Forum Confidence intervals [stats]

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 December 3rd, 2011, 06:50 AM #1 Newbie   Joined: Jan 2011 Posts: 17 Thanks: 0 Confidence intervals [stats] Hi, I've been doing some questions on confidence intervals (stats 3) and I've come along a question which I believe i have done with the correct method, however after checking the answers (at the back of the book) they seem to have a different answer. Anyway, the question is as follows: 5 masses are normally distributed with mean ? and standard deviation ?. The 5 randomly chosen masses are (grams) 124.31 125.14 124.23 125.41 125.76 Calculate the symmetric 95% confidence interval for the mean mass of the population a. assuming that ? = 0.85 b. without assuming that ? = 0.85 Ok so i can do part a easily, however for part b, my method is first finding the variance (using sum of x^2/n - mean^2) and going about finding the confidence interval in the usual manner. So my answer was [124.43, 125.50] but this isnt what is at the back of the book. Does anyone know if maybe my method is wrong? (i have find a normal variance, rather than an unbiased estimate because the sample of 5 is not classified as large) or maybe if anyone could work through and share their solution? Thanks in advance
 December 4th, 2011, 03:39 AM #2 Senior Member   Joined: Apr 2011 From: USA Posts: 782 Thanks: 1 Re: Confidence intervals [stats] It would help a whole heck of a lot if you'd show your work instead of just your answers -- I have no idea where you got those from. I get much closer to your answer using z instead of t, and by using variance instead of standard deviation, though I still can't find your answers. It would also help if you gave the answers the book is giving, just to check to see if the method I would use agrees with your book's answer. I have no idea what "sum of x^2/n - mean^2" is. Without parenthesis, not sure how to interpret that and I'm not familiar with it. I use $\frac{\sum(x\ -\ \overline{x})^2}{n\ -\ 1}$ for variance, and the square root of that for standard deviation. And I'm getting .6764 for s.d. Then I'm using $\sigma_{\overline{x}}\=\ \frac{s}{sqrt{n}}$ Then I'm using a t chart since n=5 (and there's no sigma), at 95%, d.f.=4. And I'm getting $124.97\pm{.84}\=\ 124.13\ to\ 125.81$. Are you using these methods?

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