Statistical Analysis - Science topic

It will completely depend on what uses you want to give it. There is the tool SPSS is quite useful, as for the license, it is somewhat expensive but you can find a way to install it. If you want to generate graphics, there is PowerBI or Tableau, which are ideal for data visualization and dashboard generation (They have free licenses for students). Finally, if you want something more advanced, you can try Octave (Open Source version of Matlab) or directly immerse yourself in programming with Python or R.

To examine the performance disparity across classifiers, you could do statistical tests such as ANOVA (Analysis of Variance) or paired t-tests.

Pairwise t-tests can determine which distinct classifiers have significantly different performance.

So, to check why Random Forest performs better, I implemented this in Python using the breast_cancer dataset from sklearn, which you may use in your IDS scenario.

Also, I used the accuracy metric for determining the performance of each model.

#import all this libraries

from sklearn.datasets import load_breast_cancer

from sklearn.model_selection import cross_val_score

from sklearn.ensemble import RandomForestClassifier, BaggingClassifier, AdaBoostClassifier, StackingClassifier

from sklearn.tree import DecisionTreeClassifier

from scipy.stats import ttest_rel

#load the dataset

data = load_breast_cancer()

X = data.data

y = data.target

# Initialize the classifiers

rforest = RandomForestClassifier()

bagging = BaggingClassifier(estimator=DecisionTreeClassifier())

boosting = AdaBoostClassifier(estimator=DecisionTreeClassifier())

stacking = StackingClassifier(estimators=[('rforest', rforest), ('bagging', bagging), ('boosting', boosting)], final_estimator=DecisionTreeClassifier())

# Train and evaluate models using cross-validation

rforest_scores = cross_val_score(rforest, X, y, cv=5, scoring='accuracy')

bagging_scores = cross_val_score(bagging, X, y, cv=5, scoring='accuracy')

boosting_scores = cross_val_score(boosting, X, y, cv=5, scoring='accuracy')

stacking_scores = cross_val_score(stacking, X, y, cv=5, scoring='accuracy')

# Perform paired t-tests

t_stat, rforest_bagging_pvalue = ttest_rel(rforest_scores, bagging_scores)

t_stat, rforest_boosting_pvalue = ttest_rel(rforest_scores, boosting_scores)

t_stat, rforest_stacking_pvalue = ttest_rel(rforest_scores, stacking_scores)

# Print p-values

print("Paired t-test p-values (Random Forest vs. Bagging):", rforest_bagging_pvalue)

print("Paired t-test p-values (Random Forest vs. Boosting):", rforest_boosting_pvalue)

print("Paired t-test p-values (Random Forest vs. Stacking):", rforest_stacking_pvalue)

#check if the difference in accuracy between the ensemble methods is statistically significant

if rforest_bagging_pvalue < 0.05:

print('The difference in accuracy between Random Forest vs. Bagging is statistically significant\n')

else:

print('The difference in accuracy between Random Forest vs. Bagging is not statistically significant\n')

if rforest_boosting_pvalue < 0.05:

print('The difference in accuracy between Random Forest vs. Boosting is statistically significant\n')

else:

print('The difference in accuracy between Random Forest vs. Boosting is not statistically significant\n')

if rforest_stacking_pvalue < 0.05:

print('The difference in accuracy between Random Forest vs. Stacking is statistically significant\n')

else:

print('The difference in accuracy between Random Forest vs. Stacking is not statistically significant\n')

I hope this one helps.

Thank you for clarifying, Mariam Mabrouk. In that case, I see that you have two possible ways to account for the correlated nature of the data:

1) Using GENLIN with generalized estimating equations (GEE); or

2) Using GENLINMIXED with occasions clustered within patients.

In both cases, I would likely choose a logit model.

For option 1, there are some relevant examples here:

For option 2, this book has some relevant info:

Cheers,

Bruce

Dear professor , I would like to thank you for your answer to my researchgate question .I would like to let you know the title of our paper will be "Meta-analysis of application of particle swarm optimization in geophysical data inverse problem solving". I have collected all papers in this field with this format (about 100 papers ) in 8 categories (a,b,c,d,e,f,g,h):The main applications of PSO to the geophysical inverse problem include the interpretation of: a. vertical electrical sounding (VES) (Fernández-Álvarez et al. 2006; Fernández Martínez et al. 2010a; Pekşen et al. 2014; Cheng et al. 2015; Pace et al. 2019b); b. gravity data (Yuan et al. 2009; Pallero et al. 2015, 2017, 2021; Darisma et al. 2017; Jamasb et al. 2019; Essa and Munschy 2019; Anderson et al. 2020; Essa and Géraud 2020; Essa et al. 2021);c. magnetic data (Liu et al. 2018; Essa and Elhussein 2018, 2020); d. multi-transient electromagnetic data (Olalekan and Di 2017); e. time-domain EM data (Cheng et al. 2015, 2019; Santilano et al. 2018; Pace et al. 2019c; Li et al. 2019; Amato et al. 2021); f. MT data (Shaw and Srivastava 2007; Pace et al. 2017, 2019a, c; Godio and Santilano 2018; Santilano et al. 2018) and radio-MT data (Karcıoğlu and Gürer 2019); g. self-potential data (Santos 2010; Pekşen et al. 2011; Göktürkler and Balkaya 2012; Essa 2019, 2020) and induced polarization (Vinciguerra et al. 2019); h. Rayleigh wave dispersion curve (Song et al. 2012) and full waveform inversion (Aleardi 2019). please guide me how to arrange them for you to analyse result and calculate impact effect of papers and perform our meta analysis method? I am looking forward to hearing from you as soon as possible. Best regards Reza Toushmalani

Methodologically, efficient questionnaire Luna Leoni ! Success 1

If you want to see how comparable the results from the two ELISAs are, simply plot the results of the first ELIZA against those of the second ELISA.

Another option is to make a mean-difference plot (aka "Bland-Altman plot"): plot the differences between the ELISA results against the mean of the ELISA results.

Doing a statistical test and interpreting a non-significant result as "there is no difference" or "the groups/runs/ELISAs are comparable" is logically flawed and complete nonsense. Don't do this, ever!

Hi Giulia,

could you please add more details about your "various experiments" and data collections?

it also gives a clear picture of the problem across time and the effectiveness of the intervention but this approach does not give which intervention aspect is more or less effective.

Hi Lisa Neve

Did you by any chance find the solution to this problem?

Dear Jessica Jovel Please do well to recommend my answer if helpful.

I don't have specific information on guides or tutorials for Distance Based Linear Modeling (DistLM) in Statgraphics. However, I can provide you with some general guidance on where to find such resources.

1. **Statgraphics Documentation:**

Check the official documentation or user guides provided by Statgraphics. They often include detailed explanations, examples, and tutorials. Visit the official Statgraphics website or look for documentation within the software.

2. **Statgraphics Support Center:**

Explore the support center on the Statgraphics website. They may offer additional resources, tutorials, or forums where users can share tips and solutions.

3. **Online Forums and Communities:**

Participate in forums and communities related to statistics, data analysis, and Statgraphics. Websites like Stack Overflow, Reddit (e.g., r/statistics), or specialized forums might have discussions and resources shared by users.

4. **Educational Platforms:**

Check educational platforms such as YouTube, Coursera, or Udemy for video tutorials and courses related to DistLM and statistical modeling using Statgraphics.

5. **Books and Academic Resources:**

Look for books on statistical modeling or multivariate analysis that may cover DistLM. Academic publications and journals could also be valuable sources.

6. **Consult Statisticians or Experts:**

If you're part of an academic institution, consider reaching out to statisticians or experts in the field who may have experience with DistLM in Statgraphics.

Since software tools and their documentation can change, it's a good idea to check for the most recent resources and updates. If there have been any developments or new resources released since my last update, you should be able to find them through the channels mentioned above.

Almost all of the major qualitative analysis programs allow you export "spreadsheet" data that you could analyze with Excel or relatively basic statistics software.

Beware – there are three things wrong with this advice from Carl Alexander Frisk

Why ANOVA? Sounds more like a regression problem (time-series).

The 'statistics' package can compute and typeset statistics like frequency tables, cumulative distribution functions (increasing or decreasing, in frequency or absolute count domain), from the counts of individual values, or ranges, or even the raw value list with repetitions.

Can you clarify the roles of the variables you mentioned? If one of them is a dependent variable, for example, which one is it? Thanks for clarifying.

Please clarify too what "a trait with continuous variables" means. Perhaps if you just said what the trait is (and what the continuous variables are), it would help readers to understand better. Thanks.

To analyze the effect of pH on desorption considering three replicates for each of the five pH values, you can employ a mixed-effects model in R using the nlme package.

Dear Dmitriy Pospelov

Yes, I don't mind, and I am interested in everything related to statistics because it is my specialty.

Glad to inform me of the details

Thank You.

The statistical tools should be used to give the results expected from the analysis. I didn't think you should leverage the tools to produce more reliable results. If you select the appropriate statistical test, then that should give you an accurate answer. If you use the wrong test, then the answer is not valid.

I think this is the key:

"I want to compare the two sets of data to look for any differences in level of protein in the two different brain locations over the 4 time points."

From each group of 5 mice you measure the protein concentration in two different brain locations. This is where you get the "two sets of data". And, yes, if you wish to say so, these data are paired between set 1 and set 2, because these data are from the same mice.

But when your data is complete (no missing values) it would be the simplest solution to calculate the ratio of the concentrations in the two locations for each individual mouse. So you end up with a single "set" of data, 20 concentration ratios for the 20 mice in total.

These 20 ratios are organized in 4 groups (each with 5 values from one time-point). All these ratios are statistcally independent, if there is nothing in the experiment that could serve as a common source of variance*, and can be analysed as such. These ratios are not normal distributed, so if you like to get p-values make sure to use the logarithms (log-ratios).**

* In the worst case, all mice in one group were kept in one cage, so that cage and group are fully confounded. In such a case there is no independent value in any group and you would not be able to answer your research question, because any difference you see in any particular time-point could be due to something related with that particular cage rather than with that time-point.

** Another way is to start with log-concentrations and calculate the differences of the log-conc between the two locations in each mouse. Yet another option would be to use generalized linear model of the quasi-Poisson family, but this is rarely used.

Good morning Qakathekile Moyo, that is a very long question posted to the chat. I have not read it in full, as in fact this morning I am attending a funeral at church for someone. Interestingly this church group have shown almost no interest in research as I have been trying to ask them to incorporate science & evidence into their talks for >decade & I've been kind of laughed at, mocked, or ignored [but not spat on as yet]. They are totally ignorant of science & evidence, yet continue with their prayers for this, that & the other. There is a famous quote; science without religion is blind and religion without science is lame. I may have this the wrong way around, it may be in reverse, but an important point. There is no point praying or worshipping for others, when the person saying the prayers is sedentary & overweight. There is an important point; take care of yourself firstly before worrying about others, as otherwise you are hypocritical. As I said I've not read your question posed in full, as it is too long for me to read, but I glanced quickly at it, but I truly consider that churches need to pull up their socks in terms of incorporating science & evidence into their talks, as they are way behind the times. Sadly, this person who passed, was over religious but in some ways looked as though they may have neglected their own physical health which maybe a contributing factor to their passing. I don't want to go into details as it is not the place on a chat but demonstrates that balance is necessary. Fair enough to worship if you want to, reading science is equally important.

It may be possible to go for a shared first-authorship.

Selvaraju Sivamani I have 3 factors: If I use 1 central point I'm getting 0 repetitions of the same runs (total 13 runs). Can I use this method for my experiment?

Are you just comparing the count of the cases ? Or are you measuring something else and comparing these values for positive and negative cases ?

If you just want to judge the counts, is the question whether 3 | 802 different from a proportion of 0.5 | 0.5 ? That could be done with a binomial test, though the answer is pretty obvious.

Also with the counts, you could calculate e.g. Clopper-Pearson confidence intervals for the counts ( (3/805) and (802/805).

You need to save the output in an object. In the case of lm(), the best is to save the results in a list. This is achieved automatically if you'd use lapply() instead of for().

models <- lapply(i, function(x) lm(x~d))

The trouble with the answers of Seyyed Amir Yasin Ahmadi and s. Rama Gokula Krishnan is that they are blind to two pieces of information that are critically important :

1. We don't know the data generation process. There are so many processes, univariate and multivariate, that can result in data expressed as percentages that I can't even begin to enumerate them. But without knowing what the process is we cannot begin to decide whether a transformation is needed and, if so, what it is.

2. We do not know the research question. There is no point in transforming a variable if this loses your ability to calculate a meaningful measure of effect size, for example. On the other hand, some measures of effect size are better expressed in log units.

Can I elevate this to the status of a general rule of data anallysis?

Sung-Jun Lee I would disagree with several of your recommendations:

ad 1. I would agree, a sample size of 7 per group is very low and one should not expect to find anything or rather everything. It is known that effect sizes are overestimated in small samples on the one hand, on the other hand you massively lack power to detect anything (with well behaved data). Therefore, such small samples are not recommended, since some single data points can completely alter your results and are not very reliable (just simulate truely normal data with known and controllable population parameters and you will see how different the results will be, just by chance). For power analyses it is recommended to test against the Smallest Effect Size Of Interest (SESOI). You will find lots of articles about it (e.g. from Daniel Lakens).

ad 2. normality tests are not recommended to assess normality (of the residuals), since they a) are also subject to power and won't find any deviations in small N (where deviations are more porblematic than with large N where lots of analyses are quite robust) and a non significant result is NOT evidence that the H0 of no deviation is true, since it is a conditional probability (conditioned on that H0 is true already).

Further, non-parametric tests are not recommended as a substitute for paramertric tests, especially in small samples, since a) they test different hypotheses and b) its about the distributions, which are harder to assess in small samples. Non-parametric tests are not a magical trick that your analyses suddenly work on small samples.

ad 3. and 4. This is not very helpful, since you are most likely interested in the conditional/differential, i.e. the interaction effect. For example, if both groups show a significant change from t1 to t2, you do not know anything if the experimental group changed differently, as compared to the control group. In case of a random allocation it would be possible just to compare the groups at t2, otherwise a split plot 2 factorial ANOVA (Time*Group) or a multiple regression with t2 as outcome and t1 as covariable and Group as predictor, where the latter has been shown to have greater power (see Average Treatment Effect [ATE] in the literature. Solomon Kurz did a series on this topic for example https://solomonkurz.netlify.app/blog/2023-04-12-boost-your-power-with-baseline-covariates/).

ad 5. The ANOVA seems reasonbale, but Friedman does not necessarily test what you are thinking you are testing. See above.

ad 6. I would agree, but Bonferroni might be too conservative, depending on the amount of comparisons, especially with several correlated dependent variables AND a very (too) low sample size in the first place.

Instead of only relying on inferential statistics, I would plot the data (including all data points) the trends (slopes) etc to see whats going on. Inferential statistics are not necessarily the answers to your questions, but a guide to separate signal from noise.

Fundamentals of Research Methodology for Healthcare Professionals. This book presents the fundamental principles of research methodology and supports the vital role that research plays in the improvement of health science practices.

How do you know that your multiple indicators really measure the same construct?

Your binary variables make me fear that you dichotomised a measurement scale score. Not a great idea – imagine taking a black and white picture and replacing each pixel with either white or black, depending on whether it was above or below the median brightness. Your intuition is correct. you can lose over a third of the information content of your variable as a result.

So my advice would be check your assumptions first.

Shakeel Ahmad May I know how to confirm mutagenicity?

Just like any industry there should be national or state-based reports, as per hospital industry has hospital admissions data in the AIHW reports, there should be a like document for shipping related to statistics collected ie no of ships docked, containers unloaded, weight of shipping containers, products dumped due to contamination etc.

Now coming to your table p-value in the right column of the forest plot is the p-value for the overall test of the treatment effect across all subgroups. It is calculated by combining the results of the individual studies in the meta-analysis. In this case, the p-value is 0.56, which is not statistically significant.

The p-value for the test for effect in the subgroup is the p-value for the test of the null hypothesis that the treatment effect in the subgroup is equal to zero. It is calculated using only the data from the studies in the subgroup. In this case, the p-value for the test for effect in the subgroup is 0.094035, which is statistically significant.

The two p-values are different because of the heterogeneity between the studies in the meta-analysis. The heterogeneity statistic (0.5) is very high, which indicates that there is a lot of variability in the treatment effects across studies. This variability could be due to a number of factors, such as different study designs, different populations of patients, and different treatment regimens.

When there is heterogeneity in the treatment effects across studies, it is more difficult to detect a significant overall treatment effect. This is because the variability in the treatment effects across studies can mask the true effect of the treatment.

In this case, the p-value for the overall test of the treatment effect is not statistically significant, but the p-value for the test for effect in the subgroup is statistically significant. This suggests that the treatment may be effective in the subgroup, but it is not possible to draw a definitive conclusion without further research.

It is important to note that a statistically significant p-value for the test for effect in a subgroup does not necessarily mean that the treatment is clinically effective in that subgroup. It is possible that the difference in the treatment effect is small or that it is not clinically meaningful.

To determine whether the treatment is clinically effective in a subgroup, it is important to consider the magnitude of the difference in the treatment effect and the clinical implications of that difference

Abolfazl Ghoodjani, I am not a GraphPad user, so pardon my ignorance if I am just misunderstanding the documentation. After reading the page you pointed to, I am left thinking that the "multiple t-test" method returns exactly the same results one would get by using the (single) paired t-test command multiple times. If so, there is nothing resembling the test of interaction that Wendy Baker Smemoe appears to be thinking about.

PS- Are you still at the AECRP? I could not find you on its website (https://www.mcgill.ca/painresearch/).

Can you show the source that you should use MW-U for small samples? I dont think that this would be a good advice.

The decision, which statistical model is appropriate for your data cannot be decided without further information. Maybe both, MW-U as well as t-test are not suitable. Both have assumptions, both answer different questions. You have to be clear what you want and which data generating process underlies your data.

Timothy A Ebert

Thank you for your answer Timothy. I want to compare the treatment both to the control and to each other. The thing is, some of my treatments are not statistically significant when compared to the untreated control. I originally thought it would be visually better to have just a column in the graph with the dots placed at zero for the untreated control. But from your answer I gather it is standard practice to just drop the control column and mention in the figure legend that it was not statistically significant.

Since you are going to involve a third variable that will eventually impact your results, you need to take that third variable into account and check for normality and other assumptions before you carry out your final analysis. However, while analysing the IV and DV, if the data is not found to be normally distributed, then a mediator or moderator is less likely to help ensure normality. In such a scenario, you could simply opt for non-parametric tests.

Davood Omidian Thank you for a detailed answer. I Appreciate it a lot.

I am using the GPower tool, and was wondering which of the MANOVA types are more sufficient?

MANOVA: Repeated measures, between factors

MANOVA: Repeated measures, within factors

MANOVA: Repeated measures, within-between interaction

And for the "Number of Groups" i should pick 6, and the "Number of measurements" i should pick 2 (If i am measuring privacy and view-out)?

The number of measurements does not have anything to do with the 12 different scenarios happening within one group?

Hope I described it clear enough,

Best Regards,

Louis H

E.A. Gawad I'm sorry. Your "friend" who is actually a second AI text generator? You are trying to pass off computer-generated text as your own contribution. Do you really have the competence to spot the errors in the text you posted. Where, for example, does the reference to Nunally fit? And why the absolutely vague references to basic books on SPSS? Because you don't know what you are talking about.

I repeat: you are a fraud.

First of all, I would like to say that determine the proportion is not appropriate. It should have been to estimate the proportion/prevalence (P) of event/knowledhe... say X.

If the objective is to estimate the prevalence/proportion of X in the population and we have information not about the whole population, but a fraction of population i.e. sample drawn from the target population. Descriptive statitics of sample gives you some information about the sample, but you should use the method of estimation and standerd error of estimate. Also to see whether the estimated valuue of P can be generlized for the entire population.

Rainer Duesing, your position sounds a bit like Fisher's position on Bayesian analysis--see the attached image. ;-)

With this tiny dataset your statistical power is incredibly low. But there's another problem too : December of year 1 is actually just before January of year 2. The division of the data into years masks the continuous nature of the real variable – time – and reduces any study power by a considerable amount.

So applying the Mark I eyeball test to the data, there's nothing to see. And thinking about your time variable, dichotomising a continuous variable is never a good idea.

I have several questions.

With 3 DVs, you may be considering MANOVA. My question 3 above is a question that Huberty & Morris (1989) raised in their classic article on MANOVA:

If you are considering MANOVA, see also this article:

Another option to consider, particularly if your DVs are not really quantitative variables (for which means & SDs are defensible) would be to turn things on their head by estimating a logistic regression model with gender as the outcome and with your 3 DVs as the explanatory variables. Frank Harrell described this approach in his well-known textbook on regression modeling: See the excerpt in the attached PDF. But note that it requires having a sufficient number of observations in each gender category. By the 20:1 rule of thumb Harrell describes in the Datamethods Author Checklist (link below), one would need at least 60 observations in each gender category with 3 explanatory variables in the model.

HTH.

Hello Muziwandie,

A variable in a specific model only be characterized as a moderator if it interacts with some other IV(s) as regards resultant scores on some DV.

That is, while the IV(s) may well have some direct impact on the DV (referred to as "main effect/s" in anova-type models)--as might the moderator, there must be some interaction as well. That is, mean effect sizes of the IV(s) and moderator are themselves insufficient for accurate estimation of scores on the DV. For this reason, in the presence of moderation (or interaction), one generally pays more attention to the nature of that moderation/interaction and less attention to the main effect/s.

The influence of the moderator might or might not be that of "counteracting" the influence of the IV(s) in question. Rather, it could be that resultant mean differences on the DV are simply magnified or made smaller by virtue of specific IV/moderator combinations or values (as opposed to so-called disordinal interaction effects).

Good luck with your work.

Halah Kadhim Tayyeh I did not get your question.

However, you can find a lot of information in the link below.

Although the number of individuals is small to conduct a statistical test and draw strong conclusions, you can still use the Chi-square test with null and alternative hypotheses. For more details, you may search any statistics homepage and apply the test in R or Python.

Hope it helps.

Best

Yes, there are R packages available for modeling soil erosion and sediment yield. One such package is "RUSLE2R" (Revised Universal Soil Loss Equation 2 - R). RUSLE2R is an R implementation of the Revised Universal Soil Loss Equation 2, which is a widely used empirical model for estimating soil erosion. The package allows users to calculate soil erosion based on factors such as rainfall, soil erodibility, slope, land cover, and erosion control practices.

Here's how you can install and use the "RUSLE2R" package in R:

thanks a a lot for these information

Khalid Kadhim Hello! What I mean by statistical analysis is that with a certain scoring scale, HE staining can be scored quantitatively or semi-quantitatively to see if there is a difference in HE staining structure of organs between different groups. Do you have a recommendation for a suitable scale?

You're welcome! I'm glad I could help. If you have any more questions in the future, feel free to ask. Best of luck with your study, and may God bless you too!

Sir Can share email?

Hello Mohammed,

You can try Python and R.

Best wishes.

R

.

Hello Kanhaiya,

If factor "F" (3 levels) and factor "M" (4 levels) are both between-subjects factors, _and_ you have a control group, then you will likely be best served by treating the data set as having 13 treatments (F1M1, F1M2, F1M3...F3M4, Control) (as in a one-way anova design or the equivalent regression model) and test your specific hypotheses of interest via contrasts.

For example:

Ho: No difference between (al)l treatments & control:

C1: +1, +1, +1, +1, +1, +1, +1, +1, +1, +1, +1, +1, -12

Ho: No difference due to main effect of factor "F":

C2: +1, +1, +1, +1, -1, -1, -1, -1, 0, 0, 0, 0, 0

C3: +2, +2, +2, +2, -1, -1, -1, -1, -1, -1, -1, -1, 0

(Note you need two contrasts to capture all the information association with 3 levels of a factor)

...and so on.

Good luck with your work.

At a minimum, you should be able to convert the education levels into a linear variable by recoding it as "years of schooling."

There are many packages that can be used for cytometry data. The most common are flowCore, flowStats, openCyto, and flowClust. They provide a comprehensive set of tools for preprocessing, analyzing, visualizing, and interpreting flow cytometry data. Though I didn't use any of them but my suggestions are based on some literatures of cells studies.

I assume you want to measure the level of agreement between the two tests (and not something like test the null hypothesis that there is no association). Weighted kappa is often used for this, but there are other measures that might fit your specific context.

A covariate is an independent variable of your study. For example, if you want to assess the association of cholera outbreak against eating raw vegetables, eat out in restaurants and drinking untreated water. These are covariates. But a confounder variable is the one which has not been collected during your data collection but could affect the finding of your study either in the positive or negative direction. For example, in the above study if you found that those who ate raw vegetables were more likely to have cholera illness but if you did not collect information on their travel history your finding could have been confounded with their contact with cholera patients during their travel. Eating raw vegetables might not be really the risk but rather their contact history.

If everyone in your dataset is a smoker, I'm not sure what your independent variable would be.

The statistical analysis of the relationship between fast attack and backcourt attack and the results of matches in a volleyball tournament typically involves the following steps:

It's important to note that the specific analyses and steps may vary depending on the research questions, the nature of the data, and the specific statistical techniques used.

It sounds like you could do a straightforward t-Test to compare the means on the DV between the control and experimental groups.

The scoring manual for The Athlete Burnout Questionnaire (ABQ) developed by Raedeke and Smith (2001) may be available through the original authors or published sources. Here are some suggestions on where you can look for the scoring manual:

Remember that access to scoring manuals may be subject to copyright restrictions and the policies of the authors or publishers. It is best to explore multiple avenues and contact the relevant experts or institutions for accurate information regarding the availability of the scoring manual for the ABQ.

As an AI language model, my responses are based on the information available up until September 2021, and I don't have real-time access to external resources or the ability to browse the internet. Therefore, I may not have specific information on the availability of scoring manuals for recently developed questionnaires.

To find the scoring manual for The Sports Mental Toughness Questionnaire (SMTQ), you can try the following steps:

Remember that access to scoring manuals may vary depending on copyright restrictions and the policies of the authors or publishers. It is best to explore multiple avenues and contact the relevant experts in the field for the most accurate and reliable information regarding the availability of the scoring manual for the SMTQ.

and for a little more convenience:

https://posit.co/download/rstudio-desktop/

To compare CFU (Colony Forming Unit) counts of a bacterium growing in four different culture mediums across three trials, you can use a one-way analysis of variance (ANOVA) followed by post hoc tests for multiple comparisons. Here's an outline of the analysis:

The choice of post hoc test depends on factors such as the number of comparisons, the desired level of significance, and whether you have a specific control group for comparison. These tests will provide you with adjusted p-values to determine which group differences are statistically significant.

Remember to check the assumptions of the ANOVA, including normality of residuals and homogeneity of variances, using diagnostic plots and statistical tests such as the Shapiro-Wilk test and Levene's test. If the assumptions are violated, you may need to consider alternative nonparametric tests, such as the Kruskal-Wallis test, which is the nonparametric equivalent of the one-way ANOVA.

It is also important to ensure proper experimental design, including randomization of the trials and independent sampling of CFU counts for accurate statistical analysis.

Consulting with a statistician or data analyst experienced in experimental design and analysis can provide further guidance specific to your dataset and research question.

If you have two traits to compare , and have 30 and up observations, you can use t - test, otherwise if you have only two means you can use chi square. If you have more variables, you better go to experimental designs such as factorial arrangement or a split plot, and so on. Good luck.

There are many software options available for statistical analysis, and several of them offer free versions or online access. Here are a few options:

These are just a few of the many software options available for statistical analysis. Each software has its strengths and weaknesses, so it is important to choose the one that best fits your needs and preferences.

he appropriate statistical analysis for comparing plant species traded at two or more markets would depend on the specific research question, the characteristics of the data, and the hypotheses being tested. Here are a few common statistical methods that could be used:

It is important to consult with a statistician or data analyst to determine the most appropriate statistical method for your specific research question and data characteristics.

The regression approach aptly described by Samuel Oluwaseun Adeyemo is sometimes called Analysis of Covariance, or ANCOVA. Using that as an internet search term might be helpful to you.

When assessing agreement of measurement in count data, the appropriate statistical analysis to use depends on the specific research question and the nature of the data. Here are some common statistical methods used for assessing agreement of measurement in count data:

In summary, when assessing agreement of measurement in count data, there are several statistical methods available, such as Cohen's kappa, ICC, Bland-Altman plot, Poisson regression, and chi-square test. The appropriate method to use depends on the specific research question and the nature of the data. You should consult with a statistician or data analyst to select the appropriate method for your study.

Dear Csenger Kovácsházi , great questions - and thanks for providing more insights.

First, answering to your specific questions:

- Applying the rules of logarithm, log(A) - log (B) = log(A/B). That is, the difference between two log2 values represent in fact the log2 fold-change.

Using A = 8 and B = 16, we have log2(A) = 3 and log2(B) = 4, FC = A/B = 8/16 = 0.5, and, finally, log2FC = log2(A/B) = log2(A) - log2(B) = 3 - 4 = -1 (go over this math using different numbers to verify this relationship)

- I know this wasn't a question, but, in fact, that is not generally true for this type of data. Furthermore, the biggest issue is the heterogeneity of the residuals within 'treatments' (the 3 groups you have), which is far more important for ANOVA than normality. In addition, proteomic data may include lots of zeros (BTW, how did you calculate the log2 of '0'?), which hinders this even more.

- As long as you are calculating it correctly (per Q1.1), I don't see any issues with its interpretation. In fact, I really like log2FC because of its ease of interpretation.

Q2: How to create a volcano plot for this dataset? ... So what do you recommend?

Sorry, but it is what it is... You simply can't do everything in a single plot... You will have to pick your battles... If you are interested in presenting a specific contrast, your p-value should be representing that specific contrast. For ease of visualization, you may color-code your volcano plot discriminating the significant and non-significant results based on the overall p-value of the ANOVA. Thus, while you are still plotting the -log10P of the contrast, the result is color coded based on the p-value of the ANOVA.

Now, a couple of comments...

1) The process of generating proteomics data is quite messy, and it is recommended to have a data-normalization step (not related to 'normal distribution'). That is, some samples might have greater abundance and diversity of PTNs by change during the quantification process. Others may have in fact because of the 'treatment'. Thus, it is important to differentiate these two samples and not treat them the same way. See this for reference: https://pubmed.ncbi.nlm.nih.gov/27694351/

2) The transformation of data is not always needed/required/justifiable when available methods are out there. For instance, Poisson and Negative Binomial methods are popular for proteomics data, as well as the use of Zero-Inflated models within these methods. I know that it is not easy to just learn these methods and apply them to your work... But, just a tip..

Let me know if you have more questions. Thanks, Nick

This is from Eviews

Hi,

The contingency theory of management proposes that the most effective style of management depends on the specific situation at hand. To prove this theory, researchers use various methods including: like case studies ,experiment , survey , interview , observations.

Regards,

Uday Bhale

The significant t-test for the interaction term in your model shows that the slopes of the two lines differ significantly. But at the 3 values of X that are shown in your results (x=-.856, x=0, x=.856), fitted values on the two lines do not differ significantly.

I suspect your output is from Hayes' PROCESS macro, and that -.856 and .856 correspond to the mean ± one SD. Is that right?

Why does it matter if the fitted values on the two lines do not differ significantly at those particular values of X? Your main question is whether the slopes differ significantly, is it not?

The minimum number of detections required for applying statistical models to estimate the density from camera traps for unmarked individuals depends on several factors, including the study area, the species being monitored, and the specific statistical method being used.

In general, a higher number of detections leads to more precise density estimates, allowing for better estimation of the detection probability and reducing the effects of random variation in the data. However, there is no specific minimum number of detections that is universally applicable to all situations.

As a general rule of thumb, some studies suggest that a minimum of 10-15 independent detections is required for reliable density estimation using camera trap data. However, this can vary widely depending on the specific situation and statistical method being used.

It is important to note that the number of detections required for reliable density estimation can also depend on other factors, such as the size of the study area and the spatial distribution of the animals being monitored. In some cases, collecting data over multiple seasons or years may be necessary to obtain a sufficient number of detections for reliable density estimation.

Statistical analysis of high-frequency time series data for two temperature loggers?

To perform a statistical analysis of high-frequency time series data for two temperature loggers, you can follow these steps:

Overall, statistical analysis of high-frequency time-series data for two temperature loggers involves collecting and processing data, visualizing the data, performing statistical analysis, interpreting the results, and reporting the findings.