# Calculate Mean and Standard Deviation, when given the confidence interval and sample size

How can I calculate the sample mean and sample standard deviation, given that I know the sample size and the confidence interval for the mean?

For example, with the info provided below:

• Sample size = 300
• 95% confidence interval of 5.18 < μ < 5.38

Obviously you will need to know the type of confidence interval you are dealing with, but let's suppose that this is a standard one-sample confidence interval for the mean, using the standard T-statistic as the pivotal quantity. In that case, the formula for the interval is:

$$\text{CI}(1-\alpha) = \Bigg[ \bar{x} \pm \frac{t_{n-1, \alpha/2}}{\sqrt{n}} \cdot s \Bigg].$$

Thus, if we denote the known lower and upper bounds of the interval as $$l$$ and $$u$$ respectively, then you can algebraically reverse-engineer the sample mean and sample standard deviation as:

$$\bar{x} = \frac{l+u}{2} \quad \quad \quad \quad \quad s = \frac{u-l}{2} \cdot \frac{\sqrt{n}}{t_{n-1, \alpha/2}}.$$

With the values specified in your example, you get:

#Set preliminary values
l     <- 5.18;
u     <- 5.38;
n     <- 300;
alpha <- 0.05;

#Compute sample mean and SD
crit <- qt(alpha/2, df = n-1, lower.tail = FALSE);
MEAN <- (l+u)/2;
SD   <- (u-l)*sqrt(n)/(2*crit);

#Print the values
MEAN;
[1] 5.28
SD;
[1] 0.8801386


Thus, assuming that your interval was a standard one-sample confidence interval, you must have had a sample mean $$\bar{x} = 5.28$$ and sample standard deviation $$s = 0.88$$.

• That is fantastic thank you! That pretty much all makes sense – bt75alfa Oct 6 '19 at 14:21