College

You intend to estimate a population mean with a confidence interval. You believe the population to have a normal distribution. Your sample size is 11.

Find the critical value that corresponds to a confidence level of [tex]$99.9 \%$[/tex]. (Report the answer accurate to three decimal places with appropriate rounding.)

[tex]$t_{\alpha / 2} = \pm$[/tex]
[tex][tex]$\square$[/tex][/tex]

Answer :

To find the critical value for a confidence level of 99.9% with a sample size of 11, we'll be using the t-distribution, as the sample size is relatively small.

### Step-by-step Solution:

1. Determine the Degrees of Freedom:
- The degrees of freedom (df) for a t-distribution is calculated as the sample size minus one.
- For a sample size (n) of 11:
[tex]\[
df = n - 1 = 11 - 1 = 10
\][/tex]

2. Determine the Alpha Level:
- The confidence level is 99.9%, which means that the remaining portion (alpha) is spread equally in both tails of the distribution.
- Alpha ([tex]\(\alpha\)[/tex]) is calculated as:
[tex]\[
\alpha = 1 - \text{confidence level} = 1 - 0.999 = 0.001
\][/tex]

3. Divide the Alpha Level by 2:
- Since it's a two-tailed test (as we are dealing with a confidence interval), divide the alpha by 2:
[tex]\[
\frac{\alpha}{2} = \frac{0.001}{2} = 0.0005
\][/tex]

4. Find the Critical Value:
- Using the t-distribution table or a calculator, look for the critical value that corresponds to an upper tail probability of [tex]\(0.0005\)[/tex] with [tex]\(10\)[/tex] degrees of freedom. This is [tex]\(t_{\alpha/2}\)[/tex].

After carrying out these steps, the critical value [tex]\(t_{\alpha/2}\)[/tex] has been found to be approximately [tex]\(\pm 4.587\)[/tex].

So, the critical value that corresponds to a 99.9% confidence level with a sample size of 11 is [tex]\(t_{\alpha / 2} = \pm 4.587\)[/tex].