Final answer:
For constructing a 90% confidence interval for a sample size of 64 with a known standard deviation, we use a z-score. Since the sample size is large, the normal distribution applies, and the critical z-value is 1.645.
Step-by-step explanation:
To determine whether to use a z-score or a t-score for constructing a confidence interval, we need to consider the sample size and whether the population standard deviation is known. If the sample size is large (typically n ≥ 30) and the population standard deviation is known, we use the z-distribution. If the population standard deviation is unknown and the sample size is small (n < 30), we use the t-distribution. However, if we have a large sample size, we can use the z-distribution even if the standard deviation of the population is unknown because the sample standard deviation can reliably estimate the population standard deviation.
In this case, the sample size is 64, which is greater than 30, and the population standard deviation is given as 3993 thousand dollars. Therefore, we can use the normal distribution and find the critical z-value for a 90% confidence interval. For a 90% confidence level, we have a total area of 10% in both tails of the normal distribution, or 5% in each tail (2.5% in each tail if we're considering a two-tailed test). The z-score corresponding to the tail area of 0.05 is approximately 1.645. Therefore, the critical value z is 1.645.