Help with making a two sided confidence interval

[ldlafountain]

I am working on a problem for my Statistics class and I am looking for some clarification and some direction.

The questions asks about a study that has been done on the diameter of 50 screws (both digital and vernier scale calipers). The differences in the measured diameters had the following frequency distribution.

Difference: Frequency:
-0.03 1
-0.02 3
-0.01 11
0.00 19
0.01 10
0.02 6

I am then asked to make a 90% two sided interval for the mean difference in digital and vernier readings.

So naturally I find the mean of all of the differences by adding up all the differences (by their frequencies) and I get 0.004 as my mean and 0.1148 as my standard deviation. Next I would try and find the t-value for Q(.95) (because we are using a 90% two sided confidence interval). I had to look it up online I came up with 1.677. I would plug this all into the following equation:

x-bar (plus-minus) t*(s/sqrt(n))

I have tried this multiple times and I keep obtaining the wrong intervals. According to the back of my book I should obtain the answer [-0.0023,0.0031], but I come nowhere near. Can anyone help me? If there is something wrong with my math or the way I set up the problem, please let me know. I am thankful for any help.

 

[I like Serena]

Hi Idlafountain! Welcome to MHB! :)

I can tell that your standard deviation is wrong. Looking at the data we can see that it should be about 0.01.
How did you calculate it?

 

[ldlafountain]

Honestly, I plugged all of the numbers into a standard deviation calculator online. I actually used the one from mathisfun.com. I got the mean the same way. But what could the standard deviation be. I understand how it is calculated, but I am still having issues.

 

[I like Serena]

Honestly, I plugged all of the numbers into a standard deviation calculator online. I actually used the one from mathisfun.com. I got the mean the same way. But what could the standard deviation be. I understand how it is calculated, but I am still having issues.

Perhaps a typing mistake? Perhaps there should be a zero after the decimal point?
Or else use a different calculator.

Btw, the remainder of your calculation is fine.

 

[ldlafountain]

So I double checked my calculation on standard deviation and I got 0.0116.

So to plug it all in, I should be able to use:

.0004(plus/minus) 1.677 * (0.0116/sqrt{50})

I got the 50 from my sample size and I got the 1.677 from a t-value chart. After plugging this in, I do not get the answer I a supposed to get in the back of the book. Any ideas?

 

[I like Serena]

So I double checked my calculation on standard deviation and I got 0.0116.

So to plug it all in, I should be able to use:

.0004(plus/minus) 1.677 * (0.0116/sqrt{50})

I got the 50 from my sample size and I got the 1.677 from a t-value chart. After plugging this in, I do not get the answer I a supposed to get in the back of the book. Any ideas?

Calculating what you have, I get:
$$0.0004 \pm 1.677 \cdot (0.0116/\sqrt{50}) = [-0.0023508, +0.0031508] \approx [-0.0024, +0.0032]$$

Isn't that what your book's answer is?
(That is, the book seems to have made either a rounding error, or a slight mistake in calculating the standard deviation.)

 

[Jameson]

I just wanted to add that I get the same thing as ILS (after rounding) when running this in R.

data <- c(-0.03,rep(-0.02,3),rep(-0.01,11),rep(0,19),rep(0.01,10),rep(0.02,6))
t.test(data,conf.level=0.9)

The 90% CI is $(-0.002350061,0.003150061)$ and the SD is $0.01159873$. :)