Question 1178632: College tuition: A simple random sample of 35 colleges and universities in the United States had a mean tuition of $18,702 with a standard deviation of $10,653. Construct a 95% confidence interval for the mean tuition for all colleges and universities in the United States.
Answer by CPhill(1987)   (Show Source):
You can put this solution on YOUR website! 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).**
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