# When to use a t value and when to use 1.645 for a 90% confidence interval?

The question I am working with is:

Setup a 90% confidence interval estimate for the average processing time.

I gathered the information below from the spreadsheet

$n = 27$

$\bar{X} =48.888$

Sample standard deviation $= 25.283$

$\sigma/\sqrt{n} = 4.871$

I am confused because I thought that to setup the confidence level I would use 1.645 which is a common level confidence for the 90% confidence level.

My final answer is wrong. I double checked with a excel template and instead of 1.645 the template used a t value calculated using an exel function called TINV which I am not sure how to calculate. Any help would be greatly appreciated.
You should use $\bar{X}\pm\sigma z_{1-\alpha/2}/\sqrt{n}$ where $z_{1-\alpha/2}$ is normal quantile when population standard deviation $\sigma$ is known. In your case you have only estimate $\hat\sigma$, therefore you should use $\bar{X}\pm\sigma t_{1-\alpha/2}(n-1)/\sqrt{n}$ where $t_{1-\alpha/2}(n-1)$ is student quantile with $n-1$ degrees of freedom (TINV(1-0.9,27-1)=1.706 function in Excel). So you obtain wider confidence interval - more uncertainty due to unknown standard deviation.
• Can I use the 1.645 value when the population standard deviation is known? or when it's not known, I use the t value. Is that correct? Could you describe how to calculate the 1.706 value without TINV? – Filype May 31 '12 at 7:17
• You should use 1.645 when the population standard deviation is known, if not (common situation) then use 1.706 value. There is no explicit formula to calculate $t$ values, however there are tables. Why not using build-in functions? – danas.zuokas May 31 '12 at 8:22