To find the critical value for a 90% confidence interval when the population standard deviation is unknown, we use the t-distribution rather than the z-distribution since we’re dealing with an unknown population standard deviation.

Step-by-Step Solution

1. Determine the Confidence Level and Degrees of Freedom:

   • Confidence Level, $C = 0.90$, so the significance level $\alpha = 1 - C = 0.10$.
   • Since we want the middle 90%, we need 5% in each tail of the distribution.
   • Degrees of Freedom $df = n - 1 = 75 - 1 = 74$.

2. Locate the Critical t-Value:

   • Using a t-distribution table or calculator, find the t-value that corresponds to 0.05 in the upper tail with 74 degrees of freedom.

The critical t-value for a 90% confidence interval with 74 degrees of freedom is approximately:

$t_{0.05, 74} \approx 1.6652$

Thus, the critical value for this confidence interval is 1.6652.

Would you like further details on how to use the t-table or any additional questions on this process?

Related Questions

1. How would the critical value change if the confidence level was 95% instead of 90%?
2. What happens to the critical value as the sample size increases to 100?
3. How is the calculation affected if the population standard deviation was known?
4. Why is the t-distribution used here instead of the normal distribution?
5. How would you construct the confidence interval using this critical value?

Tip:

The t-distribution approaches the normal distribution as the sample size increases, making them nearly identical for very large samples.