Prediction interval - what did I do wrong?

mnphys
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Homework Statement


A random sample of 100 automobile owners in the state of Virginia shows that an automobile is driven on average 23,500 kilometers per year with a standard deviation of 3900 kilometers. Assume the distribution of the measurements to be approximately normal. Construct a 99% prediction interval for the kilometers traveled annually by an automobile owner in Virginia.

Homework Equations



qkjwMyn.png

x-bar is the mean, z-value is self-explanatory, σ is the standard deviation and n is the sample size.

The Attempt at a Solution


Here is my attempt at a solution:

rvUuiLN.png


However, the book has the bounds as (13,075, 33,925).

I also considered trying a t-value instead of z-value, but...
a) the book uses z-values for problems like this
b) the book never, ever asks for a t-value with n > 30
c) the t-value I do get if I try it that way doesn't work anyway.

I did this backwards based on the book's answer. With an interval of 10425, standard deviation of 3900, and n=100, the z/t value you would need is 2.66 - this doesn't work as a z-value, and as a t-value it would only work with a sample size of 60 or so. Please help, I'm lost!
 

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mnphys said:

Homework Statement


A random sample of 100 automobile owners in the state of Virginia shows that an automobile is driven on average 23,500 kilometers per year with a standard deviation of 3900 kilometers. Assume the distribution of the measurements to be approximately normal. Construct a 99% prediction interval for the kilometers traveled annually by an automobile owner in Virginia.

Homework Equations



View attachment 215551
x-bar is the mean, z-value is self-explanatory, σ is the standard deviation and n is the sample size.

The Attempt at a Solution


Here is my attempt at a solution:

View attachment 215552

However, the book has the bounds as (13,075, 33,925).

I also considered trying a t-value instead of z-value, but...
a) the book uses z-values for problems like this
b) the book never, ever asks for a t-value with n > 30
c) the t-value I do get if I try it that way doesn't work anyway.

I did this backwards based on the book's answer. With an interval of 10425, standard deviation of 3900, and n=100, the z/t value you would need is 2.66 - this doesn't work as a z-value, and as a t-value it would only work with a sample size of 60 or so. Please help, I'm lost!

The value of ##z_{\alpha/2} ## is closer to 2.576 than to your 2.575, so the computed interval should be very slightly larger than yours (13,404 → 33,596). Furthermore, in this problem the t and the z are noticeably different (##t_{\alpha/2}(99) = 2.626##), so that gives a still wider interval, but still not as wide as that in the book.

I don't think you made a substantial error, so I would say that the book's answer is wrong. That type of issue is unfortunate, but is not uncommon.
 
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Ok. I was kinda wondering if maybe the book was wrong in this case. Thanks for your reply!
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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