Question

You plan on exposing 9 strains of bacteria (labelled A - I) to a mutagen. You hypothesize that mutagen A induces more mutations than mutagen B. Following each exposure, you count the mutations observed in the genome compare to the wild strain genotype. Figure out what kind of statistical test would be appropriate for this study.

Number mutations observed in genome
Mutation    A B    C    D E    F    G H I
Mutagen A 52 17 45 40 20 30 10    15    41

Mutagen B 59 15 63 40 17 45 32 20 40

What is your null and alternate hypothesis?  What are your statistical results?  Should you reject, or fail to reject, your null hypothesis? Why?

Answer 1

SAMPLE 1	SAMPLE 2	difference , Di =sample1-sample2	(Di - Dbar)²
52	59	-7.000	0.049
17	15	2.000	77.049
45	63	-18.000	125.938
40	40	0.000	45.938
20	17	3.000	95.605
30	45	-15.000	67.605
10	32	-22.000	231.716
15	20	-5.000	3.160
41	40	1.000	60.494

	sample 1	sample 2	Di	(Di - Dbar)²
sum =	270	331	-61.000	707.556

mean of difference ,    D̅ =ΣDi / n =   -6.7778
std dev of difference , Sd =    √ [ (Di-Dbar)²/(n-1) =    9.4045

Ho :   µd=   0
Ha :   µd >   0
      
std error , SE = Sd / √n =    9.4045   / √   9   =   3.1348      
                          
t-statistic = (D̅ - µd)/SE = (   -6.778   -   0   ) /    3.1348   =   -2.1621
                          
Degree of freedom, DF=   n - 1 =    8                  
  
p-value =        0.9687 [excel function: =t.dist.rt(t-stat,df) ]               
Conclusion:     p-value>α , Do not reject null hypothesis                    

there is no evidence that mutagen A induces more mutations than mutagen B