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How to compute a confidence interval for a population mean using z-scores (in R)

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If we have a set of data that seems normally distributed, how can we compute a confidence interval for the mean? Assume we have some confidence level already chosen, such as $\alpha=0.05$.

We will use the normal distribution, which assumes either that we know the population standard deviation, or we have a large enough sample size (typically at least 30). If neither of these is true in your case, then you can use $t$-scores instead; see how to compute a confidence interval for a population mean.

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When applying this technique, you would have a series of data values for which you needed to compute a confidence interval for the mean. But in order to provide code that runs independently, we create some fake data below. When using this code, replace our fake data with your real data.

We include the population standard deviation below, assuming it is known. See the notes at the end for what to do if you do not know the population standard deviation in your situation.

alpha <- 0.05       # replace with your chosen alpha (here, a 95% confidence level)
pop.std <- 250      # replace with your population standard devation, if known
data <- c( 435,542,435,4,54,43,5,43,543,5,432,43,36,7,876,65,5 ) # fake

# Compute the sample mean, as an estimate for the population mean.
sample.mean <- mean( data )

# The margin of error then has the following formula.
z.score <- qnorm( alpha / 2, lower.tail=FALSE )
moe <- pop.std * z.score / sqrt( length( data ) )

# The confidence interval is centered on the mean with moe as its radius:
c( sample.mean - moe, sample.mean + moe )
[1]  91.3362 329.0167


  1. If you do not have the population standard deviation, but your sample is large enough (typically at least 30), you can approximate the population standard deviation with the sample standard deviation, using the code pop.std <- sd( data ). If your sample is not that large, then consider using a different technique instead; see how to compute a confidence interval for a population mean.
  2. The solution above assumes that the population is normally distributed, which is a common assumption in introductory statistics courses, but we have not verified that assumption here.

Content last modified on 24 July 2023.

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Contributed by Nathan Carter (