Answer:
We want to construct a confidence interval at 99% of confidence, so then the significance level would be [tex]1-0.99 =0.01[/tex] and the value of [tex]\alpha/2 =0.005[/tex]. 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:
[tex] z_{\alpha/2}= \pm 2.576[/tex]
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:
[tex] \alpha=0.01, n =26, \sigma = 28.9[/tex]
We want to construct a confidence interval at 99% of confidence, so then the significance level would be [tex]1-0.99 =0.01[/tex] and the value of [tex]\alpha/2 =0.005[/tex]. 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:
[tex] z_{\alpha/2}= \pm 2.576[/tex]
And we can use the following excel code:
"=NORM.INV(0.005,0,1)"