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I'm having trouble finding the proper $z$ score so that I can find the $99\%$ confidence interval. $\bar{x} = 6.01231$. with an $s$ of $1.96833$ and $n$ of $26$, and I got $2.575$ for $z_\frac{\alpha}{2}$.

It's not right for some reason and I can't figure out why.

The confidence interval formula I was using is $\bar{x}\pm Z_\frac{\alpha}{2}s/\sqrt{n}.$

I think it assumes $n$ is big enough which is generally over an $n$ of $30$ I think?

I tried using the $t$ table as well but not sure if I did it right, I got 2.576 but that wasn't right either -- I took the $(1-.99)/2=0.005$ value in the t table with the df of infinity and got $2.576$.

Followup question: I had a similar question earlier with an $n$ of $35$ and looking for a confidence interval of $94\%$, how do I use the $t$ table for those values that aren't listen on it such as that $0.03$? Also I had assumed $n$ of $35$ would've been large enough to just use the $z$ table but apparently not?

Glen_b
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Spencer
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  • Is there a way to compute it on non-programmable calculators? I know my professor doesn't let us use those for exams so I'm using a Casio fx-991MS and can't figure it out. For a question just worth one mark it seems cruel and unusual for my prof to make us do this interpolation. Does it make any difference if he provided us with a mu of 189 while the n is 35? Also I realized for my first question I used df as 26 (the same as n) to get the answer and I'm thinking now I was supposed to do n-1 when getting the t – Spencer Apr 23 '15 at 10:43
  • There are ways to compute t-values on a calculator but you won't be expected to use them; interpolation is done because it's faster/easier. You may have been taught some other way that's less accurate. – Glen_b Apr 23 '15 at 10:55
  • I've edited information from your comments into your question to give a question that's feasible to give some answers on. I've also edited my comments into an answer, and deleted most of the comments. – Glen_b Apr 23 '15 at 10:59

1 Answers1

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A CI based on $Z$ is normally used when $\sigma$ is known rather than estimated from the sample; when you estimate $\sigma$, you'd normally use a $t$-table.

Some people will approximate $t$ by $z$ when the sample size is large.

Note that $t$ with df of $∞$ is just the z-table. You need to consider how you work out the df for a t.

(Edit: you're right, it's not $n$, but $n-1$.)

Followup question: Use interpolation. See here.

Whether some particular $n$ is large enough to use $z$ in place of $t$ for some purpose might depend on all manner of things - different people use different criteria, but somewhere in the region of 60-80 seems okay to me (I don't understand why people use 30 so often, it doesn't generally seem accurate enough in the tail for me).

Personally I rarely use the $z$ when I can use a $t$, unless df is very large or I'm doing it in my head. Computers give exact t-values, so why approximate?

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Glen_b
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