Question 1178632
Here's how to construct the 95% confidence interval for the mean tuition:

**1. Identify the given values:**

* Sample mean (x̄) = $18,702
* Sample standard deviation (s) = $10,653
* Sample size (n) = 35
* Confidence level = 95%

**2. Determine the appropriate test statistic:**

* Since the population standard deviation is unknown and the sample size is relatively small (n < 30), we will use the t-distribution.

**3. Find the t-value:**

* Degrees of freedom (df) = n - 1 = 35 - 1 = 34
* For a 95% confidence interval and df = 34, the t-value (tα/2) is approximately 2.032 (using a t-table or calculator).

**4. Calculate the standard error (SE):**

* SE = s / √n
* SE = $10,653 / √35
* SE ≈ $1,801.76

**5. Calculate the margin of error (ME):**

* ME = tα/2 * SE
* ME = 2.032 * $1,801.76
* ME ≈ $3,660.00

**6. Calculate the confidence interval:**

* Lower bound = x̄ - ME = $18,702 - $3,660.00 = $15,042.00
* Upper bound = x̄ + ME = $18,702 + $3,660.00 = $22,362.00

**Therefore, the 95% confidence interval for the mean tuition for all colleges and universities in the United States is approximately ($15,042, $22,362).**