Question

If you were constructing a 99% confidence interval of the population mean based on a sample of n 1) = 25 where the standard deviation of the sample S = 0.05, the critical value of t will be

Answer 1

Answer:

We want a confidence of 99% so then the significance level would be 1%

The degrees of freedom are given by:

$df = n-1= 25-1=24$

And the critical value using the significance level $\alpha=0.01$ and $\alpha/2 =0.005$ and the critical value would be:

$t_{\alpha/2}= \pm 2.797$

Step-by-step explanation:

We know the following info :

$s= 0.05$ represent the standard deviation from the sample

$n = 25$ represent the sample size selected

We want a confidence of 99% so then the significance level would be 1%

The degrees of freedom are given by:

$df = n-1= 25-1=24$

And the critical value using the significance level $\alpha=0.01$ and $\alpha/2 =0.005$ and the critical value would be:

$t_{\alpha/2}= \pm 2.797$