What critical value should he used to construct a 90% confidence interval for a mean value n=16, s=10.5, and the population appears to be normally distributed? Round to three decimal places

To find the critical value for a 90% confidence interval with n=16, we first need to find the degrees of freedom, which is n-1.

Degrees of freedom (df) = n - 1 = 16 - 1 = 15.

Next, we need to find the critical value from the t-distribution table for a two-tailed test with a confidence level of 90% and 15 degrees of freedom.

Looking up in the t-distribution table, the critical value with a confidence level of 90% and 15 degrees of freedom is approximately 1.753.

Therefore, the critical value to construct a 90% confidence interval for a mean value with n=16 is 1.753.

\boxed{1.753}