Medical Statistics - Science topic

Hi,

The study starts with a Cohort A and on Follow up if Ef<40 then it will be in Group B. This Shift suggests that the survival decreases (Failure to be in Group A) i.e Survival Analysis is applicable. Since factors affecting the survival would be examined, then Cox Proportional Hazards Model is applicable. Survival curves are cumulative curves.

Use one of the following.

A- odds ratio,

B- relative risk,

C- risk differences.

Put group on the rows and outcomes on the columns.

My preintervention values in group a and group b are 83.78±38.16 and 216 ±121.73 and post-intervention values are 36.20±29.75 and 35.40±14.71.In this scenario,how do we calculate the percentage reduction in values in both group A and Group B ? Please Help

If your interest is in linear trend, then plot x-axis as year and Y-axis as % of patients then calculate the trend and identify slope values.

Rearrange the data monthly and you may repeat the same monthly-wise.

Then you will have 10 slopes values for both sets ..then attempt t-test based on slope. if you have subject details maybe you can calculate subject-wise slope as well.

best!!

Daniel Wright

sorry for the vague term, I used simple term to explain to someone and forgot to edit it in this post (and English is not my first language, sorry!). I want to find the correlation (or prove there is a relationship) between my variables. My hypothesis is that when Y1 is normal, visuospatial tests are normal and as we go to displaced and ruptured, the visuospatial tests are getting worse (probably linearly). After, if we find a correlation between the integrity of the tract and visuospatial skills, I want to show that if we ´´repair’´ the tract during surgery, the visuospatial skills are reversible and the patient will score better on the test. I have some background in stats, but not in biostats and the statistician I contacted isn’t either…

David Eugene Booth

exactly, in the litterature those 3 tracks have some unclear link to visuospatial abilities, so that’s why we chose them. They should be independent (meaning that when only y1 is rupture for example, there is a decrease in visuospatial score, same for y2 and y3 taken alone). But, if y1 and y2 are ruptured, I want to show that there is even worse score

Yes, you can. The beta value and correlation value are exactly the same when the standard deviations of both variables are the same.

If we know the standard deviations of both variables and the beta value, we can easily calculate the correlation coefficient. Please check out the example below

You can run an ANCOVA (analysis of covariance), i.e. a regression:

Post = Pre + Treatment/Group (in that order). The assumption is that there is no interaction between 'Pre*Group'. i.e. the heterogeneity of slopes.

If this assumption is not met, you can use the Johnson-Neyman technique to remediate.

If you want to validate the device for all 3 diseases in one step, then you should also consider this in the sample size calculation. You might want to look for multiple co-primary endpoints.

Best Matthias

I apologize for the delay of response. Yes, I would encourage you to follow PI's advice. My input is based on my training in statistical science, but by all means, I do not claim to be an expert in your field. Please take information below as a mere suggestion for alternative approach to answer your research question.

Having said that, here are my elaborated version of the previous response: it is my understanding that your interest is on evaluating 'slopes' at each time point by group. For instance, your 'none' group shows positive slope between 'pre-treatment' and '0-1 month', whereas the other two groups show negative slope.

You can run linear or generalized linear mixed effect model to quantify time-specific, age- and sex-adjusted slopes at the individual level, and extract those values and save them as data. You would then have individual-level slope information at multiple time points, which in turn multiple comparison tests can be used to evaluate slopes at each time point by group. For instance based on the graph you shared, your 'none' group slope would be statistically different to the other two groups for obvious reasons (positive slope vs. negative slope).

As you take this longitudinal 'piece-wise' regression approach, the benefits are: 1) you can control for auto-correlation for repeated observations; 2) you have a secondary dataset that can be manipulated further (for tables and graphs); 3) you can also make your time variable as continuous and predict individual's outcome based on the formula you built. (This would be a possible solution for your unequal time interval problem).

In my training, to do any kind of meaningful trend analysis, you need more intensive time points. You may consider doing a simple change for the two periods and run a simple binomial hypothesis testing, but the problem with this approach would be centering the sample mean, which can be tricky. Hope information were not overwhelming. Once again, take these information as a potential alternative approach. Good luck.

Please see this

Dear Shaun,

Yes, if you have enough data then the empirical density estimates will be very helpful. If you know enough about the process(es) by which T was generated this may be enough on its own. In other cases one might make decisions by comparing various models.

Take care,

Occasionally authors encounter a situation where data for the same outcome are presented in some studies as dichotomous data and in other studies as continuous data. For example, scores on depression scales can be reported as means or as the percentage of patients who were depressed at some point after an intervention (i.e. with a score above a specified cut-point). This type of information is often easier to understand and more helpful when it is dichotomized. However, deciding on a cut-point may be arbitrary and information is lost when continuous data are transformed to dichotomous data.

There are several options for handling combinations of dichotomous and continuous data. Generally, it is useful to summarize results from all the relevant, valid studies in a similar way, but this is not always possible. It may be possible to collect missing data from investigators so that this can be done. If not, it may be useful to summarize the data in three ways: by entering the means and standard deviations as continuous outcomes, by entering the counts as dichotomous outcomes and by entering all of the data in text form as ‘Other data’ outcomes.

There are statistical approaches available which will re-express odds ratios as standardized mean differences (and vice versa), allowing dichotomous and continuous data to be pooled together. Based on an assumption that the underlying continuous measurements in each intervention group follow a logistic distribution (which is a symmetrical distribution similar in shape to the normal distribution but with more data in the distributional tails), and that the variability of the outcomes is the same in both treated and control participants, the odds ratios can be re-expressed as a standardized mean difference according to the following simple formula (Chinn 2000):

.

The standard error of the log odds ratio can be converted to the standard error of a standardized mean difference by multiplying by the same constant (√3/ π = 0.5513). Alternatively standardized mean differences can be re-expressed as log odds ratios by multiplying by π/√3 = 1.814. Once standardized mean differences (or log odds ratios) and their standard errors have been computed for all studies in the meta-analysis, they can be combined using the generic inverse-variance method in RevMan. Standard errors can be computed for all studies by entering the data in RevMan as dichotomous and continuous outcome type data, as appropriate, and converting the confidence intervals for the resulting log odds ratios and standardized mean differences into standard errors (see Chapter 7, Section 7.7.7.2).cutually, cochrane hab

Dear Chuang Liu,

Kindly explain the idea behind the "nested-test" as mentioned by you.

Also, mention the types of hypothesis that can be tested by this method along with the test statistic, with the probability distribution followed by it, used in the method.

Statistics Corner: A guide to appropriate use of Correlation coefficient in medical research. M.M Mukaka. Malawi Medical Journal; 24(3): 69-71 September 2012

0.90-1.00 Very high ; 0.70-0.90 High; 0.50-0.70 Moderate ; 0.30-0.50 Low ; 0.00-0.30 Negligible

Kindly see the following RG links and PDF attachments.

For those who are interested, the Statalist thread can be viewed here:

Please see the following RG link and PDF attachment.

Amr Muhammed , yes, to be more precise the weights will be not exactly 1/2 but will depend on the groups sizes. But in the mentioned table (and many other similar papers) this SD is referred to a difference: difference mean = 5.0, difference SD = 20.1.... Probably it is some kind of conventional notation for a united sample SD but it seems to be too confusing.

The new research methodology and statistics book of our group is in the stage of publication and we expect your questions and contributions.

Which statistical tests should be used when analyzing, Factors Associated with Stunting among Children (0-24 Months).

1. Birth order

2. Education (parents)

3. Hb (Mother)

4. Breast Feeding

5. Complementary feeding

6. Gender

Thanks a lot in advance.

Regards,

Here is some Stata Code for explanation:

Hello there,

Please find my answer in the steps given here:

Step #1:

Compute the overall mean score for the items of that variable of interest

Step #2:

Make cutoff points using a calculator (Maximum- Minimum / n). How? In case of a five-point Likert scale, you may, for example, have assigned the value (1) for Strongly Disagree, (2) Slightly Disagree, (3) Neutral, (4) Slightly Agree, (5) Strongly Agree. So in order to make cutoff points, you've to do this simple math, as per the formula stated above: (5-1/3). Upon calculation you will get this value (1.33), right! This is the interval value.

* "Highest" refers to the highest score of the given Likert scale (5, in our example)

* "Lowest" refers to the lowest (1)

* "n" refers to the number of CATEGORIES you intend to create

Step #3:

Do the math for the three categories (Low, Mid & High). How? Just add up the score of your interval value to the three category, as in the following:

Step #4:

If you are using SPSS, enter these category-related values by navigating through "Transform", and "Recode into Different Variable". Here you will need to recode the Mean score we've obtained in Step# 1. After you give it a new Name and Label, click on "Old and New Values". A new window will pop up. Select "Range", and then insert the values of the categories given above. You should give a value for each range (i.e. 1 for Range no 1), in the New Value empty field. Then click "Add". Follow the same routine for the other range values of your categories, 2 & 3. Click "Continue", and then "Ok".

Step #5:

Go to the Data Editor page and scroll down to find your new created variable. You'll need to click on Value to make labels. Please add 1 in the Value field, and in the Label field the word "Low", and thus 2 for (Mid), 3 for (High). Click "Add" each time you make a new try, and then "Ok".

That's all.

Cheers,

Ahmed

Please see my paper:

Hello Dr. Greg Camilli.

I am working on running EFA for dichotomous data for 2000 observations. I read your previous comment in the on-going discussion (which was really helpful).

I am curios that whether using "ml" in "fa" is a good way while dealing with dichotomous data. I am a new learner of IRT and EFA, just want to make sure that I understand its application in a refined way.

Tatyana,

I'd use R. You need to install the psych package, as previously noted. Create a data file x of dichotomous responses and then:

out <- tetrachoric(x,y=NULL,correct=TRUE,smooth=TRUE,global=TRUE,

weight=NULL,na.rm=TRUE, delete=TRUE)

mla <- fa(out$rho,nfactors=Q,covar=TRUE,rotate="oblimin",fm="ml")

Although it's very simple. this approach is time consuming with large data sets, and it no so good in the presence of missing data--not to mention that factoring tetrachorics is a very old technology). A  modern approach would be obtained with the software flexMIRT or IRTPRO.

You can use the meta package in R.

First step, perform the metaprop command in an object (e.g. metaprop_results). The metaprop function needs three arguments: the number of events per study (event), the study population (n), the dataframe in which they are stored (df).

metaprop_results <- metaprop(event = your_events, n = your_n, data = df)

Second step plot your results as forest plot

forest(metaprop_results)

The statistical approach to be applied, in this research project, depends upon the type of results (or values) of the trend that are available.

If the values of the trend are available then the test of significance of the difference between the trend of CRP and the trend of WBC can be applied.

This will enable to know whether there is any significant difference between the trend of CRP and the trend of WBC.

However, if the dependence of one on the other (between the trend of CRP and the trend of WBC) is supposed to be known then regression analysis can be applied.

Look like the data description fits into an interrupted time-series analysis

As the biology of a patient has some influence on each of your results, those estimates are in a way dependent on each other. Let's say a subject has low health - this will cause his antibody results be lower in both results compared to someone who is healthy. I assume that the test you used depends on random samples (at least) because of that matter.

My suggestion is to split your subjects sample randomly and use one part for diagnostic test 1, the second part of the sample for test 2. Then compare those results as you did. (be sure to have a large enough sample and to split the sample randomly or else the effect will be biased)

I hope this helps you a little.

If you are very new to medical statistics, you may register the course 'Introduction to Systematic Review and Meta-analysis' for free in Coursera. The course is taught by Bloomberg School of Public Health at Johns Hopkins University. It will help you to get some understanding on how to perform and interpret meta-analysis.

Thanks.

Why not use culture or PCR as a gold standard for calculations?

Nawaf is predicting one measure from another using a regression coefficient (b) as a "conversion" factor.  WLS should be used with the heteroscedasticity in the error structure.  I hope I have now stated that more clearly.  

In a scientific sense, transparency is great. Do science without political or social consequences, a pure search for fact/truth. However, few people work that way.

In a social/political sense I am not so certain that this is a good idea. In an open system, the mistakes that you make can haunt you forever as author or as reviewer. It is hard enough reading the published literature, but adding all the reviewer comments and author responses is too much. The only thing that I can see happing with making drafts and reviews public is that people with an axe to grind will be the ones to sift through to find trouble.

I agree that sometimes the flaws are more interesting. Yet multiple analyses are "confusing" and an "over-analysis of the data." The more frequent comment is to make the methods shorter, get rid of extraneous material, and keep it simple.

/*(For NFHS-3)

For Underweight (-SD), Stunting (-SD), Wasting (-SD)*/

use IAKR52FL

gen haz06=hw70

replace haz06=. if hw70>=9996

gen waz06=hw71

replace waz06=. if hw71>=9996

gen whz06=hw72

replace whz06=. if hw72>=9996

(For -2SD)

gen below2_haz = ( haz06 < -200)

replace below2_haz=. if haz06==.

gen below2_waz = ( waz06 < -200)

replace below2_waz=. if waz06==.

gen below2_whz = ( whz06 < -200)

replace below2_whz=. if whz06==.

(For -3SD)

gen below3_haz = ( haz06 < -300)

replace below3_haz=. if haz06==.

gen below3_waz = ( waz06 < -300)

replace below3_waz=. if waz06==.

gen below3_whz = ( whz06 < -300)

replace below3_whz=. if whz06==.

label variable haz06 "Length/height-for-age Z-score (stunting)"

label variable waz06 "Weight-for-age Z-score (Underweight)"

label variable whz06 "Weight-for-length/height Z-score (Wasting)"

label variable below2_haz "Stunting (-2SD)"

label variable below3_haz "Stunting (-3SD)"

label variable below2_waz "Underweight (-2SD)"

label variable below3_waz "Underweight (-3SD)"

label variable below2_whz "Wasting (-2SD)"

label variable below3_whz "Wasting (-3SD)"

save IAKR52FL, replace

good question I follow

I agree with Michael S. Martin.

Using R is the best option as suggested by Carmen , it can do everything but one thing is that it is not so easy.

One other option you can go through is Epiinfo , it is specially designed for Medical scholars and is free as well.

I hope it helps.

Dear Dr Zurakowski,

One of the most applied approaches in multiple failure-time framework are extensions of the Cox's regression model (all of them are proportional hazards models). According with Kelly and Lim (2000), there have been proposed several extensions which seek to address to the most varies particularities of each case study. For reccurent event analysis, I recommend to you apply the Prentice, Williams and Peterson (PWP) model or the Andersen and Gill (AG) model. The PWP model is suggested to be applies when the risk of occurrence of the following events is affected by the previous ones and the AG model is applies when the events reveals that they have the same risks of occurence. Read the following paper of Amorin and Cai (2015) to easily understand the implementation of this two models: https://www.ncbi.nlm.nih.gov/pmc/articles/PMC4339761/

Further information can be found on my researchgate profile: https://www.researchgate.net/profile/Ivo_Sousa-Ferreira

Best regards,

Ivo Sousa-Ferreira.

Dear Spencer:

 May I say that your question leaves out details that would enable one to provide a cogent response to you.  However, let’s assume that all subjects are already undergoing treatment with standard of care for depression and will remain on such treatment(s) during the study.  If that is the case, then it will depend on what it is you are doing during the intervening period between recruitment and randomization.  Another consideration or question is when you consent subjects and what you tell them this period is to be used for.  There are a number of other points of consideration that come to mind, but in deference to pithiness in this forum I shall stop with the ones I’ve highlighted above.  Good luck

 Cheers,

Chuke

Please let me know if the following references/sites are helpful to you in your quest:

1.  Why Propensity Scores Should Not Be Used for Matching - Gary King

https://gking.harvard.edu/files/gking/files/psnot.pdfDec 16, 2016 ... We show that propensity score matching (PSM), an enormously popular ... proximates random matching which, we show, increases imbalance ...

2.  An Introduction to Propensity Score Methods for Reducing the ...

https://www.ncbi.nlm.nih.gov/pmc/articles/PMC3144483/Jun 8, 2011 ... Propensity score matching entails forming matched sets of treated and untreated subjects who share a similar value of the propensity score ...

3.  Propensity Score Matching: Definition & Overview - Statistics How To

http://www.statisticshowto.com/propensity-score-matching/Feb 3, 2017 ... Statistics Definitions > Propensity Score Matching What is a Propensity Score? A propensity score is the probability that a unit with certain ...

4.  Propensity-Score Matching (PSM)

http://cega.berkeley.edu/assets/cega_events/31/Matching_Methods.pptPropensity score matching: match treated and untreated observations on the estimated probability of being treated (propensity score). Most commonly used.

5.  Understanding Propensity Scores

http://bayes.cs.ucla.edu/BOOK-09/ch11-3-5-final.pdf11.3.5 Understanding Propensity Scores. The method of propensity score ( Rosenbaum and Rubin 1983), or propensity score matching (PSM), is the most ...

Dennis

Dennis Mazur

I don't know if there some formula to convert risk ratio to Cohen's D, but certainly you can convert odds ratio to Cohen's D. And, i suppouse if you have an 2x2 table you can calculated both: a risk ratio and a odds ratio. So i recomend you to use odds ratio. 

Micheal Borenstein's book "Introduction to Meta-Analysis" (2010) give the formulas for converting odds ratio to cohen's d and viceversa in the chapeter 7. I attach the link to that chapeter.

Also, there are sites when you can do the calculations automatically, i'll post you one link that i found useful.

I hope i was helpful.

What T tests do you know?

I'm not clear on what information you have, but if you have an incidence rate (i.e., cases per person-time) then you can assume a Poisson distribution.  Suppose you have 150 cases over 20,000 person-years.  Your incidence rate (IR) would be cases/(total pt):  150/20,000 or 75 cases per 10,000 person-years.  The variance for the IR is cases/((total pt)^2):  150/(20000^2) or 3.75 x e-7.  Your standard error the square root of this:  sqrt( 3.75 x e-7) or 0.00061237.   Thus your 95% CI for the IR (0.0075)  would range from 0.0063 to 0.0087.  So if you only have the incidence rate and total person-time, you can work your way back to an SE.

If you know the number of incident cases, then the SE for a Poisson random variable is:  sqrt(cases). (For me, this is the easier approach).   For the above problem we have 150 cases (ignore the person-time for the moment).  The 95% CI for the number cases is:  150 - 1.96*(sqrt(150)) or 126 as the lower bound, and 150 + 1.96*(sqrt(150)) or 174 for the upper bound.  From here, you can the 95% CI for the incidence rate by dividing the lower and upper bounds for the cases by the person-time.  126/20000 or 0.0063, and 174/20000 or 0.0087.   These agrees well with the previous IR estimates.

Paul

Of the 44 how many stopped taking the tasks, and how many continued. The imbalance may be not so worrisome unless a lot of students did not continue the task.

Coming from a clinical quality improvement background, my projects tend to be pilot or exploratory studies. I typically don't run power calculations beforehand. Still, I suggest when presenting results that a limitation might be the sample size.

 Hi Aleksandra, how's it going? 

I have the same exact question about the prevalence index and bias index with more than 2 observers and with variables that contain more than 2 categories. 

Did you get some additional thoughts about this question? 

Thanks!

no you cannot do that. In case of case-control study your outcome will be binary that is diseased and nondiseased. Plus the variable you are matching on cannot be studies.

Dear Tang

If I correctly understand the design of the two studies you quoted: The investigators started with identifying cases of Colo-rectal cancer and measure the RNA expression. In the mean time they measured the RNA expression in a comparative group who did not have Colo-rectal cancer. Thus both the disease status and the assumed risk factors were measured at the same time. This is a clear cross-sectional comparative study. Now, it is assumed that the RNA expression preceded the appearance of cancer , i.e. RNA expression could be a risk factor for Colo-rectal cancer. The analysis of data then will be like analysis of case -control studies. No intervention was made by the investigators. This means that you consider both of these two studies as case-control ones in your classification. In standard epidemiological text books you will find support to my view> Keep in your mind that in teaching epidemiology, we tend to make sharp distinction among various designs. In real life, overlaps  and modification do exist and feasible. For example from cohort studies we can make nested case-control studies, from cross sectional studies we can measure incidence rates and  also make a case-control sub-design. 

With best regards

Cross Validated

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up vote4down votefavorite

Are different p-values for chi-squared and z test expected for testing difference in proportions?

r chi-squared p-value binomial proportion

I'm trying to test the difference in proportions using the z test method and chi-squared method, but am getting very different answers. Is that normal?

My data:

CI CII

Male 205 102

Female 83 39

Calculating the z score I get 0.25 which should correlate to a p-value of 0.4013. Calculating the chi-squared score I get 0.0626 correlating to a p-value of 0.8025.

I read that the z-score requires some assumptions (probability of success is ~0.5 and n is high). Is this violating those? Or is it just the nature of these different approaches that gives very different answers with the same meaning (no evidence of difference).

I'm certainly open to miscalculations, but I've re-checked. If this behaviour isn't normal I'll recheck again!

Here are my calculations in R.

> r1 <- 205

> r2 <- 102

> n1 <- 288

> n2 <- 141

> (p1 <- r1/n1)

[1] 0.7118056

> (p2 <- r2/n2)

[1] 0.7234043

> (common.proportion <- (r1+r2)/(n1+n2))

[1] 0.7156177

> (se.pooled <- sqrt(common.proportion*(1-common.proportion)*(1/n1+1/n2)))

[1] 0.0463676

> (zscore <- (p1-p2)/se.pooled)

[1] -0.2501466

>

> # chi-squared

> prop.test(c(205,102), c(288,141), correct = FALSE)

2-sample test for equality of proportions without continuity

correction

data: c(205, 102) out of c(288, 141)

X-squared = 0.0626, df = 1, p-value = 0.8025

alternative hypothesis: two.sided

95 percent confidence interval:

-0.10208385 0.07888645

sample estimates:

prop 1 prop 2

0.7118056 0.7234043

share improve this question

askedMar 13 '15 at 4:13

Tom

542●1●7●14

editedMar 13 '15 at 17:33

gung

87.5k●23●194●398

  

 

A related issue. If I'm presenting confidence intervals around each proportion calculated using the binomial distribution, but then comparing them and presenting a p-value using chi-squared, that seems a bit wrong. I might have overlapping CIs, but then a p-value that's less than 0.05. Am I thinking correctly? – Tom Mar 13 '15 at 4:29

3

 

Tom, can you show your math? These two tests should give very similar results for your sample sizes (especially if making the same choice about continuity corrections). – Alexis Mar 13 '15 at 5:19

  

 

Thanks, @Alexis. I've added the math in now. – Tom Mar 13 '15 at 7:24

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up vote7down voteaccepted

Very simple: both the z test and the contingency table χ2χ2 test are two tailedtests, but you have got the one-sided pp-value for your z test statistic. That is for H0:p1−p2=0H0:p1−p2=0, the pp-value = P(|Z|≥|z|)P(|Z|≥|z|), but your reported pp-value is only P(Z≤z)P(Z≤z).

Notice that 0.4013×2≈0.80250.4013×2≈0.8025. Easy!

share improve this answer

answeredMar 13 '15 at 17:01

Alexis

10.8k●2●31●70

editedMar 13 '15 at 19:05

5

 

And the square of the Z-score is (−0.2501466)² = 0.06257, which equals the test statistic X-squaredfrom the prop.test() output. – Karl Ove Hufthammer Mar 13 '15 at 18:55

1

 

Thank you for that "easy" answer! (Which prompted a good schooling on one-tailed tests. And now @KarlOveHufthammer you're sending me down another schooling. If x-squared is simply the z score squared, why do we even have it? And why isn't it called z-squared? (Obviously, not to be answered here. I have a lot to learn!) – Tom Mar 16 '15 at 6:29

2

 

@Tom The reason that it’s called X-squared in the R output, is that the X is really an ASCII interpretation of the greek capital letter chi (Χ) (the lowercase version of this letter looks like this: χ). And it’s a chi-squared test, which is used in lots of other situations. That said, the test(s) should never be used for comparing binomial proportions, as they have terrible statistical properties. See stats.stackexchange.com/questions/82720/… – Karl Ove Hufthammer Mar 16 '15 at 16:42

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I wouldn't draw a conclusion based on a p-value at all.

If you designed an experiment to specifically test the diabetes-association of one particular SNP because you had some prior hypothesis why this should play a role, then a large p-value would tell you that your data is insufficient to show that association (your data is inconclusive). If the p-value is small, then this is only the very first step towards an interpretation that should appreciate the rest of the biological knowledge about the role of this SNP in the pathology of diabetes and the estimated effect size.

If you did a screening, p-values (corrected or not) are essentially only a proxy to rate the results according to their "statistical signal-to-noise ratio". A good ratio (= a small p-value) can be the result of a strong signal or the result of a small noise. I would use the results only to generate a ranked list to highlight the "top candidates" (according to thier "statistical signal-to-noise ratio" or p-value). I would see it only as an experiment-internal feature, not to generalize. The aim of a screening is to identify the most promising candidates for further studies. This essentially means: a screening is done to generate hypotheses, not to test them.

And please do not confuse statistical significance (p-values) with effect sizes. A tiny p-value does not (neccesarily) indicate a strong association, and a strong association may give a large p-value. These are two different things only indirectly linked via the model (what cofactors are considered), the sample size (what one can experimentally control) and the variance (what one cannot control).

Hi Mari

There is an article that discuss this issue in details and may help you. The article name is 'Should Meta-Analyses of Interventions Include Observational Studies in Addition to Randomized Controlled Trials? A Critical Examination of Underlying Principles'.

Good luck!

Hi Saraswati

Peter's advice is excellent. If you decided to power up for a medium effect, then you would need 64 patients in each study arm for a 2-arm trial with 80% power and alpha=0.05. If as Peter suggested, you go for a 3-arm trial, then you would need 69 patients per study arm to allow for (say) two pairwise comparisons and with alpha set to 0.04 to adjust for multiple comparisons. Although choosing a medium effect size might seem a bit arbitrary, a difference of 10 on a 100-point scale (eg the SF-36) has often been found to be of clinical importance. Further, a 100-point scale is likely to have an SD of around 20. This gives an effect size of 0.5 (or medium), which is what we have powered for.

Adrian

Could you give a little more information:

Do you have the table of survival rates over times in that 2 year interval or is it just a pair of times 2 years apart?

If it is the former, you have the ability to work out the p-value for the log rank test.  In computing the log rank test, you need the information needed for the hazard ratio anyway if my understanding is correct.

If it is the latter, I would expect that the p-value wouldn't mean anything.  You need the observed survival information and then you need to compare it to expectation.  You can calculate expectation simply using methods in the following:

Medical Statistics such as this are not my first field, so I may be off base, but I cannot see that the p-value matters if you have the survival data itself given what I know about the hazard ratio's equation.

There are two (or more) options here:

i) Convert the coding of your variable. If it were 0/1, change it to 1/0. This requires some more efforts.

ii) AFAIR, SPSS has an option in the box when "Categorical" button is hit. This option indicates that which category is the reference(First or Last). You can simply swap it to get the desired HR.

You can use logistic regression, but for obtaining odds ratio it is better to categorized age variable. If gender is matched across the groups the gender variable would be non significant.

Thank you.

Thank you all for your respective inputs! 

Hi Bruce

I am glad you find it useful

The mvmeta in Stata is quite straightforward to use. There also other commands (including a mvmeta) in R.

I attach the link of a review paper for multivariate meta-analysis with guidance on how to perform it in various software

This problem can be solved by using the technique of functional data analysis, where no distributional assumption is required. 

Thank you Dianatinasab for the reference.

I wonder whether we can use the given formula in the paper to do the sample size calculation for cross-over design or not?

given formula in the paper --> n= 2 ((a+b)^2) x Variance)/ ((mean group 1- mean group 2)^2)

In case anyone comes across the same question: I ended up using standard deviations which can be compared easily (e.g., using LeveneTest). The SD correlates with the a_wg and related measures of dispersion above .90 (Roberson, Sturman, & Simons, 2007)

since you want to study the effect of multiple risk factors and you are measuring before and after cycling, the design is experimental - not randomized and not observational with One group of cyclists, since your comparison group is the same cyclists before event. For analysis if you can dichotomize cTn elevation and define what is considered illness or not, then a conditional logistic regression would be enough. If not a repeated measures analysis also known as longitudinal data analysis should be done.

Thanks for the answers. This isn't for my work in fact, but a student project which is being undertaken over the summer, hence not having many details to share yet. Useful advice, thank you.

Hi,

Basically, there are two approaches to sample size calculation, precision-based (relevant if you are trying to estimate the proportion, mean difference etc) and power-based (relevant if you are trying to test a hypothesis; how small a difference is it important to detect and with what degree of certainty). See file attached for more details.

If you are unable to find previous studies, you can consider conducting a small pilot trial and use the estimates from that to calculate your ideal sample size.

Good luck!

You can always guess the standard deviation by asking someone who knows the area to tell you what would be an unexpectedly high and an unexpectedly low value. Divide this range by four. 

I got interested in this and used to ask, say, obstetricians what would be a high and a low fetal heart rate. They produced more or less a range of four SD. 

Thank you for your answers that has been very helpful!

Based on the test results you have provided, you do not have evidence to conclude that publication bias exists. That may be because there is no publication bias or you do not have the statistical power to detect it.

Thank you so much dear Mr. Weaver.

It is very unlikely that 43 items can be converted into 12 reliable and valid variables. Use the exploratory factor analysis (Dimension Reduction / Factor) to determine the number of factors (variables) in your data set. You might also want to analyze your items scores to determine the consistency between your items (Scale / Reliability Analysis).

Thanks a lot for all your answers I used the age <60 and >60  as an examples my question was about any 2 descriptive variables thanks again you were very helpful

A simple and efficient approach is to use models for discrete-time, grouped data, as described in

Prentice, R. L. & Gloeckler, L. A. Regression analysis of grouped survival data with application to breast cancer data; Biometrics, 1978, 34, 57-67

You can use marginal (eg binomial logistic regression) or random-effect models. We have implemented some models for grouped data in the package aods3 for R 'on CRAN). A case study in

This approach has the advantage of accounting for competing risks when the time step is short. Also, when using a complementary log-log link and a small time step, estimates are equivalent to those obtained with continuous time (Cox) models.

Should not be unless technical fault on formula setup up at excel ...not spss. Thanks

Regards

JOE