Question 1184230
Here's how to calculate the lower end of the 95% confidence interval for the math scores:

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

Sum the scores: 88 + 93 + 93 + 94 + 90 + 90 + 90 + 100 + 92 + 90 + 84 + 88 + 95 + 93 + 88 + 96 + 100 = 1624
Number of scores (n) = 17
x̄ = 1624 / 17 ≈ 95.53

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

* Find the squared difference of each score from the mean:
    (88-95.53)²=56.7, (93-95.53)²=6.4, (93-95.53)²=6.4, (94-95.53)²=2.3, (90-95.53)²=30.6, (90-95.53)²=30.6, (90-95.53)²=30.6, (100-95.53)²=19.9, (92-95.53)²=12.5, (90-95.53)²=30.6, (84-95.53)²=133.0, (88-95.53)²=56.7, (95-95.53)²=0.3, (93-95.53)²=6.4, (88-95.53)²=56.7, (96-95.53)²=0.2, (100-95.53)²=19.9
* Sum of squared differences: 56.7 + 6.4 + 6.4 + 2.3 + 30.6 + 30.6 + 30.6 + 19.9 + 12.5 + 30.6 + 133.0 + 56.7 + 0.3 + 6.4 + 56.7 + 0.2 + 19.9 = 523.2
* Variance (s²): 523.2 / (17-1) ≈ 32.7
* s = √32.7 ≈ 5.72

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

Since the sample size is small (n < 30), we use the t-distribution. For a 95% confidence level and 16 degrees of freedom (n-1 = 17-1 = 16), the t-value is approximately 2.120. 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 = 2.120 * (5.72 / √17)
Margin of Error = 2.120 * (5.72 / 4.123)
Margin of Error ≈ 2.94

5. **Calculate the Confidence Interval:**

* Upper limit = x̄ + Margin of Error = 95.53 + 2.94 ≈ 98.47
* Lower limit = x̄ - Margin of Error = 95.53 - 2.94 ≈ 92.59

Therefore, the lower end of the 95% confidence interval is approximately 92.59.