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QUESTION:   An experiment was conducted to ascertain which of two bacteria, X or Y, would grow best using lactose as a substrate. Four cultures each of Bacterium X and Bacterium Y were grown and produced the biomass results (in milligrams) shown in the table below. Which bacterium, X or Y, had a higher mean biomass when grown using lactose as the substrate, and was the difference significant?

Bacterium X

Bacterium Y

Flask A

501

233

Flask B

468

270

Flask C

523

280

Flask D

460

249

ANSWER:

In order to compare the means for X and Y, we will use Student’s t-test. Essentially, the t-test compares the difference between the two means to the overall variation in the data—the standard deviation of the difference between the means. To compute the necessary t statistic, we must first find the means for X and Y, the variance for X and for Y (and by extension overall variance), and the mean difference between X and Y. We then look up this t value in a table for the t-distribution with 6 degrees of freedom (nx + ny – 2) and two-tails,  giving the probability P that the mean difference between X and Y is due to chance. The calculations are shown below.

Bacterium X

Bacterium Y

Σx

1952

1032

First we calculate the sum of the values
for X and Y

n

4

4

n is simply the number of samples for
each condition;
in this case the number of flasks of
each bacterium

x̄

488

258

Next, we calculate the mean biomass
for X and for Y

Σx2

955154

267590

Σx2 signifies the sum of the values
squared for X and
Y respectively

(Σx)2

3810304

1065024

(Σx)2 signifies the totals for X and Y
(respectively) squared.

(Σx)2/n

952576

266256

Σd2

2578

1334

Σd2= Σx2-(Σx)2/n , where Σd2 is the
sum of the differences
between X and Y values (respectively)
squared

σ2

859.3

444.7

σ2 = Σd2/(n-1) , where σ2 is the variance
of X and Y
(respectively)

σd2

326

σd2 = σx2/nx + σy2/ny , where σd2 is
the variance of the
difference between the mean for X
and that for Y

σd

18.1

t

12.7

t = (x̄x - x̄y)/σd

P

0.00001436

From the calculations above, the probability that we would observe the mean difference found between X and Y purely by chance is 0.001436%. That value is lower than the conventional 5% and 1% thresholds for significant and highly significant data respectively. Thus, we can confidently claim that Bacterium X grew better than Bacterium Y using lactose as a substrate!

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