To find the critical value $t_{\alpha/2}$ for an 89.1% confidence level with a sample size of 17, we need to use the following information:

1. Confidence Level: 89.1%, which means $\alpha = 1 - 0.891 = 0.109$.
2. Degrees of Freedom: For a sample size of 17, the degrees of freedom $df = n - 1 = 17 - 1 = 16$.

Step 1: Calculate $\alpha/2$

Since $\alpha = 0.109$, we divide this by 2: $\alpha/2 = 0.109 / 2 = 0.0545$

Step 2: Find the Critical $t$-Value

Now, we need the critical value $t_{\alpha/2}$ for 16 degrees of freedom and a cumulative probability of $1 - 0.0545 = 0.9455$ (since 89.1% confidence level implies that the central 89.1% is covered, leaving 5.45% in each tail).

Using a $t$-distribution table or statistical software, we find: $t_{0.0545, 16} \approx 1.754$

Answer:

The critical value $t_{\alpha/2}$ is approximately: $t_{\alpha/2} = \pm 1.754$

Would you like further details or have any questions?

Related Questions

1. How would the critical value change if the confidence level were increased to 95%?
2. What is the significance of the degrees of freedom in finding $t$-values?
3. How does a larger sample size affect the critical $t$-value?
4. Why do we use a $t$-distribution instead of a normal distribution here?
5. How can one interpret the meaning of a confidence interval with this $t$-value?

Tip:

For uncommon confidence levels, remember that dividing $\alpha$ by 2 and using the cumulative probability in the $t$-distribution are key to finding precise critical values.