2.
by rule of huge error 
not discardable as a huge error
by Dixon test
t 
t = .507 from table
Therefore it can be discarded.
f = 4
D = 0.6 so the uncertainty in the difference (0.477) is smaller than the difference between the means--difference is significant at 95% CL
| 9 . | |||||||||
Site |
mean |
s |
n |
t(95%) |
t(80%) |
t(90%) |
ts/Ö n (95%) |
ts/Ö n (80%) |
ts/Ö n (90%) |
| b | 0.64 |
0.23 |
5 |
2.776 |
1.533 |
2.132 |
0.286 |
0.158 |
0.219 |
| a | 1.125 |
0.22 |
4 |
3.182 |
1.638 |
2.353 |
0.350 |
0.180 |
0.259 |
| d | 1.32 |
0.32 |
5 |
2.776 |
1.533 |
2.132 |
0.397 |
0.219 |
0.305 |
| c | 2.2 |
0.25 |
3 |
4.303 |
1.886 |
2.92 |
0.621 |
0.272 |
0.421 |
| e | 2.3 |
0.28 |
4 |
3.182 |
1.638 |
2.353 |
0.445 |
0.229 |
0.329 |
| A | Overall Mean= | 1.407 |
|||||||
| B | Pooled estimate | of s= |
0.265 |
||||||
| C | Confidence interval = Mean +/- ts/ Ö n (from above) | ||||||||
| D | At 90% CL, the confidence intervals of sites c and e overlap. | ||||||||
| Sites a and d also overlap | |||||||||
(note that calculations such as these, with lots of repitition are most easily solved using a spreadsheet program)
10. Since the value of s is known, the method shown in sec 1.4.4 is used. Unfortunately, the table of areas of the standard error curve (z values) is not in your text.
(95% of points fall within 1.96 s of the mean, so:)
For 95% CL, z = 1.96, and the limits are
mean ± 1.96(s /n)
5.5 ± 1.96(.5/1) = 5.5 ± .975 for 1 replicate
5.5 ± 1.96(.5/5) for 5 replicates
5.5 ± 1.96(.5/15) for 15 replicates
(z = 2.57 at 99% level and 3.291 at 99.9%)
| 12. | ||||||||||||||||
| b. | conc |
mV |
||||||||||||||
1 |
-54 |
|||||||||||||||
2.1 |
-28 |
|||||||||||||||
3.03 |
2.8 |
|||||||||||||||
4 |
32.2 |
|||||||||||||||
5.05 |
65.8 |
|||||||||||||||
| Slope = | 29.9 |
|||||||||||||||
| Intercept = | -87 |
|||||||||||||||
c. 7.8 = 29.9 x -87
x = 3.17
d. Calibration sensitivity is 29.9 mv/ng/ml
13. a. From the plotted data, slope = 0.066, = calibration sensitivity
b. Detection limit
Sm = .03+ (3 x .008) = .054 (smallest det. signal)
Cm = (.054 - .03)/.066 = 0.36 microgram/liter (this is the LOD)
c. RSD
For Point 1 (0.0084/.03)x 100 = 26%
For Point 4 (.015 / 1.2) x 100 = 1.17% (Note RSD is higher for lower conc points)