1
$\begingroup$

Stats newb here, I have to determine if two time series are really different instead of being part of the same population with noise in the samples.

The data is a comparison between two algorithms ctr by day, a control one (a) and an experimental (b)

Plot of algorithms comparison

a      b
1 3.6162 3.6808
2 3.8967 4.0155
3 4.0669 4.2945
4 4.3680 4.4321
5 4.0558 4.2071
6 3.9234 3.9131
7 3.7467 3.9533

We can see the mean is bigger in b (4.070914 vs 3.953386) But are they statistically significant? To see that im doing ANOVA to get the p-value and compare to the alpha of 0.05 as far as i know, if it is smaller then the null hypothesis H0 of being equals is false.

The problem is when i do the oneway.test i get a p-value really big, am i doing something wrong?

data
        x name
1  3.6162    a
2  3.6808    b
3  3.7467    a
4  3.8967    a
5  3.9131    b
6  3.9234    a
7  3.9533    b
8  4.0155    b
9  4.0558    a
10 4.0669    a
11 4.2071    b
12 4.2945    b
13 4.3680    a
14 4.4321    b

 oneway.test(x~name, data = data)

    One-way analysis of means (not assuming equal variances)

data:  x and name
F = 0.77477, num df = 1.00, denom df = 11.97, p-value = 0.3961 

Thanks a lot!

$\endgroup$
1
  • $\begingroup$ You lost a lot of information when you ignored the paired nature of the data. If that's an unfamiliar concept, please see the related threads on this site. $\endgroup$
    – whuber
    Feb 10, 2016 at 15:53

1 Answer 1

3
$\begingroup$

As the comment pointed out, the pairing of observation is crucial in this type of data. Instead of treating the data into 2 groups, we introduce a variable d to denote the difference between group 1 and group 2 for each observation.

Then the t-statistics is calculated by

formula
(source: fao.org)

To do this in r, just use the following

a<-c(3.6162,3.8967,4.0669,4.368,4.0558,3.9234,3.7467)
b<-c(3.6808,4.0155,4.2945,4.4321,4.2071,3.9131,3.9533)

t.test(a,b,paired=T)

Then you can see the output,

Paired t-test

data:  a and b
t = -3.6651, df = 6, p-value = 0.01052
alternative hypothesis: true difference in means is not equal to 0
95 percent confidence interval:
 -0.19599321 -0.03906393
sample estimates:
mean of the differences 
         -0.1175286
$\endgroup$
0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.