Q27.6-3ITD
Question
The variation found in a set can be estimated by the standard deviation, s:
\({s_x}\, = \sqrt {\frac{1}{{n - 1}}} \sum {({x_i}} - \,\overline x {)^2}\)
Calculate the standard deviation for each treatment.
Treatment | Dose (mg/kg) | Log of number of colonies | Mean (\(\overline x \,\)) | \({x_i}\, - \,\overline x \) |
Control | - | 9.0,9.5,9.0,8.9 | 9.1 | (-0.1), 0.4, (-0.1), (-0.2) |
Vancomycin | 1.0 | 8.5,8.4,8.2 | 8.36 | 0.14, 0.04, (-0.16) |
| 5.0 | 5.3,5.9,4.7 | 5.3 | 0, 0.6, (-0.6) |
Teixobactin | 1.0 | 8.5,6.0,8.4,6.0 | 7.22 | 1.28, (-1.22), 1.18, (-1,22) |
| 5.0 | 3.8,4.9,5.2,4.9 | 4.7 | (-0.9), 0.2,0.5, 0.2 |
Step-by-Step Solution
VerifiedThe standard deviation for each treatment is as follows:
- Control: 0.270
- Vancomycinat 1mg/kg: 0.152
- Vancomycin at 5mg/kg: 0.6
- Teixobactin at 1mg/kg: 1.14
- Teixobactin at 5mg/kg: 0.616
In order to calculate the standard deviation for control, values are substituted in the following equation:
\(\begin{aligned}{l}{s_c}\, &= \sqrt {\frac{1}{{n - 1}}} \sum {({x_i}} - \,\overline x {)^2}\\\,\,\,\, &= \sqrt {\frac{1}{{4 - 1}}} \sum {{{({{( - 0.1)}^2} + {{0.4}^2} + {{( - 0.1)}^2} + {{( - 0.2)}^2})}^{}}} \\\,\,\,\, &= \sqrt {\frac{1}{3}} \times 0.22\\\,\,\,\, &= \sqrt {0.073} \,\\\,\,\,\, &= 0.270\end{aligned}\)
Thus, the standard deviation for control is 0.270.
In order to calculate the standard deviation for vancomycin at 1 mg/kg, values are substituted in the equation given:
\(\begin{aligned}{l}{s_{v1\,}} &= \sqrt {\frac{1}{{n - 1}}} \sum {({x_i}} - \,\overline x {)^2}\\\,\,\,\, &= \sqrt {\frac{1}{{3 - 1}}} \sum {({{(0.14)}^2} + {{0.04}^2} + {{( - 0.16)}^2})} \\\,\,\,\, &= \sqrt {\frac{1}{2}} \times 0.0468\\\,\,\,\, &= \sqrt {0.0234} \,\\\,\,\,\, &= 0.152\end{aligned}\)
Thus, the standard deviation for vancomycin at 1mg/kg is 0.152.
In order to calculate the standard deviation for vancomycin at 5 mg/kg, values are substituted in the equation given:
\(\begin{aligned}{l}{s_{v5\,}} &= \sqrt {\frac{1}{{n - 1}}} \sum {({x_i}} - \,\overline x {)^2}\\\,\,\,\, &= \sqrt {\frac{1}{{3 - 1}}} \sum {({{(0)}^2} + {{0.6}^2} + {{( - 0.6)}^2})} \\\,\,\,\, &= \sqrt {\frac{1}{2}} \times 0.72\\\,\,\,\, &= \sqrt {0.36} \,\\\,\,\,\, &= 0.6\end{aligned}\)
Thus, the standard deviation for vancomycin at 5mg/kg is 0.6.
In order to calculate the standard deviation for teixobactin at 1 mg/kg, values are substituted in the equation given:
\(\begin{aligned}{l}{s_{t1\,}} &= \sqrt {\frac{1}{{n - 1}}} \sum {({x_i}} - \,\overline x {)^2}\\\,\,\,\, &= \sqrt {\frac{1}{{4 - 1}}} \sum {({{(1.28)}^2} + {{( - 1.22)}^2} + {{1.18}^2} + {{( - 1.22)}^2})} \\\,\,\,\, &= \sqrt {\frac{1}{3}} \times 6.006\\\,\,\,\, &= \sqrt {2.002} \,\\\,\,\,\, &= 1.14\end{aligned}\)
Thus, the standard deviation for teixobactin at 1 mg/kg is 1.14.
In order to calculate the standard deviation for teixobactin at 1 mg/kg, values are substituted in the equation given:
\(\begin{aligned}{l}{s_{t5\,}} &= \sqrt {\frac{1}{{n - 1}}} \sum {({x_i}} - \,\overline x {)^2}\\\,\,\,\, &= \sqrt {\frac{1}{{4 - 1}}} \sum {{{({{(0.9)}^2} + {{(0.2)}^2} + {{0.5}^2} + 0.2)}^2})} \\\,\,\,\, &= \sqrt {\frac{1}{3}} \times 1.14\\\,\,\,\, &= \sqrt {0.38} \,\\\,\,\,\, &= 0.616\end{aligned}\)
Thus, the standard deviation for teixobactin at 5 mg/kg is 0.616.