I've got a data frame with 36 variables and 74 observations. I'd like to make a two sample paired ttest of 35 variables by 1 grouping variable (with two levels).
For example: the data frame contains "age" "body weight" and "group" variables.
Now I suppose I can do the ttest for each variable with this code:
t.test(age~group)
But, is there a way to test all the 35 variables with one code, and not one by one?
Sven has provided you with a great way of implementing what you wanted to have implemented. I, however, want to warn you about the statistical aspect of what you are doing.
Recall that if you are using the standard confidence level of 0.05, this means that for each t-test performed, you have a 5% chance of committing Type 1 error (incorrectly rejecting the null hypothesis.) By the laws of probability, running 35 individual t-tests compounds your probability of committing type 1 error by a factor of 35, or more exactly:
Pr(Type 1 Error) = 1 - (0.95)^35 = 0.834
Meaning that you have about an 83.4% chance of falsely rejecting a null hypothesis. Basically what this means is that, by running so many T-tests, there is a very high probability that at least one of your T-tests is going to provide an incorrect result.
Just FYI.
An example data frame:
dat <- data.frame(age = rnorm(10, 30), body = rnorm(10, 30),
weight = rnorm(10, 30), group = gl(2,5))
You can use lapply:
lapply(dat[1:3], function(x)
t.test(x ~ dat$group, paired = TRUE, na.action = na.pass))
In the command above, 1:3 represents the numbers of the columns including the variables. The argument paired = TRUE is necessary to perform a paired t-test.
Related
I need to run a 2-sample independent t-test, comparing Column1 to Column2. But Column1 is in DataframeA, and Column2 is in DataframeB. How should I do this?
Just in case relevant (feel free to ignore): I am a true beginner. My experience with R so far has been limited to running 2-sample matched t-tests within the same data frame by doing the following:
t.test(response ~ Column1,
data = (Dataframe1 %>%
gather(key = "Column1", value = "response", "Column1", "Column2")),
paired = TRUE)
TL;DR
t_test_result = t.test(DataframeA$Column1, DataframeB$Column2, paired=TRUE)
Explanation
If the data is paired, I assume that both dataframes will have the same number of observations (same number of rows). You can check this with nrow(DataframeA) == nrow(DataframeB) .
You can think of each column of a dataframe as a vector (an ordered list of values). The way that you have used t.test is by using a formula (y~x), and you were essentially saying: Given the dataframe specified in data, perform a t test to assess the significance in the difference in means of the variable response between the paired groups in Column1.
Another way of thinking about this is by grabbing the data in data and separating it into two vectors: the vector with observations for the first group of Column1, and the one for the second group. Then, for each vector, you compute the mean and stdev and apply the appropriate formula that will give you the t statistic and hence the p value.
Thus, you can just extract those 2 vectors separately and provide them as arguments to the t.test() function. I hope it was beginner-friendly enough ^^ otherwise let me know
EDIT: a few additions
(I was going to reply in the comments but realized I did not have space hehe)
Regarding the what #Ashish did in order to turn it into a Welch's test, I'd say it was to set var.equal = FALSE. The paired parameter controls whether the t-test is run on paired samples or not, and since your data frames have unequal number of rows, I'm suspecting the observations are not matched.
As for the Cohen's d effect size, you can check this stats exchange question, from which I copy the code:
For context, m1 and m2 are the group's means (which you can get with n1 = mean(DataframeA$Column1)), s1 and s2 are the standard deviations (s2 = sd(DataframeB$Column2)) and n1 and n2 the sample sizes (n2 = length(DataframeB$Column2))
lx <- n1- 1 # Number of observations in group 1
ly <- n2- 1 # # Number of observations in group 1
md <- abs(m1-m2) ## mean difference (numerator)
csd <- lx * s1^2 + ly * s2^2
csd <- csd/(lx + ly)
csd <- sqrt(csd) ## common sd computation
cd <- md/csd ## cohen's d
This should work for you
res = t.test(DataFrameA$Column1, DataFrameB$Column2, alternative = "two.sided", var.equal = FALSE)
I have the following data
Species <- c(rep('A', 47), rep('B', 23))
Value<- c(3.8711, 3.6961, 3.9984, 3.8641, 4.0863, 4.0531, 3.9164, 3.8420, 3.7023, 3.9764, 4.0504, 4.2305,
4.1365, 4.1230, 3.9840, 3.9297, 3.9945, 4.0057, 4.2313, 3.7135, 4.3070, 3.6123, 4.0383, 3.9151,
4.0561, 4.0430, 3.9178, 4.0980, 3.8557, 4.0766, 4.3301, 3.9102, 4.2516, 4.3453, 4.3008, 4.0020,
3.9336, 3.5693, 4.0475, 3.8697, 4.1418, 4.0914, 4.2086, 4.1344, 4.2734, 3.6387, 2.4088, 3.8016,
3.7439, 3.8328, 4.0293, 3.9398, 3.9104, 3.9008, 3.7805, 3.8668, 3.9254, 3.7980, 3.7766, 3.7275,
3.8680, 3.6597, 3.7348, 3.7357, 3.9617, 3.8238, 3.8211, 3.4176, 3.7910, 4.0617)
D<-data.frame(Species,Value)
I have the two species A and B and want to find out which is the best cutoffpoint for value to determine the species.
I found the following question:
R: Determine the threshold that maximally separates two groups based on a continuous variable?
and followed the accepted answer to find the best value with the dose.p function from the MASS package. I have several similar values and it worked for them, but not for the one given above (which is also the reason why i needed to include all 70 observations here).
D$Species_b<-ifelse(D$Species=="A",0,1)
my.glm<-glm(Species_b~Value, data = D, family = binomial)
dose.p(my.glm,p=0.5)
gives me 3.633957 as threshold:
Dose SE
p = 0.5: 3.633957 0.1755291
this results in 45 correct assignments. however, if I look at the data, it is obvious that this is not the best value. By trial and error I found that 3.8 gives me 50 correct assignments, which is obviously better.
Why does the function work for other values, but not for this one? Am I missing an obvious mistake? Or is there maybe a different/ better approach to solving my problem? I have several values I need to do this for, so I really do not want to just randomly test values until I find the best one.
Any help would be greatly appreciated.
I would typically use a receiver operating characteristic curve (ROC) for this type of analysis. This allows a visual and numerical assessment of how the sensitivity and specificity of your cutoff changes as you adjust your threshold. This allows you to select the optimum threshold based on when the overall accuracy is optimum. For example, using pROC:
library(pROC)
species_roc <- roc(D$Species, D$Value)
We can get a measure of how good a discriminator Value is for predicting Species by examining the area under the curve:
auc(species_roc)
#> Area under the curve: 0.778
plot(species_roc)
and we can find out the optimum cut-off threshold like this:
coords(species_roc, x = "best")
#> threshold specificity sensitivity
#> 1 3.96905 0.6170213 0.9130435
We see that this threshold correctly identifies 50 cases:
table(Actual = D$Species, Predicted = c("A", "B")[1 + (D$Value < 3.96905)])
#> Predicted
#> Actual A B
#> A 29 18
#> B 2 21
I'm new to R and am trying to run a t-test for two means. I keep getting the error is.atomic is not TRUE. I know I need to make my data atomic, but I haven't found a way online.
I've ran code to check that the data is recursive and did a as.data.frame(mydata).
titanic_summary <- data.frame(Outcome = c("Survived", "Died"),
Mean_Age = c(28.34369, 30.62618),
N = c(342, 549),
Total_Missing = c(52, 125))
titanic_summary
Run a stats test (two sample T-test)
str(titanic_summary)
as.data.frame(titanic_summary)
is.atomic(titanic_summary)
is.recursive(titanic_summary)
titanic_test <- titanic_summary %>%
t.test(Outcome~Mean_Age)
Error in var(x) : is.atomic(x) is not TRUE
t.test does not work the way you seem to think. To avoid that particular error, you could instead use something like titanic_test <- t.test(Mean_Age ~ Outcome, data = titanic_summary) but that would just give you different errors, which comes down to the real question:
You presumably want to see whether there may be a relationship between age and survival, i.e. whether the difference in average ages of 2.28249 is significant but you will need the individual ages or some other additional information about dispersion for this
If you do use the detailed dataset then I suspect that what you really want is something like this:
library(titanic)
titanic_test <- t.test(Age ~ Survived, data = titanic_train)
which would give (for the Kaggle selected training set used in the titanic package)
> titanic_test
Welch Two Sample t-test
data: Age by Survived
t = 2.046, df = 598.84, p-value = 0.04119
alternative hypothesis: true difference in means is not equal to 0
95 percent confidence interval:
0.09158472 4.47339446
sample estimates:
mean in group 0 mean in group 1
30.62618 28.34369
I have a question about the sign of t in a paired-sample t-test using different data structures, but the same data. I know that the sign doesn't make a difference in terms of significance, but, it does generally tell the user if there have been decreases over time or increases over time. So, I need to make sure that the code I provide produces the same results OR, is explained correctly.
I have to explain the results (and code) as an example we're giving users of our software, which uses R (Rdotnet within a C# program) for statistics. I'm unclear as to the proper order of variables in both methods in R.
Method 1 uses two matrices
## Sets seed for repetitive number generation
set.seed(2820)
## Creates the matrices
preTest <- c(rnorm(100, mean = 145, sd = 9))
postTest <- c(rnorm(100, mean = 138, sd = 8))
## Runs paired-sample T-Test just on two original matrices
t.test(preTest,postTest, paired = TRUE)
The results show significance and with the positive t, tells me that there has been a reduction in the mean difference from preTest to PostTest.
Paired t-test
data: preTest and postTest
t = 7.1776, df = 99, p-value = 1.322e-10
alternative hypothesis: true difference in means is not equal to 0
95 percent confidence interval:
6.340533 11.185513
sample estimates:
mean of the differences
8.763023
However, most people are going to get their data not from two matrices, but, from a file with values for BEFORE and AFTER. I will have these data in a csv and import them during a demo. So, to mimic this, I need to create data frame in the structure that users of our software are used to seeing. 'pstt' should look like the dataframe I have after I import a csv.
Method 2: uses a flat-file structure
## Converts the matrices into a dataframe that looks like the way these
data are normally stored in a csv or Excel
ID <- c(1:100)
pstt <- data.frame(ID,preTest,postTest)
## Puts the data in a form that can be used by R (grouping var | data var)
pstt2 <- data.frame(
group = rep(c("preTest","postTest"),each = 100),
weight = c(preTest, postTest)
)
## Runs paired-sample T-Test on the newly structured data frame
t.test(weight ~ group, data = pstt2, paired = TRUE)
The results for this t-test has the t negative, which may indicate to the user that the variable under study has increased over time.
Paired t-test
data: weight by group
t = -7.1776, df = 99, p-value = 1.322e-10
alternative hypothesis: true difference in means is not equal to 0
95 percent confidence interval:
-11.185513 -6.340533
sample estimates:
mean of the differences
-8.763023
Is there a way to define explicitly which group is the BEFORE and which is the AFTER? Or, do you have to have the AFTER group first in Method 2.
Thanks for any help/explanation.
Here is the full R program that I used:
## sets working dir
# setwd("C:\\Temp\\")
## runs file from command line
# source("paired_ttest.r",echo=TRUE)
## Sets seed for repetitive number generation
set.seed(2820)
## Creates the matrices
preTest <- c(rnorm(100, mean = 145, sd = 9))
postTest <- c(rnorm(100, mean = 138, sd = 8))
ID <- c(1:100)
## Converts the matrices into a dataframe that looks like the way these
data are normally stored
pstt <- data.frame(ID,preTest,postTest)
## Puts the data in a form that can be used by R (grouping var | data var)
pstt2 <- data.frame(
group = rep(c("preTest","postTest"),each = 100),
weight = c(preTest, postTest)
)
print(pstt2)
## Runs paired-sample T-Test just on two original matrices
# t.test(preTest,postTest, paired = TRUE)
## Runs paired-sample T-Test on the newly structured data frame
t.test(weight ~ group, data = pstt2, paired = TRUE)
Since group is a factor, the t.test will use the first level of that factor as the reference level. By default factor levels are sorted alphabetically to "AFTER" would come before "BEFORE" and "postTest" would be come before "preTest". You can explicitly set reference level of a factor with relevel().
t.test(weight ~ relevel(group, "preTest"), data = pstt2, paired = TRUE)
Ok straight to the question. I have a database with lots and lots of categorical variable.
Sample database with a few variables as below
gender <- as.factor(sample( letters[6:7], 100, replace=TRUE, prob=c(0.2, 0.8) ))
smoking <- as.factor(sample(c(0,1),size=100,replace=T,prob=c(0.6,0.4)))
alcohol <- as.factor(sample(c(0,1),size=100,replace=T,prob=c(0.3,0.7)))
htn <- as.factor(sample(c(0,1),size=100,replace=T,prob=c(0.2,0.8)))
tertile <- as.factor(sample(c(1,2,3),size=100,replace=T,prob=c(0.3,0.3,0.4)))
df <- as.data.frame(cbind(gender,smoking,alcohol,htn,tertile))
I want to test the hypothesis, using a chi square test, that there is a difference in the portion of smokers, alcohol use, hypertension (htn) etc by tertile (3 factors). I then want to extract the p values for each variable.
Now i know i can test each individual variable using a 2 by 3 cross tabulation but is there a more efficient code to derive the test statistic and p-value across all variables in one go and extract the p value across each variable
Thanks in advance
Anoop
If you want to do all the comparisons in one statement, you can do
mapply(function(x, y) chisq.test(x, y)$p.value, df[, -5], MoreArgs=list(df[,5]))
# gender smoking alcohol htn
# 0.4967724 0.8251178 0.5008898 0.3775083
Of course doing tests this way is somewhat statistically inefficient since you are doing multiple tests here so some correction is required to maintain an appropriate type 1 error rate.
You can run the following code chunk if you want to get the test result in details:
lapply(df[,-5], function(x) chisq.test(table(x,df$tertile), simulate.p.value = TRUE))
You can get just p-values:
lapply(df[,-5], function(x) chisq.test(table(x,df$tertile), simulate.p.value = TRUE)$p.value)
This is to get the p-values in the data frame:
data.frame(lapply(df[,-5], function(x) chisq.test(table(x,df$tertile), simulate.p.value = TRUE)$p.value))
Thanks to RPub for inspiring.
http://www.rpubs.com/kaz_yos/1204