Q4ITD - Campbell Biology

Question

The standard error (SE), which indicates how greatly the mean would likely vary if the experiment was repeated, is calculated as:\({\rm{SE}} = \,\frac{s}{{\sqrt n }}\). 

As a rough rule of thumb, if an experiment were to be repeated, the new mean typically would lie within two standard errors of the original mean (that is, within the range of \(\overline x \, \pm \,2{\rm{SE}}\)). Calculate \(\overline x \, \pm \,2{\rm{SE}}\)for each treatment, determine whether these ranges overlap, and interpret your results.

 

Treatment	Dose (mg/kg)	Log of number  of colonies	Mean (\(\overline x \,\))	\({x_i}\, - \,\overline x \)	Standard deviation (s)	SE
Control	-	9.0,9.5,9.0,8.9	9.1	(-0.1), 0.4, (-0.1), (-0.2)	0.270	0.135
Vancomycin	1.0	8.5,8.4,8.2	8.36	0.14, 0.04, (-0.16)	0.152	0.087
	5.0	5.3,5.9,4.7	5.3	0, 0.6, (-0.6)	0.6	0.346
Teixobactin	1.0	8.5,6.0,8.4,6.0	7.22	1.28, (-1.22), 1.18, (-1,22)	1.14	0.57
	5.0	3.8,4.9,5.2,4.9	4.7	(-0.9), 0.2,0.5, 0.2	0.616	0.308

Step-by-Step Solution

Answer

For each treatment, \(\overline x \, \pm \,2{\rm{SE}}\)calculated is as follows:

• Control: 8.83 to 9.37
• Vancomycin (1mg/kg): 6.62 to 10.1
• Vancomycin (5mg/kg): 4.61 to 5.99
• Teixobactin (1mg/kg): 6.08 to 8.36
• Teixobactin (5mg/kg): 4.08 to 5.31

 

The ranges of the antibiotics, both at low and high doses, overlap with each other.

Step 1: Two standard errors of the original mean for control

The two standard errors of the original mean for control is calculated by \(\overline x \, \pm \,2{\rm{SE}}\).

\(\begin{aligned}{l}\overline x \, + \,2{\rm{SE}}\,{\rm{ = }}\,{\rm{9}}{\rm{.1  + }}\,{\rm{2 }} \times \,{\rm{0}}{\rm{.135}}\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{\rm{ = }}\,{\rm{9}}{\rm{.37}}\end{aligned}\)

\(\begin{aligned}{l}\overline x \, - \,2{\rm{SE}}\,{\rm{ = }}\,{\rm{9}}{\rm{.1  - }}\,{\rm{2 }} \times \,{\rm{0}}{\rm{.135}}\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{\rm{ = }}\,8.83\end{aligned}\)

Thus, the new mean would lie within the range of 8.83 to 9.37.

Step 2: Two standard errors of the original mean for vancomycin (1mg/kg)

The two standard error of the original mean for vancomycin (1mg/kg) is calculated by\(\overline x \, \pm \,2{\rm{SE}}\).

\(\begin{aligned}{l}\overline x \, + \,2{\rm{SE}}\,{\rm{ = }}\,8.36{\rm{   + }}\,{\rm{2 }} \times \,{\rm{0}}{\rm{.087}}\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{\rm{ = }}\,10.1\end{aligned}\)

\(\begin{aligned}{l}\overline x \, - 2{\rm{SE}}\,{\rm{ = }}\,8.36{\rm{   - }}\,{\rm{2 }} \times \,{\rm{0}}{\rm{.087}}\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 6.62\end{aligned}\)

Thus, the new mean would lie within the range of 6.62 to 10.1.

Step 3: Two standard errors of the original mean for vancomycin (5mg/kg)

The two standard error of the original mean for vancomycin (5mg/kg) is calculated by\(\overline x \, \pm \,2{\rm{SE}}\).

\(\begin{aligned}{l}\overline x \, + \,2{\rm{SE}}\,{\rm{ = }}\,5.3{\rm{   + }}\,{\rm{2}} \times \,{\rm{0}}{\rm{.346}}\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{\rm{ = }}\,5.99\end{aligned}\)

\(\begin{aligned}{l}\overline x \, - 2{\rm{SE}}\,{\rm{ = }}\,5.3\,{\rm{ - }}\,{\rm{2}} \times \,{\rm{0}}{\rm{.346}}\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 4.61\end{aligned}\)

Thus, the new mean would lie within the range of 4.61 to 5.99.

Step 4: Two standard errors of the original mean for teixobactin (1mg/kg)

The two standard error of the original mean for teixobactin (1mg/kg) is calculated by\(\overline x \, \pm \,2{\rm{SE}}\).

\(\begin{aligned}{l}\overline x \, + \,2{\rm{SE}}\,{\rm{ = }}\,7.22{\rm{   + }}\,{\rm{2}} \times \,0.57\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{\rm{ = }}\,8.36\end{aligned}\)

\(\begin{aligned}{l}\overline x \, - 2{\rm{SE}}\,{\rm{ = }}\,7.22\,{\rm{ - }}\,{\rm{2}} \times \,{\rm{0}}{\rm{.57}}\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 6.08\end{aligned}\)

Thus, the new mean would lie within the range of 6.08 to 8.36.

Step 5: Two standard errors of the original mean for teixobactin (5mg/kg)

The two standard error of the original mean for teixobactin (5mg/kg) is calculated by\(\overline x \, \pm \,2{\rm{SE}}\).

\(\begin{aligned}{l}\overline x \, + \,2{\rm{SE}}\,{\rm{ = }}\,4.7{\rm{   + }}\,{\rm{2}} \times \,0.308\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{\rm{ = }}\,5.31\end{aligned}\)

\(\begin{aligned}{l}\overline x \, - 2{\rm{SE}}\,{\rm{ = }}\,4.7\,{\rm{ - }}\,{\rm{2}} \times \,{\rm{0}}{\rm{.308}}\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 4.08\end{aligned}\)

Thus, the new mean would lie within the range of 4.08 to 5.31.

Step 6: Overlapping of ranges

At a low dose (1mg/kg), the range for vancomycin was 6.62 to 10.1, and for teixobactin was 6.08 to 8.36. Similarly, at a high dose, the range for vancomycin was 4.61 to 5.99, and for teixobactin was 4.08 to 5.31.

Thus, the ranges of the two antibiotics overlap both at a low dose and at a high dose.