# Stats help on pearson's and spearman's

#### Sarah20

##### New Member
Can someone please tell from the information given below, which equation (Pearson's r or Spearman's) would be most accurate and why?

Pearson' s r= 0.906477
Spearman's = r=0.9779

Comparing two APGAR results from two doctorss gave a mean difference and standard deviation of:

Mean = -0.1875
SD = 0.981
t= 2.04
n=30 (16 from doctor 1 and 16 from doctor 2)
the degrees of freedom here are 16 + 16 - 2 = 30
After calculating the 95 % CI I came to a result of (1.3085, 1.6835)

The question is, which equation is most accurate and why? Thanks everyone!

#### jamesmartinn

##### Member
Pearson's r should only be used on data that follow a linear relationship, or at least a rough linear relationship. Examine your scatterplot for this. Spearman's I believe is an alternative for ranked data, ie ratings from a likert scale, etc.

#### Sarah20

##### New Member
my prof said to compare both answers to see which one is more accurate. if not, he said to just calculate the 95% CI. the other part of the questino was to discuss if the difference between the two sets of apgar results is significant at the 5% level and 15 DOFreedom and also if the scores are siginificant at the 1% level at 15 DOFreedom. we don't need to use a scatterplot

#### Dragan

##### Super Moderator
Can someone please tell from the information given below, which equation (Pearson's r or Spearman's) would be most accurate and why?

Pearson' s r= 0.906477
Spearman's = r=0.9779

Comparing two APGAR results from two doctorss gave a mean difference and standard deviation of:

Mean = -0.1875
SD = 0.981
t= 2.04
n=30 (16 from doctor 1 and 16 from doctor 2)
the degrees of freedom here are 16 + 16 - 2 = 30
After calculating the 95 % CI I came to a result of (1.3085, 1.6835)

The question is, which equation is most accurate and why? Thanks everyone!

Given the information you have provided it appears you have a bigger problem on your hands.

More specifcally, given the degrees of freedom (30), your t-statistic is based on a two-independent samples t-test (Right?).

And, it also appears that the Pearson correlation between the two doctors data points (n1 + n2 = 32) is r= .906477 (Right?).

Before I go any further, can you tell me that my assumptions are correct. If not then you should present you information more clearly.

#### Sarah20

##### New Member
Yes both of your assumptions are correct.

#### Dragan

##### Super Moderator
Yes both of your assumptions are correct.

Okay, one problem you have is that you've violated the assumption underling the two-independent samples t-test because the data points are correlated. In short, if X and Y are independent then this implies that the correlation between X and Y is zero.

That said, because the doctors' scores are so highly correlated (either based on Pearson or Spearman - makes no difference), the appropriate statistic to compute is the dependent (or paired) samples t-test. This is equilanent to a one-sample t-test on the difference scores (say, X = Doctor1 - Doctor2 scores). Under a null hyposthesis, it is assumed that the population mean of the difference scores is zero.

The degrees of freedom in this case would be N - 1 = 15 because you will have 16 difference scores.

Is this clear?

#### Sarah20

##### New Member
Yes thank you very much! Can you please tell me if I'm on the right track?
SDeviation=1
Sample average xbar= 0.2
SEM=SD/root n= 1/4=0.25
t value from table = 2

so the 95% CInterval= xbar +/- t * SEM
= 0.2+2*0.25
=0.7 for the upper limit
= 0.3 for lower limit

OR doc 1's number:
Mean 6.6875
Standard Deviation 2.272

doc 2's numbers:
mean=6.875
SD= 1.857

so the difference between the 2:
mean= -0.1875
sd= 0.981

#### Dragan

##### Super Moderator
Yes thank you very much! Can you please tell me if I'm on the right track?
SDeviation=1
Sample average xbar= 0.2
SEM=SD/root n= 1/4=0.25
t value from table = 2

so the 95% CInterval= xbar +/- t * SEM
= 0.2+2*0.25
=0.7 for the upper limit
= 0.3 for lower limit

OR doc 1's number:
Mean 6.6875
Standard Deviation 2.272

doc 2's numbers:
mean=6.875
SD= 1.857

so the difference between the 2:
mean= -0.1875
sd= 0.981

It appears you are the right track. What I find odd is the standard deviation of 0.981 (where your mean difference is -0.1875 above). Double check this. This should be the standard deviation of the difference scores. I'm not saying you're wrong, here, but I just have my doubts because the correlation between the two doctors scores is so high (r>.9).

A couple of minor notes: You're critical t-value is 2.131 df = 15. Also, the lower limit of your confidence interval should be negative (-0.3).

#### Sarah20

##### New Member
To obtain a mean difference and SD difference, I simply subtratced doc 1 from doc 2. Thats how I got -0.1875 with an SD 0.981.

#### Dragan

##### Super Moderator
To obtain a mean difference and SD difference, I simply subtratced doc 1 from doc 2. Thats how I got -0.1875 with an SD 0.981.
The SD of .981 looks good - I checked it another way. That is, you can get the standard error of the estimate (what you call SEM) as follows:

Standard Error of the difference scores is equal to:

Sqrt [ (2.272 /4 )^2 + (1.857 /4)^2 - 2*(.906477)*(2.272/4)*(1.857/4) ]

which is close to the value of .981/4 that you used,

and where r = 0.906477 is the Pearson correlation above.