Question

Find the critical value needed to construct a 95% CI of a population mean with an unknown standard deviation and a sample size of 39. Use correct notation and show all workneeded to justify your answer.

Answer 1

Answer:

$t_{\alpha/2}=-2.02, t_{1-\alpha/2}=2.02$

Step-by-step explanation:

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".  

The margin of error is the range of values below and above the sample statistic in a confidence interval.  

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".  

In order to find the critical value we need to take in count that we are finding the critical value for the population mean and without the standard deviation, so on this case we need to use the t distribution. Since our interval is at 95% of confidence, our significance level would be given by $\alpha=1-0.95=0.05$ and $\alpha/2 =0.025$. We need to calculate the degrees of freedom are given by:

$df=n-1=39-1=38$

We can find the critical value using the following excel codes:

"=T.INV(0.025;38)" or "T.INV(1-0.025;38)"

And the critical values would be given by:

$t_{\alpha/2}=-2.02, t_{1-\alpha/2}=2.02$