Continuity correction when using normal as approximation for binomial

[songoku]

TL;DR Summary

I have learnt how to do continuity correction when the value of random variable is integer, such as P(X < 5) changes to P(X < 4.5) when the distribution changes from binomial to normal

What if the value of X is not integer, such as P(X < 1.2)?

a) Will the continuity correction be P(X < 1.2 - 0.5) = P(X < 0.7)?

or

b) Will the continuity correction be P(X < 1.2 - 0.05) = P(X < 1.15)?

or

c) Something else?

Thanks

 

[pbuk]

Do you understand why we have to make the continuity correction for an integer (i.e. discontinuous) variable?

 

[songoku]

Do you understand why we have to make the continuity correction for an integer (i.e. discontinuous) variable?

Because binomial distribution is discrete distribution so to change it to normal distribution (continuous distribution) there should be adjustment. I am thinking like changing a line of x = 4 (discrete) to a box where the left vertex of the box is 3.5 and right vertex is 4.5 so that the box can touch the other box made from line x = 3 and x = 5 (becoming continuous distribution)

 

[pbuk]

So how does that apply when you have a line at 1.2?

 

[songoku]

So how does that apply when you have a line at 1.2?

It means I have to know the location of other lines. I am trying to form a hypothetical question regarding this but I just can't think of one.

So if the location of other lines are 1.1 and 1.3, P(X < 1.2) will be P(X < 1.15) and P(X > 1.2) will be P(X > 1.25)

But if the location of other lines is not in regular intervals, such as one is at 1.1 and the other is at 1.4, then P(X < 1.2) will be P(X < 1.15) and P(X > 1.2) will be P(X > 1.3)?

Thanks

 

[pbuk]

If your data is discrete so that it is only possible for a 'line' (whatever that is) to take values of 1.1, 1.2, 1.3, 1.4 etc. then a continuity correction may make sense, but if it the data are by nature continuous but there just happen to be some gaps then it would not make sense.

 

[songoku]

If your data is discrete so that it is only possible for a 'line' (whatever that is) to take values of 1.1, 1.2, 1.3, 1.4 etc. then a continuity correction may make sense, but if it the data are by nature continuous but there just happen to be some gaps then it would not make sense.

I understand. I haven't encountered such questions. All the practice questions are about integers so my query is only due to curiosity.

Is it not possible to have discrete data with irregular intervals, such as 1.1 , 1.3 , 1.4 , 1.9?

Thanks

 

[pbuk]

Is it not possible to have discrete data with irregular intervals, such as 1.1 , 1.3 , 1.4 , 1.9?

Of course it is, but the continuity correction is not about what values the data actually take, rather it is about what values the data can possibly take.

 

[songoku]

Of course it is, but the continuity correction is not about what values the data actually take, rather it is about what values the data can possibly take.

So is it correct to say that if the data is 1.1 , 1.3 , 1.4 and 1.9 the continuity correction for P(X < 1.3) is P(X < 1.2) and for P(X > 1.3) is P(X > 1.35)?

What if for P(X < 1.5)? Since the midpoint of 1.4 and 1.9 is 1.65, would the continuity correction for P( X< 1.5) be P(X > 1.65)?

Thanks

 

[pbuk]

So is it correct to say that if the data is 1.1 , 1.3 , 1.4 and 1.9 the continuity correction for P(X < 1.3) is P(X < 1.2) and for P(X > 1.3) is P(X > 1.35)?

This would only be the case if the data was binned with variable width bins including (1.1, 1.3) and (1.3, 1.4).

What if for P(X < 1.5)? Since the midpoint of 1.4 and 1.9 is 1.65, would the continuity correction for P( X< 1.5) be P(X > 1.65)?

This would only be the case if the data was binned with variable width bins including (1.4, 1.9).

But unless there was a good reason for binning the data you would get a better result by using individual samples.

It is good to be curious, but I think you have got lost down a rabbit hole; it's time to move on.

 

[songoku]

Thank you very much pbuk