R standard deviation how to#

R standard deviation how to#

To summarize: In this article, I have explained how to avoid the “warning message in cor(X) : the standard deviation is zero” in the R programming language.

ggplot2 Warning Message – Use of data$X is discouraged.

Warning Message: Number of items to Replace is not Multiple of Length.

Warning Message in R: argument is not numeric or logical: returning NA.

Warning Message in read.table: Incomplete Final Line Found by readTableHeader.

In addition, you might have a look at the related articles on this homepage: I illustrate the R programming codes of this tutorial in the video: Have a look at the following video on the Statistics Globe YouTube channel. Maybe you have performed a previous data manipulation step wrong, or maybe your data is just not suited for the calculation of a correlation coefficient. In case this warning message occurs, you should investigate why one of your variables contains only one specific value. Note that updating the data as shown in this example usually doesn’t make sense from a theoretical viewpoint.

If you ignore the 95% confidence clause and reframe the question to make a guess to compensate for how ambiguous it was, then the answer is: The standard deviation is 3.6.Cor (x2, y2 ) # Apply cor function # 0.6546537Īs you can see, a valid output is returned and no warning messages are shown.

It is either incomplete,mis-transcribed, or whoever wrote it seems to be a little confused.

The question you posted doesn't make any sense.

> qnorm(.05,mean=32,sd=3.64774)Ģ6 or less is what the 5 % of customers that saved the least got. 95 cumulative probability, mean is 32 and we assert that the StDev is 3.64774. Let's verify that we got this right: > qnorm(.95,mean=32,sd=3.64774) # Verify in R that the stdev we calculated is correct. The mean + 1.644854 standard deviations is 38 (95% of customers save no more than this)ģ8 - 32 = 6 (this is equal to 1.644854 StDev): Algebraically that's written: Then the question should be would be worded like this: 'We know that in 95% of all cases saved no more than $38 and 5% of customers saved $26 or less.' In this case we could drop the standard error term altogether and we would then only be worried about the standard deviation and mean of the population: That whoever wrote the question does not understand the proper usage of the phrase 95% confident or was exhausted when they wrote it.

We are talking about individual probabilities that a given observation falls at or below a given value, not that we are 95% confident that the average observation does, and 3. the 95% confident clause is junk information 2.

If you re-frame the question and assume that 1. It looks like the 95% confidence clause (and absence of a mention of a sample mean) is meant to throw you off, but in reality, it just makes the person asking you this question appear to be confused as to what question they are asking you. We are kind of out of luck if this is where they were going with this. The problem you posted simply didn't tell us enough information to solve this with the assumption this was a confidence interval from a sample. There's a problem here that this rhetorical exercise was meant to point out: We still have 2 variables. If I didn't screw up my mental algebra at midnight without paper in hand. We could attempt to solve algebraically for StDev by putting the upper bound formula on the left side of the = sign and put 38 which is the upper bound on the right side:ģ2 + (1.644854 * StDev / sqrt(sample size)) = $38. Upper bound: 32 + 1.644854 * StDev / sqrt(sample size) # we will use this below Lower bound: 32 - 1.644854 * StDev / sqrt(sample size)

Let's say we try to solve for the st dev. (if this were a confidence interval problem). $26 we now know is 1.644854 standard errors below the sample mean if his 95% confidence implies that this is a sample mean and $38 we now know is 1.644854 standard errors above the estimated mean from their sample mean. Now we knew that the money saved is between: $26 and $38. You said you want to do this in R, so use qnorm() > qnorm(.95) First let's calculate the z score on the normal distribution table which corresponds with a. Let's try to rationalize out how we can or can't solve this, and we will discover some problems with the wording of this problem as we go: It appears that whoever wrote this problem is confused and doesn't know if they are asking a sample mean confidence interval problem '95% confident' or a simple population normal distribution problem.