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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 was:

We are 90% confident that the average processing time is between 40.8 and 56.9 days.

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.

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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.

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  • $\begingroup$ 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? $\endgroup$ – Filype May 31 '12 at 7:17
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    $\begingroup$ 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? $\endgroup$ – danas.zuokas May 31 '12 at 8:22
  • $\begingroup$ I am studying for an exam and wanted to know how to do it by hand. I am assuming then that we will be given this table to work with, Thanks so much for your help. $\endgroup$ – Filype May 31 '12 at 8:46
  • $\begingroup$ The t test is robust to departures from normality but its strict validity depends on the observations being normally distributed. It critical value is larger than for the standard normal because the estimation of standard deviation leads to a symmtric distribution that has heavier tailed than the normal distribution. As Danas points out to get the cutoff requires calculating the tail probabilities for the appropriate t distirbution which requires the TINV function or a table for the t distribution. $\endgroup$ – Michael R. Chernick May 31 '12 at 11:16

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