Homework Answers
For completing table for practice :
1. Number of data points = 5
2. Mean = sum of all samples/number of samples
= (79.7 + 81.2 + 80.5 + 80.9 + 79.9)/5
= 80.44%
3. Standard deviation:
| # sample | mean - sample = xi | (mean - sample value)2 = (xi)2 |
| 79.7 | 80.44 - 79.7 = 0.74 | (0.74)2 = 0.5476 |
| 81.2 | 80.44 - 81.2 = -0.76 | (-0.76)2 = 0.5776 |
| 80.5 | 80.44 - 80.5 = -0.06 | (-0.06)2 = 0.0036 |
| 80.9 | 80.44 - 80.9 = -0.46 | (-0.46)2 = 0.2116 |
| 79.9 | 80.44 - 79.9 = 0.54 | (0.54)2 = 0.2916 |
Standard deviation = [sum of (xi)2/number of samples]1/2
= [(0.5476 + 0.5776 + 0.0036 + 0.2116 + 0.2916)/5]1/2
= 0.3264 %
4. Degrees of freedom = n-1
= 5-1 = 4
5. t-value at 95% confidence limit :
value from table at 95% C.L. for 4 degrees of freedom = 2.776
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