Iam doing a project looking at two ways to scan a knee and comparing them against surgery they had prior to the scans however it is only in patients who have had surgery is this a conventient sample ? and what tests can i use. I have to wrie a bit in my dissertation plan ??? :mad:

Jill,

Are you slelcting x number of patients from N patients who underwent surgery? If the selection is random I think it's a random sample.

The scans before and after surgery for each patient is dependent. You can use a paired t-test to analyze the data, (or a Wilcoxon Signed Rank Test if the population is not normally distributed).

Let me know if you have further questions.

Thanks very much for your help. Iam sure I'll be in touch again
JIll

The tests to be used actually depend on what kind of data you're looking at - in other words, how are you quantitatively comparing the scans?

That will determine what type of test is appropriate.

I have a provisional ethical acceptance now however Iam having trouble with my sample size. It is a paied design. I have a limited number of patients ie.40 and am finding this hard to justify. Any suggestions ??

Jill

Jill,

How much statistical power and how large an effect size are you looking for? Is the sample of n=40 representative of your study population? How are the samples collected and are you doing any randomization? Sorry for all these questions, but I need to know these information in order to justify any sample size. I actually have many more questions. :)

Woops! Jin snuck in there while I was editing / away from my computer!:D

Jill,

Just so I get your study straight in my head:

40 patients who have had knee surgery
20 will be assigned at random to 1 of 2 different scan methods
you want to do a before-after comparison for each of the two groups of 20

Well, yes, it is a convenience sample, but so are lots of biomedical studies - I challenge anyone to find me a single biomedical study that is a purely random sample of the population...

Anyway, a few questions:
- could you get a sample (20 - 40) of similar aged people who have not had surgery? if so, why? I guess I need to understand your study more.
- what's the problem with n=40? especially with a paired design, any statistical test with n=20 and matched pairs should be a pretty powerful test....
- with the scans - exactly what are you "measuring" or evaluating - i.e., what is your dependent variable?

John: Yes convenient samples are common in biomedical studies, and n=40 could very well be sufficient.

Jill: Please provide more details regarding your study design. ie how patients are paired and assigned treatment. etc.

The sample n=40 is limited due to the number of patients who have had previous surgery within the hospital. The sample is collected randomly from the computer system storing the images and reports

From n=40 All patients will have both types of scan following which thay will have a 'gold standard' op with a camera to determine the definative result

Each of the scans has a yes tear or no tear answer which will be the variables

I hypothesize MR arthrography is significantly more accurate at detecting specific meniscal pathology than plain MR alone in patients who have had previous meniscal surgery.

Hi there

I hope this makes sense and thanks very much for your input iam definately a medic and not a statistician

Jill

Jill,

If you were a statistician, you wouldn't be here, and a lot of other people wouldn't be here, and Jin would never get this site to where he wants it to be:)

Anyway-

Now I see where the sample size issue comes into play. It looks like you're basically comparing the proportion (percentage) of accurate diagnoses between scan method, and you really need a large sample size for proportions.

If you think that one method will be definitively better than the other, then maybe you will still be able to detect a statistical difference, however, if you think that the two methods will be somewhat close, then there may be a problem with statistical power.

Do you have any idea, approximately, what the accuracy rates may be for the two methods, or what the difference may be?

With n=40, and worst case accuracy is around 50% (i.e., no better than flipping a coin), then the standard error of a proportion will be:

SEp = sqrt(p*q/n)
where p=rate of correct diagnoses
q = rate of incorrect diagnoses = 1-p
n = sample size

so, in this example, SEp = sqrt(.5*.5/40) = .079 or approx 8%

Now for a 95% confidence level, you'll need to multiply this % by 1.96, so that brings it up to .155 or 15.5%
- in order to detect a difference between methods, the difference would need to be at least 15-16%, but this is a "worst case" estimate

If the actual diagnosis accuracy rates are higher, then the 15-16% would drop off a bit.....

Cheers John

The accuracy rate without dye have been reported to be between 50-80% and with dye the accuracy increased to 88%–92%.

Jill

Then n=40 should be enough......

what about sensitivity and specificity i think the problem is on one of the papers i have read the expected prevelence was 68% from a past paper without dye had a sensitivity of 86%, specificity of 67%, PPV of 83%, NPV of 71%, with dye sensitivity 90%, specificity 78%, PPV 90%, NPV 78%
Jill

jill,

how many participants were in that study?

you may not have enough participants to analyze sens, spec, PPV, NPV for each method, because those %'s are pretty close, although the method w/dye appears to be slightly better in all 4 respects.

however, if you want to just make a basic statement about diagnosis accuracy, then you should be OK.

JohnM

there were 104 patients in the other study. just worried with a 68% expected prevelence (n=40) that only leaves around 13 patients who wont have tears ??? I just need to some how write something that will keep the ethics comitee happy using some sort of sample size formula statistical power etc.. any ideas
Jill

Jill,

If you go to the Examples section on this site, I have a post on Sample Size Determination.

Look at the one regarding estimation of a percentage, with a specified margin of error, at a certain level of confidence.

Good luck.

John

Hi John

So if i want to use the past average of around 65% without dye to estimate the actual percentage to within +/- 3 percentage points, at a 95% confidence

n >= ((p*q)/d^2) * Z^2

n >= ((0.65* 0.35)/(0.03^2)) * 1.96^2

n >= (.228/.0009) * 3.8416

n >= 97

Therefore i would need 97 patients. Is this what you ment for me to do ??

Jill

Yes, but check your math - you should get n >= 972

Like I mentioned before, if you're willing to go with a simple comparison of accuracy -

no dye --> 50-80%, or avg 65%
with dye --> 88-92% or avg 90%

then you could set the margin of error higher, possibly as high as 15%, and reduce n to 39

Ok

Got the correct result (i dont believe it iam so stressed i cant even do simple maths now!!!!) If i just want to use a simple comparison of accuracies at 65% and 90% is it possible to set this out in the mathamatical form to get n=39 as i cant achieve this ?? sorry for being such a hassle

Jill

Yes.

I did make a math error before, so I'll correct it here: you'll need to "settle" for 90% confidence rather than 95%.

n >= ((p*q)/d^2) * Z^2

Assuming accuracy(no dye) = .65 and accuracy(dye) = .90, let's work with the .65, since %'s closer to 50% have larger standard errors, and this will give you a "safer" estimate of the minimum sample size.

Since .90 - .65 = .25, we split the difference to figure out the largest margin of error we can tolerate, which would be approx 12%, so

n >= ((.65 * .35)/.12^2) * 1.282^2

n >= 26

so n = 40 should be enough to detect a significant difference in accuracy at the 90% confidence level

(If I plug in 1.96 for z, using 95% confidence, then n becomes 61, which is, unfortunately, too large)

i cant thank you enough

Jill

My pleasure - good luck with your thesis.