Question 1184230: Using the 95% confidence level math scores 88, 93, 93, 94, 90,90, 90, 100, 92,90, 84, 88, 95, 93, 88, 96, 100 you would expect the lower end of the confidence interval to be approximately what value?
Answer by CPhill(1959)   (Show Source):
You can put this solution on YOUR website! 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.
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