Confidence Intervals

A confidence interval (CI) gives an “interval estimate” of an unknown population parameter such as the mean. It gives us the probability that the parameter lies within the stated interval (range). The precision or accuracy of the estimate depends on the width of the interval.

For us to define a 100(1 – α)% confidence interval for θ, we must specify two random variables θ1(X) and θ2(X) such that P(θ1(X) < θ < θ2(X)) = 1 – α.

The most common value for α is 0.05, which leads us to a 95% confidence interval. As such, P(θ1(X) < θ < θ2(X)) = 0.95 specifies θ1(X) and θ2(X) such that there is a 95% chance of finding the true value of θ in the interval. Alternatively, we could say that 5% of the realizations of such intervals would not contain the true value of θ.

The Importance of a Confidence Interval

It is not always possible to know the exact values of the population mean or the population standard deviation since most populations are too large making data collection from every subject in the population not only economically unviable but very time-consuming. This makes confidence intervals very important.

They can also be used to predict the value of a given parameter. We usually assume that the underlying random variable has a normal distribution. In this regard, the central limit theorem (the assertion that most distributions tend to adopt a normal distribution when n is large) is a very important tool.

The following statements are true for any random variable that assumes a normal distribution:

  1. 50% CI: Approximately 50% of all observations fall in the range μ ± (2/3)σ
  2. 68% CI: Approximately 68% of all the observations fall in the range μ ± σ
  3. 95% CI: Approximately 95% of all observations fall in the interval μ ± 2σ
  4. 99% CI: Approximately 99% of all the observations fall in the interval μ ± 3σ

Note: The words “interval” and “range” have been used interchangeably in this context.

Example: Confidence interval

A financial analyst encounters a client whose portfolio return has a mean yearly return of 24% and a standard deviation of 5%. Assuming a normal distribution, compute a 50% confidence interval for the expected return.


$$ \begin{align*}
\text{The CI} & = \left\{ 0.24 – \cfrac {2}{3} * 0.05, 0.24 + \cfrac {2}{3} * 0.05 \right\} \\
& = \left\{ 0.207, 0.273 \right\} \\
\end{align*} $$


Repeat the question above assuming the client is interested in finding a 99% confidence interval for the expected return.

A. {0.08, 0.49}

B. {0.09, 0.39}

C. {0.09, 0.9}


The correct answer is B.

For a 99% CI, 99% of all the observations fall in the interval μ ± 3σ.


$$ \begin{align*} \text{the CI} & = \left\{ 0.24 – 3 * 0.05, 0.24 + 3 * 0.05 \right\} \\
& = \left\{ 0.09, 0.39 \right\} \\
\end{align*} $$

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Determine the probability that a normally distributed random variable lies inside a given interval.


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