Question 1184231
Here's how to calculate the lower end of the 95% confidence interval for the writing scores:

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

Sum the scores: 84 + 80 + 90 + 90 + 98 + 94 + 91 + 86 + 90 + 91 + 90 = 1034
Number of scores (n) = 11
x̄ = 1034 / 11 = 94

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

* Find the squared difference of each score from the mean:
    (84-94)²=100, (80-94)²=196, (90-94)²=16, (90-94)²=16, (98-94)²=16, (94-94)²=0, (91-94)²=9, (86-94)²=64, (90-94)²=16, (91-94)²=9, (90-94)²=16
* Sum of squared differences: 100 + 196 + 16 + 16 + 16 + 0 + 9 + 64 + 16 + 9 + 16 = 478
* Variance (s²): 478 / (11-1) = 47.8
* s = √47.8 ≈ 6.91

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

Since the sample size is small (n < 30), we use the t-distribution.  For a 95% confidence level and 10 degrees of freedom (n-1 = 11-1 = 10), the t-value is approximately 2.228.  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 = 2.228 * (6.91 / √11)
Margin of Error = 2.228 * (6.91 / 3.317)
Margin of Error ≈ 4.63

5. **Calculate the Confidence Interval:**

* Upper limit = x̄ + Margin of Error = 94 + 4.63 ≈ 98.63
* Lower limit = x̄ - Margin of Error = 94 - 4.63 ≈ 89.37

Therefore, the lower end of the 95% confidence interval is approximately 89.37.