Statistics (Gray)
- Page ID
- 279704
In-class Exercises, Class 2
Name: _________________________
1) How many significant figures in each of the following numbers?
- 82.059
- 0.0003
- 200300
- 300.0
2) Write the numbers above in scientific notation maintaining the same number of significant figures.
3) To how many significant figures should each answer be rounded?
- 40.5 / 1020.2 = 0.039698
- 0.002 kg + 98.3 kg = 98.3002 kg
- 1001 cm – 20.86 cm = 980.14 cm
4) Calculate the answer to the correct number of significant figures.
- 102.259 =?
- What is the pH if the [H+] = 7.245 x 10-6 M
In-class Exercises, Class 3
Name: _________________________
1) Find the absolute and percent relative uncertainty and express each answer with the correct number of significant figures.
- 9.23 (±0.03) + 4.21 (±0.02) – 3.26 (±0.06) = ?
- 91.3 (±1.0) * 40.3 (±0.2) / 21.1 (±0.2) = ?
- [6.2 (±0.2) – 4.2 (±0.1)] / 9.43 (±0.05) =?
2) Why is the data reported below incorrect? Report the data correctly.
5.4359 M ± 0.00671 M
In-class Exercises, Class 4
Name: _________________________
1) For the following data set, calculate the 99% confidence interval.
Sample |
Value |
---|---|
1 |
3.0541 |
2 |
2.9845 |
3 |
3.0512 |
4 |
2.99584 |
2) A standard reference material is certified to contain 94.6 ppm of an organic contaminant in soil. Your analysis gives values of 98.6, 98.4, 97.2, 94.6 and 96.2 (\(\bar{x}\) = 97.0, s = 1.655). Do your results differ from the expected result at the 95% confidence level? If you made one more measurement at 94.5, would your conclusion change (new \(\bar{x}\) = 96.583, s = 1.798)?
In-class Exercises, Class 5 and 6
Name: _________________________
1) Given the following data, are the precisions of the two datasets significantly different at the 95% confidence level? What is the tcalc for these two means? Are the two means significantly different at the 95% confidence level?
\(\bar{X}_1\) = 80.34 s=0.0548 N=4
\(\bar{X}_2\) = 80.46 s=0.2793 N=5
2) Using both the Grubbs and the Q-test, can the last data point be excluded at the 95% CI?
Data |
---|
0.1503 |
0.1505 |
0.1496 |
0.1493 |
0.1496 |
0.1497 |
0.1507 |
0.1617 |
\(\bar{x}\) = 0.1514
s = 0.004181
In-class Exercises, Class 7
Name: _________________________
1) Match the formula below to the statistical method.
Methods:
- Mean
- Standard deviation
- Variance
- Confidence intervals
- Relative standard deviation
- t-test, case 1
- t-test, case 2; variances are equal
- t-test, case 2; variances are not equal
- compare 2 variances
- reject an outlier
\[\text{Degrees of freedom}=\dfrac{(s_1^2/n_1 + s_2^2/n_2)^2}{\dfrac{(s_1^2/n_1)^2}{n_1-1}+\dfrac{(s_2^2/n_2)^2}{n_2-1}} \nonumber\] |
|
\[s_{pooled}=\sqrt{\dfrac{s_1^2(n_1-1)+s_2^2(n_2-1)}{n_1+n_2-2}} \nonumber\] |
\[s = \sqrt{\dfrac{\sum_{i=1}^{N}(X_i-\bar{X})^2}{N-1}} \nonumber\] |
\[=100\times\dfrac{s}{\bar{x}} \nonumber\] |
\[=s^2 \nonumber\] |
\[\mu=\bar{x}\pm \dfrac{ts}{\sqrt{N}} \nonumber\] |
\[|\mu-\bar{x}|>\dfrac{ts}{\sqrt{N}} \nonumber\] |
\[t_{cal}=\dfrac{\bar{x}_1-\bar{x}_2}{s_{pooled}}\sqrt{\dfrac{n_1n_2}{n_1+n_2}} \nonumber\] |
\[t_\text{calculated}=\dfrac{|\bar{x}_1-\bar{x}_2|}{\sqrt{s_1^2/n_1 + s_2^2/n_2}} \nonumber\] |
\[\bar{X}=\dfrac{\sum_{i=1}^{N}X_i}{N} \nonumber\] |
\[Q=\dfrac{gap}{range} \nonumber\] |
\[F_{cal}=\dfrac{s_1^2}{s_2^2} \nonumber\] |
\[G_{calc} = \dfrac{|questionable\: value - \bar{x}|}{s} \nonumber\] |
Contributors and Attributions
- Sarah Gray, Stockton University (sarah.gray@stockton.edu)
- Sourced from the Analytical Sciences Digital Library