Construct the confidence interval for the population mean mu. c = 0.90​, x = 16.9​, s = 9.0​, and n = 45. A 90​% confidence interval for mu is:______.

 To construct a 90% confidence interval for the population mean (μ) when the population standard deviation is not known and the sample size is less than 30, we use the t-distribution. However, when the sample size is large (n greater than 30), the distribution of the sample mean can be approximated by a normal distribution due to the Central Limit Theorem.

The formula for the confidence interval for a population mean using the t-distribution is:

CI = x̄ ± (t * (s/√n))

where: - CI is the confidence interval - x̄ is the sample mean - t is the t-score associated with the desired confidence level - s is the sample standard deviation - n is the sample size

Since the sample size in this case is 45 (which we consider enough to use the normal approximation), and we are creating a 90% confidence interval, we need to find the critical value for the t-distribution with n-1 degrees of freedom (44 in this case) that corresponds to a 90% confidence level. You would typically use a t-distribution table or statistical software to find this value.

For a 90% confidence interval, the t-score with 44 degrees of freedom is approximately 1.684 (this value may slightly differ based on the t-distribution table).

Plugging in the values we have: x̄ = 16.9 s = 9.0 n = 45

CI = 16.9 ± (1.684 * (9/√45))

First, calculate s/√n: s/√n = 9/√45 ≈ 9/6.7082 ≈ 1.3416

Now, calculate the margin of error (t * (s/√n)): ME = 1.684 * 1.3416 ≈ 2.2600

Finally, calculate the confidence interval: Lower limit = x̄ - ME = 16.9 - 2.2600 ≈ 14.64 Upper limit = x̄ + ME = 16.9 + 2.2600 ≈ 19.16

Therefore, the 90% confidence interval for the population mean μ is approximately (14.64, 19.16).