Consider the function represented by the table.
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Answer:
Step-by-step explanation:
Hello!
The variable of interest is:
X: fitness of a line of E. coli grown on an acidic environment.
n= 6 E. coli lines
Recorded fitness for each line: 1.24, 1.22, 1.23, 1.24, 1.18, 1.09
The relative fitness of 1 indicates that both bacteria types are equally fit.
A relative fitness larger than 1 indicates that the acid-evolved line is more fit than the parental line kept at neutral pH when both are grown in acidic conditions.
Meaning that if the average fitness of the E. coli lines grown on an acidic environment is greater than 1 then they are better adjusted to live in acidic conditions, symbolically: μ > 1
The statistic hypotheses are:
H₀: μ ≤ 1
H₁: μ > 1
α: 0.05
Assuming that the variable has a normal distribution you have to apply a one-sample t-test:
X[bar]= 1.20
S= 0.06
The p-value for this test is 0.0002
Since the p-value= 0.0002 is less than α:0.05 the decision is to reject the null hypothesis.
Then at a 5% significance level, there is significant evidence to conclude that the bacteria evolved in acidic pH are better adapted to acidic conditions.
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