Do one of the following, as appropriate. (a) Find the critical value z Subscript alpha divided by 2, (b) find the critical value t Subscript alpha divided by 2, (c) state that neither the normal nor the t distribution applies. Confidence level 99%; nequals26; sigma equals 28.9; population appears to be normally distributed.

We want to construct a confidence interval at 99% of confidence, so then the significance level would be and the value of . And for this case since we know the population deviation is not appropiate use the t distribution since we know the population deviation and the best quantile assuming that the population is normally distributed is given by the z distribution.

And if we find the critical value in the normal standard distribution or excel and we got:

And we can use the following excel code:

"=NORM.INV(0.005,0,1)"

Step-by-step explanation:

For this case we have the following info given:

We want to construct a confidence interval at 99% of confidence, so then the significance level would be and the value of . And for this case since we know the population deviation is not appropiate use the t distribution since we know the population deviation and the best quantile assuming that the population is normally distributed is given by the z distribution.

And if we find the critical value in the normal standard distribution or excel and we got:

And we can use the following excel code:

"=NORM.INV(0.005,0,1)"

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