You intend to estimate a population mean with a confidence interval. You believe the population to have a normal distribution. Your sample size is 18.

Question

While it is an uncommon confidence level, find the critical value that corresponds to a confidence level of 97.1%.
(Report answer accurate to three decimal places with appropriate rounding.)

ta/2 =

Answer 1

To find the critical value $t_{\alpha /2}$ corresponding to a confidence level of 97.1% with a sample size of 18, you'll need to follow these steps:

1. Determine the significance level $\alpha$: $$\alpha =1-\text{confidence level}=1-0.971=0.029$$ Then, $\alpha /2$ is: $$\alpha /2=0.029/2=0.0145$$

2. Determine the degrees of freedom: Since the sample size $n=18$, the degrees of freedom $df$ is: $$df=n-1=18-1=17$$

3. Find the critical value $t_{\alpha /2}$: You will look for the $t$ value that corresponds to $0.0145$ in the upper tail with $df=17$. This can be obtained from a t-distribution table or using statistical software.

Using statistical tables or software, we find:

• For $df=17$ and $\alpha /2=0.0145$, the critical value is approximately $t_{0.0145}$.

After searching through a t-table or using software, the closest value found for $t_{0.0145}$ with 17 degrees of freedom is approximately 2.624.

Thus, $$t_{\alpha /2}\approx 2.624$$

Final answer rounded to three decimal places: $$t_{\alpha /2}=2.624$$