Question 1184225
Here's how to calculate the lower end of the 95% confidence interval for the given scores:

1. **Calculate the Sample Mean (x̄):**

Sum of scores = 97 + 93 + 95 + 97 + 97 + 94 + 92 + 92 + 98 + 92 + 93 + 94 + 95 + 96 + 96 + 96 = 1520
Number of scores (n) = 16
x̄ = 1520 / 16 = 95

2. **Calculate the Sample Standard Deviation (s):**

* Find the squared difference of each score from the mean: (97-95)²=4, (93-95)²=4, (95-95)²=0, (97-95)²=4, (97-95)²=4, (94-95)²=1, (92-95)²=9, (92-95)²=9, (98-95)²=9, (92-95)²=9, (93-95)²=4, (94-95)²=1, (95-95)²=0, (96-95)²=1, (96-95)²=1, (96-95)²=1
* Sum of squared differences = 4 + 4 + 0 + 4 + 4 + 1 + 9 + 9 + 9 + 9 + 4 + 1 + 0 + 1 + 1 + 1 = 60
* Variance (s²) = 60 / (16-1) = 60 / 15 = 4
* s = √4 = 2

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

Since the sample size is small (n < 30), we use the t-distribution. For a 95% confidence level and 15 degrees of freedom (n-1 = 16-1 = 15), the t-value is approximately 2.131. You can find this using a t-table or a calculator with statistical functions.

4. **Calculate the Margin of Error:**

Margin of Error = t * (s / √n)
Margin of Error = 2.131 * (2 / √16)
Margin of Error = 2.131 * (2 / 4)
Margin of Error = 2.131 * 0.5
Margin of Error ≈ 1.07

5. **Calculate the Confidence Interval:**

* Upper limit = x̄ + Margin of Error = 95 + 1.07 ≈ 96.07
* Lower limit = x̄ - Margin of Error = 95 - 1.07 ≈ 93.93

Therefore, the lower end of the 95% confidence interval is approximately 93.93.