Biostatistical Methods - Science topic

Daniel Ruivo Marques, pardon me for the late response. Thank you for your guidance.

Le signe négatif en mathématique en général et en statistique en particulier a toute son importance notamment au niveau des commentaires des résultats (exemple: la corrélation positive s'oppose à la corrélation linéaire négative). Les signes sont à respecter.

It sounds like you have a lot of interesting questions to investigate! Here are some suggestions on how to approach your analysis:

For all these analyses, it would be useful to calculate alpha and beta diversity metrics such as Shannon diversity index and Bray-Curtis dissimilarity, and to visualize the data using ordination plots. Additionally, you may want to use differential abundance analysis to identify specific microbial taxa that are differentially enriched or depleted between treatments, time points or genotypes. There are several R packages available for performing these analyses, including vegan, nlme, and DESeq2.

Good luck with your analysis!

I am interested

I feel, that the people who posted answers completely ignored my first response. Each assay is based on its own chemistry, designed for certain goals, and used in different conditions. You need to understand well what you are measuring. Otherwise, you put into a basket apples, oranges, pears and unknown fruits and want to compare your and someone's else baskets. How you can say which basket is better? My point is that the RACI does not have any chemical sense.

Hi,

Look at the distribution of scores whether they are skewed. Correct for the skew by some transformation and again look for the distribution. You may be able to do further analysis. Conversely, frequencies in the categories can be compared.

Here are a few references on the Psychometric properties and its use in some studies of the scale:

Morin CM, Belleville G, Bélanger L, Ivers H. The Insomnia Severity Index: psychometric indicators to detect insomnia cases and evaluate treatment response. Sleep. 2011 May 1;34(5):601-8. doi: 10.1093/sleep/34.5.601

Schulte, T., Hofmeister, D., Mehnert-Theuerkauf, A. et al. Assessment of sleep problems with the Insomnia Severity Index (ISI) and the sleep item of the Patient Health Questionnaire (PHQ-9) in cancer patients. Support Care Cancer 29, 7377–7384 (2021). https://doi.org/10.1007/s00520-021-06282-x

Yusufov M, Zhou ES, Recklitis CJ. Psychometric properties of the Insomnia Severity Index in cancer survivors. Psychooncology. 2019 Mar;28(3):540-546. doi: 10.1002/pon.4973

Ohayon MM. Epidemiology of insomnia: what we know and what we still need to learn. Sleep Med Rev. 2002 Apr;6(2):97-111. doi: 10.1053/smrv.2002.0186

Okajima I, Miyamoto T, Ubara A, Omichi C, Matsuda A, Sumi Y, Matsuo M, Ito K, Kadotani H. Evaluation of Severity Levels of the Athens Insomnia Scale Based on the Criterion of Insomnia Severity Index. Int J Environ Res Public Health. 2020 Nov 26;17(23):8789. doi: 10.3390/ijerph17238789

Gagnon C, Bélanger L, Ivers H, Morin CM. Validation of the Insomnia Severity Index in primary care. J Am Board Fam Med. 2013 Nov-Dec;26(6):701-10. doi: 10.3122/jabfm.2013.06.130064

Kraepelien M, Blom K, Forsell E, Hentati Isacsson N, Bjurner P, Morin CM, Jernelöv S, Kaldo V. A very brief self-report scale for measuring insomnia severity using two items from the Insomnia Severity Index - development and validation in a clinical population. Sleep Med. 2021 May;81:365-374. doi: 10.1016/j.sleep.2021.03.003

Massimo Sivo very nice of you that you want to try to help your colleague, however, as was mentioned earlier, you should understand experimental design very well before you have sufficient information to construct an appropriate statistical model (that can than provide meaningful insights). To understand experimental design of a clinical trial it is not enough to understand some statistics, you should also understand some medicine. E.g. From what Shahnawaz Ahmad Wani described one can by no means understand what was actually measured, how, and why. It is super challenging to sample 5 patients with appropriate controls to make any kind of biologically meaningful inference about the population of patients. Tons of confounders and no power to take them into account. From how the problem was presented I can image such (confounder) data was not collected. Using such data and shoving them into e.g. mixed model to account for repeated measures (and dependence of some parameters) will most likely give meaningless results. Even if everything was done right but you did not account for a biologically meaningful confounder... it would still be meaningless. For example, Shahnawaz Ahmad Wani told us he is measuring some kind of biochemical parameter from the blood from female patients. Many parameters in the blood change with the part of the menstrual cycle, and it is not clear whether the authors analysed this. Without accounting for such a serious confounder the study makes no sense. Furthermore, how would you construct a model around the dependence of venous blood and "tissue blood". The variables are most definitely related, but we cannot imagine the nature of their dependence without having much more information. So, before asking for data, you should really ask for the explanation of the design to make sure all the confounders were ruled out the the level of design as this is the only scenario in which several lines of code or several clicks in some statistical software will help you in any way to make some kind of inference.

Please be mindful about this kind of stuff, uncertain and misleading conclusions can be very dangerous in medicine and it is healthier for the community to talk about them than to just generate some numbers from the data.

Best,

jan

The observational study whether cross sectional , prospective .Also retrospective study will help , it depends what type of error .Using trigger tool for specific errors can help too

Hi,

As there are only 18 units and a single dose study without repetition of treatments, Whynot go for simple randomization procedure with random number generator into the three groups (A,B,C) one of the groups is similar to other that gives a 2:1 ratio. Groups A,B and C could themselves be randomized.

Here is a good book on this subject.

Lachin, John M.; Rosenberger, William F (Wiley series in probability and statistics) Randomization in clinical trials: theory and practice [2 ed.] John Wiley & Sons,2016.

Just type in your data and do it with Excel just as you have in the past. There's no major difference with what you did in the past. Best wishes David Booth

Kindly follow the SPSS structure for determining the critical value here: It is pretty simple and intuitive

Thank you!

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.

You might try looking at amphiban/reptile studies, as many have been done at various altitudes for populations in vernal pools. I don't have such a paper on hand, but you can search for that, and they all contain some type of corrleation factor. Joell

Actually, the log transformation in the figure I attached was done according to the formula: log((#1+1)/(#2+1)). Just later, I thought that I added "1" to my values to be able to carry out log transformation (not to eliminate zero values). So I considered that maybe, it would be more correct to add "1" to adjusted values just before the transformation.

Thanks again :) Jochen Wilhelm

Thank u so much for both your answer and suggestion David Eugene Booth

This probably does not help you, but I thought that I would have a look at the original Student (Gossett) paper of 1918 as the test was specifically designed for (very) small samples:

"if our sample be small, we have two sources of uncertainty: (1) owing to the “error of random sampling” the mean of our series

of experiments deviates more or less widely from the mean of the population,

and (2) the sample is not sufficiently large to determine what is the law of

distribution of individuals. It is usual, however, to assume a normal distribution,

because, in a very large number of cases, this gives an approximation so close

that a small sample will give no real information as to the manner in which

the population deviates from normality: since some law of distribution must

he assumed it is better to work with a curve whose area and ordinates are

tabled, and whose properties are well known. This assumption is accordingly

made in the present paper, so that its conclusions are not strictly applicable to

populations known not to be normally distributed; yet it appears probable that

the deviation from normality must be very extreme to load to serious error. " My emphasis

" Section X. Conclusions

1. A curve has been found representing the frequency distribution of stan-

dard deviations of samples drawn from a normal population.

2. A curve has been found representing the frequency distribution of the

means of the such samples, when these values are measured from the mean of

the population in terms of the standard deviation of the sample.

3. It has been shown that the curve represents the facts fairly well even

when the distribution of the population is not strictly normal." Again my emphasis.

The are several examples with a sample size below 10 in the paper.

When I used to teach this stuff (1st year geography students), I would demonstrate the Fisher Randomization and permutation test for very small numbers as the students could do this by hand and thereby see the underlying logic of the test. I would show that you could permute the data of the two variables under the null hypothesis of no difference and see how extreme a result you could get 'by chance' and then compare the observed value to this; no normality assumptions were needed in coming to some sort of judgement.

Try Correlation matrix on R Programming. Try corrplot( ) package in R

The calculation of the sample size depends on the variance in the results. Certain software such as G power can help you calculate the sample size based on the mean difference, and the variance between the groups.

Yes, I think Tami Barry's answer is good, thank you very much

The plastic dosage seems to be a quantitative variable, and I hope you are free to choose the amount. If the functional form of the relationship between plastic dosage (X) and compressive strength (Y) are not known, I would use a dynamic range as large as possible, from tiny amounts to huge amounts of X, and I would choose as many different values of X as practically feasable, being more or less logarithmically spaced over the range of X. It may be even worth using only relatively few different values of X (5-10 or so) in a first round just to get an idea what range of X might be interesting, and then focus on this range in a second round.

Thank you a lot for your help!

You need normalization

there are many different ways to estimate the sample size and you can select the suitable one for your research.

Good Morning,

I've never used SPSS and the last time I used SAS for survival analysis was 20 years ago when I was still in University. It is relatively simple to set up a Kaplan Meier analysis in R using the Survival package. Granted, you have to learn a bit of code, but there are lots of resources available online to walk you through it. On the positive side, once you get your code working, you can re-use it, and, while it is more work at the beginning, you are forced to think about and understand your data and each step in the analysis, which leads to a better understanding of the outcomes. A good introduction tutorial on survival analysis using R can be found here: https://www.emilyzabor.com/tutorials/survival_analysis_in_r_tutorial.html

Good Luck!

You may try this by using Chart Builder.

NEW FILE.

DATASET CLOSE ALL.

GET FILE "C:\SPSSdata\bankloan.sav".

DATASET NAME raw.

* OMS.

DATASET DECLARE logistic.

OMS

/SELECT TABLES

/IF COMMANDS=['Logistic Regression'] SUBTYPES=['Variables in the Equation']

/DESTINATION FORMAT=SAV NUMBERED=TableNumber_

OUTFILE='logistic' VIEWER=YES

/TAG = 'logistic'.

LOGISTIC REGRESSION VARIABLES default

/METHOD=ENTER age employ address income debtinc

/PRINT=CI(95)

/CRITERIA=PIN(0.05) POUT(0.10) ITERATE(20) CUT(0.5).

OMSEND TAG = ["logistic"].

DATASET ACTIVATE logistic.

COMPUTE Vfilter = Var2 NE "Constant".

FILTER by Vfilter.

VARIABLE LABELS Var2 "Variable".

* Chart Builder.

GGRAPH

/GRAPHDATASET NAME="graphdataset" VARIABLES=Var2 MAXIMUM(Upper)[name="MAXIMUM_Upper"]

MINIMUM(Lower)[name="MINIMUM_Lower"] MEAN(ExpB)[name="MEAN_ExpB"] MISSING=LISTWISE REPORTMISSING=NO

/GRAPHSPEC SOURCE=INLINE.

BEGIN GPL

SOURCE: s=userSource(id("graphdataset"))

DATA: Var2=col(source(s), name("Var2"), unit.category())

DATA: MAXIMUM_Upper=col(source(s), name("MAXIMUM_Upper"))

DATA: MINIMUM_Lower=col(source(s), name("MINIMUM_Lower"))

DATA: MEAN_ExpB=col(source(s), name("MEAN_ExpB"))

COORD: rect(dim(1,2), transpose())

GUIDE: axis(dim(1), label("Variable"))

GUIDE: axis(dim(2), label("Odds Ratio & 95% CI"))

SCALE: linear(dim(2), include(0))

ELEMENT: interval(position(region.spread.range(Var2*(MINIMUM_Lower+MAXIMUM_Upper))),

shape(shape.ibeam))

ELEMENT: point(position(Var2*MEAN_ExpB), shape(shape.circle))

END GPL.

Good luck

Hao Tran With Poisson regression use predicted lambdas , you need to convert them to predicted probabilities to get at least 1, which would be: 1 - exp( - lambda).

Good luck.

Many mathematical models and projections were done regarding the covid-19 pandemic. They predicted that it will last so many months or weeks etc. some advocated complete lockdowns and then projections were done accordingly.

But to the best of my knowledge, none of the projections were proven right. Now only time will tell how long its course will run. It will depend upon the post vaccination scenario and emergence of new strains, probably with antigenic drift only. We will come to know more clearly about the pathophysiology of this virus in due course of time as more evidence is coming day by day...So I think we still don't know how long the pandemic will last.

It sounds like you already tested your hypothesis with the linear model and used up your p value. So, you can't do more p-value tests.

As far as choosing a model, that isn't a question for the stats folks, but is related to the theory you have (stats folks can say how to implement or approximate the model).

You can do it with the glst command in STATA for a single study. For more information: https://www.stata.com/meeting/nordic-and-baltic13/abstracts/materials/se13_orsini.pdf

Mahbubur Rahman This article is best explaining your request using SPSS.

Step by Step Explanation.

https://docs.wfp.org/api/documents/WFP-0000022418/download/

Cohen's d isn't a test. It is a measure of effect size.

It allows you to express the difference between two groups in terms of the naturally-occurring variation in the thing you are measuring. The variation is measured by using information from both groups and pooling it as the pooled standard deviation.

The trouble with Cohen's d is that people tend to convert it to tee-shirt sizes – small, medium, large. This seems very vague when you go to the bother of doing all those calculations, somehow. And studies looking at typical values of d in different research areas suggest that it's not appropriate to have the same definitions of small, medium and large for all disciplines. I have a reference somewhere that I can post if I locate it!

How can I found out X ( microorganisms concentration- biomass concentration) and S ( substrate concentration) from blackwater characteristic data? I mean, COD or VSS does represent these?

and in case if you time between the PM2.5 exposure until the time they got TB, then it's to use COX regression and represent it through KM curve as well.

I would recommend using Revman. but you still could use SPSS for metanalysis.

.

your expected mortality rate is a confounding variable in your analysis

the adjustment method depends a lot of the data but you can have a look at the following thread

https://www.researchgate.net/post/What_is_the_best_method_to_deal_with_confounding_Stratification_or_matching

if you have a good overlap between the score distribution in both groups (despite different means), you could go for stratification, although it may rule out too small samples

.

Graphic representations represent the normal probability of model residuals. It is useful for determining the pattern of residuals. A higher concentration of points around the diagonal also means supporting the claim of approximate linear dependence or accepting the assumption of normally arranged residuals.

I'm not sure where you found that t tests do not take into account standard errors of means. T-tests are based on the t statistic, which is the quotient of the mean difference and the standard error of the difference.

Do you mean that the t-test only take into account variability between growth curves of individual cells, but not the variability of each curve? T-tests assume that individual data point (in your case growth curve slopes) comes from a single measurement. One underlying assumption of t-tests is that measurement error is distributed normally, so the average measurement error will approach (or be) zero as the number of data point increases.

Hi Louis Violeau ,

Attached you'll find it. However, I recommend:

1. Watching the Dr. Cumming's video explaining how to analyze heterogeneity in terms of fixed/random effects and diamond ratio (this is extended in his book): https://www.youtube.com/watch?v=_bB2k-1Hv9E

2. Running the "metaforest" or "mc.heterogeneity" function within R. Here you have the recent version of the package 'meta' (https://cran.r-project.org/web/packages/meta/meta.pdf) and a very nice hands-on guide (https://bookdown.org/MathiasHarrer/Doing_Meta_Analysis_in_R/).

Hope this help.

You can use a Mann-Whitney U test for non parametric data

Dear Rubina,

I am working on analyzing polymorphisms on CYP3A5 gene. The percentaje of the AA (CYP3A51*/1*) genotype in the population (50 patients) is 72%, of Ab (CYP3A51*/3*) genotype is 28% and of bb genotype (CYP3A53*/3*)is 0.

I am trying to calculate Hardy Weinberg equilibrium on STATA but I don't know how do I have to introduce the data....¿Do I have to generate a variable named, for example, "Genotype" and put the results of the genotyping analysis as categorical results, I mean: "AA", "Ab" and "bb"? (Patient 1: AA, Patient 2: Ab, and so on.....)¿Which is the second variable to which I have to do the comparison in order to obtain a p value? I am very confused.

Please If anyone could help me, I would really appreciate it! Thanks!

Yes do use the chi-square test. In attachment an R script for this, with Post Hoc tests.

Finally I got the answer and is... just do not provide the allele frequencies , the analysis will run fine, reassuring this, here is a Github manual of how to perform an EWAS meta-analysis (https://github.com/ammegandchips/meta_EWAS/blob/master/metal.md) . As you can see the parameters regarding the frequencies and genomic control are off.

I agree with Jochen. You can do a linear mixed model with treatment and as fixed factor and experiment as random factor. This is easily done in R with the lmer (package lme4):

model <- lmer(confluence ~ Treatment + (1|experiment). data=data).

If treatment is significant, you can do follow up pairwise comparisons to ascertain which treatment is having the strongest effect on your cell confluence.

Good luck!

Even though this is an older question, I would like to add that the aov function is used for balanced designs (the results can be hard to interpret without balance ).

Please see the "Note" here: https://www.rdocumentation.org/packages/stats/versions/3.6.2/topics/aov

I think you can show an association of the clinical variable with your disease state using linear/logistic regression and calculate Odd Ratio (OR). In addition, you may carry on same for other test, and compare their ORs. It does not need to test sensitivity, precision, and accuracy as like diagnostic tool (i.e., ROC-AUC). It would be supportive for follow-up experiments.

Hi Sir....As discussed above, you can always revisit the effect size (Cohen's d) taken for the sample size calculation. Taking a very large effect size may yield a very low sample size. using G power software will help you to calculate precise effect size and sample size for the your study.

Thanks

http://environmentalcomputing.net/introduction-to-mvabund/

SOLUTION: Use Wan's method (2014) ~820 cited. Superior to Hozo's method. Hereby the PDF and Excel.

As I understand, the cells were randomly distributed over the 6 different concentrations, but each cell was treated for 12, 24 and 48 hours. If this is correct, then you have a Between (6 levels of Concentration) x Within (3 levels of Time (hours)) design. First run a factorial repeated measures ANOVA. If the there are sign. effects (of Concentration, Time and/or the interaction) then you can run Post Hoc tests. Before running these tests, check the distributions of the measures on the treated cells with histograms. If they obviously do not look normal, then run a non-parametric test such as the f1.LD.f1 from the parLD R package (https://www.jstatsoft.org/article/view/v050i12 )

I suppose you normalize the images of the cells like in this stdudy: https://www.ncbi.nlm.nih.gov/geo/query/acc.cgi?acc=GSM1224512

In the two way table D+ D-

E+ a b

E- c d

disease odds ratio = (a/b )/(c/d) = ad/bc

exposure odds ratio = (a/c)/(b/d) = ad/bc

Nice link, Bruce. Bland and Altman have a great series about statistics in bmj. It's certainly worth reading their other articles, too.

Today I would add that we do have statistical modelling techniques available to everyone that allow using proper distributional assumptions. Transforming the response should be avoided. I had some references showing that but I don't recall where I put them :( Here is one paper I found in a haste:

Unfortunately, the authors alsow fall back on the ancient suggestion to use "non-parametric" approaches.

There are some papers I found saying that log-transformations of count data is inappropriate (e.g. )

The example given in the bmj artice should, as I see it, be analyzed using a gamma-GLM with log-link. The response should be the latencies (Tab1) or the concentrations (Tab2) and the baseline values should be used as covariate.

In the Tab.1 data set it is quite obvious that the 11th record is "outlying". The interesting question is: why that? Is that value trustworthy? If yes, we should take it seriously, and in this case we must admit that the variability in the data is so large that we should not confidenctly conclude that the treatment reduces the AB concentration (p=0.3). If the value is in fact problematic and one can show that there was a mistake (e.g. the after-value was 1.22 instead of 12.2), the data very clearly supports a decrease in the concentration (p<0.000001 when removing #11 or when changing it to 1.22).

A sign test (giving p = 0.004) is obscuring that there might be a problem with the data. In this case it is ok since changing the (possibly erroneous) value to 0 would still give p = 0.02 (using values > 12.2 would not change the result anyway). However, there are many zeros and ties, so "exact" p-values can not be calculated (I don't have any clue how big the error may be).

Classify the data in a two-way table i.e. a table that is composed of the two different techniques as two rows and the five different conditions as five columns.

Then apply Analysis of Variance technique. You will be able to compare the difference between two techniques as well as the difference among the five conditions.

Ignoring the columns and considering the observations under the two rows, you can apply t test for comparing the two techniques. However, in this approach, you will not be able to compare the five conditions.

Hello Robert, there are extensions of recursive partitioning and trees for longitudinal and clustered data. They essentially include a mixed model element into the algorithm. I have used the RE-EM algorithm in the past (see DOI: 10.1007/s10994-011-5258-3 and DOI: 10.1016/j.csda.2015.02.004). There are also binary partitioning for continuous longitudinal data (DOI: 10.1002/sim.1266) and mixed-effect random forest (DOI: 10.1080/00949655.2012.741599). Implementations can be found in R packages: REEMtree, longRPart2, MixRF.

If you harvest during exponential growth clearly that may play a part in your interpretation of experimental results. However that was not the issue. All cells were of the same cell type to within a small number of spontaneous mutations perhaps. The only difference between the plates is that one plate contains hormone while the other does not. The inoculate for each paired plate comes from the same individual. I assume growth curves were obtained for cells with and without hormone so you know if hormone has an effect on cell growth and what if any that effect is so if such an effect exists you may consider it with the rest of the experimental results. Given all this unless some strange effect happens that requires explanation. This is statistically a paired t test. Any effect on the growth is known so any correction necessary can be made, such as was suggested by Dr. Ebert. Again when all necessary corrections, transformations and what have you are finished. The comparison is the paired t-test. This is exactly how we did chemical mutagenisis studies in Herman E. Brockman's lab many years ago. See the link for publications from some in the group with Materials and Methods sections. I apologize for my poor description but this was 50 years ago for me. Mea Culpa.

https://www.google.com/search?q=Herman+E+Brockman+Neurospora+yeast+mutagenisis&rlz=1C1CHBF_enUS847US847&oq=Herman+E+Brockman+Neurospora+yeast+mutagenisis&aqs=chrome..69i57j33.47880j0j8&sourceid=chrome&ie=UTF-8

Again, I see no reason why a poorly remembered something or other affects either the statistics or the biology. As always the assumptions of the test should be checked. Best, David Booth

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.

In R the model could be formulates like

model = glmer(rates ~ implants*vesicles + (1|donor), family=Gamma)

where rates are the rate constants detemined, implants and vesicles are binary factors (yes/no) and donor is a factor identifying the subject (used a random intercept factor in the model).

You are interested in the interaction of implants and vesicles. The difference-in-difference statistic can be obtained with

summary(model)

and you can get a p-value with

anova(model, test="Chisq") or drop1(model, test="chisq")

No need to clubbed the data. find the correlation between the variables not the groups. For example in group one variables height and weight and similarly in group two. so we will apply the correlation analysis between the variables height and weight. we will check how much correlated height and weight to each other.

if the data is normal then use Pearson correlation otherwise Spearsman. After getting the correlation value ( r or rho) then simply check in which group the variables are much correlated ( in control or case group).

Dear Emmanuel Nene Odjidja,

If you familiar with R, then you can try with R survival package for cox regression model. Please follow the link: https://rviews.rstudio.com/2017/09/25/survival-analysis-with-r/. I hope the above link would be helpful for you.

Hey, Evan

Absolutely, low < med < high ---> ordered scale (Likert-type)

Thanks for the information. I have used the psmatch packaged in STATA to perform propensity score matching but I don´t known how can I get 5 controls for 1 case. Can anyone help me with syntax for STATA??

Dear Idah Chatindiara

Thank you for your comment

In fact, since 2015, things changed about heteroskedastic dealing, and now it has become almost mandatory (so I think that it was why you was asked by peer reviewer). Happily, the last versions of SPSS integrate it in cox regression through sandwich estimators and, more important, HC in general linear models.

http://www-01.ibm.com/support/docview.wss?uid=swg21477323

Hope it helps,

Kind regards,

Hello,

If I well undertood your question, I think here you can use the Bayes Theorme, based on clinic litterature if you have about this specific test. we can consider the ultrasound results as a priori information about the clinic test, and the Gold-standard one as the real data (x).

I can reformulate your problem as :

If your test results come back positive when using ultrasound method, what are your chances that children actually have the disease (with G-S method) ?

CHELLAI. F

Hello Abdulbaset,

In order for SPSS (which is what I presume you are using for analysis) to compute and report a Box M test (multivariate homogeneity of covariance matrices test, a test that is not needed if you are strictly sticking to univariate repeated measures designs), you need:

1. Two or more independent groups (e.g., at least one between-subjects factor).

2. Scores on two or more measures per case.

3. SOME variance for each measure within each group.

4. The scores must not be linearly dependent (e.g., score2 can't be a simple function of score1).

Conditions #1, #2, #3 don't impact whether a MANOVA can be executed or not; hence, you can get results (regardless of whether they are meaningful!). If condition #4 holds, SPSS will omit at least one variable from the MANOVA.

Good luck with your work!

Thanks for fruitful information

Hello, linearity important to have more reliable answer.

linearity of sandwich assay curve for protein sample detection is called linear regression.

Example for calculation by non-linear curve is linked below and attached.

Hope this info will help you.

Good luck!

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

Following answers

Risk ratio is only for dichotomous variables. Continuous variables can be compared with t tests, anova or correlation/regression if normal in distribution, or non parametric tests, such as Wilcoxon, if non parametric