# Confidence Interval: Sample Mean

This lesson describes how to construct a confidence interval around a sample mean, x.

## Estimation Requirements

The approach described in this lesson is valid whenever the following conditions are met:

- The sampling method is simple random sampling.
- The sampling distribution is approximately normally distributed.

Generally, the sampling distribution will be approximately normally distributed if any of the following conditions apply.

- The population distribution is normal.
- The sampling distribution is symmetric, unimodal, without outliers, and the sample size is 15 or less.
- The sampling distribution is moderately skewed, unimodal, without outliers, and the sample size is between 16 and 40.
- The sample size is greater than 40, without outliers.

## The Variability of the Sample Mean

To construct a confidence interval for a sample mean, we need to know the variability of the sample mean. This means we need to know how to compute the standard deviation or the standard error of the sampling distribution.

- Suppose
*k*possible samples of size*n*can be selected from a population of size*N*. The standard deviation of the sampling distribution is the "average" deviation between the*k*sample means and the true population mean, μ. The standard deviation of the sample mean σ_{x}is:σ

where σ is the standard deviation of the population,_{x}= σ * sqrt{ ( 1/n ) * ( 1 - n/N ) * [ N / ( N - 1 ) ] }*N*is the population size, and*n*is the sample size. When the population size is much larger (at least 10 times larger) than the sample size, the standard deviation can be approximated by:σ

_{x}= σ / sqrt( n ) - When the standard deviation of the population σ is unknown,
the standard deviation of the sampling distribution
cannot be calculated. Under these
circumstances, use the standard error.
The standard error (SE) provides an unbiased estimate of the
standard deviation. It can be calculated from the equation below.
SE

where_{x}= s * sqrt{ ( 1/n ) * ( 1 - n/N ) * [ N / ( N - 1 ) ] }*s*is the standard deviation of the sample, N is the population size, and*n*is the sample size. When the population size is much larger (at least 10 times larger) than the sample size, the standard error can be approximated by:SE

_{x}= s / sqrt( n )

Note: In real-world analyses, the standard deviation of the population is seldom known. Therefore, the standard error is used more often than the standard deviation.

## Alert

The Advanced Placement Statistics Examination only covers the "approximate" formulas for the standard deviation and standard error. However, students are expected to be aware of the limitations of these formulas; namely, the approximate formulas should only be used when the population size is at least 10 times larger than the sample size.

## How to Find the Confidence Interval for a Mean

Previously, we described how to construct confidence intervals. For convenience, we repeat the key steps below.

- Identify a sample statistic. Use the sample mean to
estimate the population mean.
- Select a confidence level. The confidence level describes the
uncertainty of a sampling
method. Often, researchers choose 90%, 95%, or 99% confidence
levels; but any percentage can be used.
- Find the margin of error. Previously, we showed
how to compute the margin of error.
- Specify the confidence interval. The range of the confidence
interval is defined by the
*sample statistic*__+__*margin of error*. And the uncertainty is denoted by the confidence level.

In the next section, we work through a problem that shows how to use this approach to construct a confidence interval to estimate a population mean.

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## Test Your Understanding

**Problem 1**

Suppose a simple random sample of 150 students is drawn from a population of 3000 college students. Among sampled students, the average IQ score is 115 with a standard deviation of 10. What is the 99% confidence interval for the students' IQ score?

(A) 115 __+__ 0.01

(B) 115 __+__ 0.82

(C) 115 __+__ 2.1

(D) 115 __+__ 2.6

(E) None of the above

**Solution**

The correct answer is (C). The approach that we used to solve this problem is valid when the following conditions are met.

- The sampling method must be simple random sampling. This condition is satisfied; the problem statement says that we used simple random sampling.
- The sampling distribution should be approximately normally distributed. Because the sample size is large, we know from the central limit theorem that the sampling distribution of the mean will be normal or nearly normal; so this condition is satisfied.

Since the above requirements are satisfied, we can use the following four-step approach to construct a confidence interval.

- Identify a sample statistic. Since we are trying to estimate
a population mean, we choose the sample mean
(115) as the sample statistic.
- Select a confidence level. In this analysis, the confidence level
is defined for us in the problem. We are working with a 99%
confidence level.
- Find the margin of error. Elsewhere on this site, we show
how to compute the margin of error when the sampling
distribution is approximately normal. The key steps are
shown below.

- Find standard deviation or standard error. Since we do not
know the standard deviation of the population, we cannot compute the
standard deviation of the sample mean; instead, we compute the standard
error (SE). Because the sample size is much smaller than the
population size, we can use the "approximate" formula for the
standard error.
SE = s / sqrt( n ) = 10 / sqrt(150) = 10 / 12.25 = 0.82

- Find critical value. The critical value is a factor used to
compute the margin of error. Because the standard deviation
of the population is unknown, we express the critical
value as a
t score
rather than a
z score.
To find the critical value, we take these steps.

- Compute alpha (α): α = 1 - (confidence level / 100) = 1 - 99/100 = 0.01
- Find the critical probability (p*): p* = 1 - α/2 = 1 - 0.01/2 = 0.995
- Find the degrees of freedom (df): df = n - 1 = 150 - 1 = 149
- The critical value is the t score having 149 degrees of freedom and a cumulative probability equal to 0.995. From the t Distribution Calculator, we find that the critical value is 2.61.

Note: We might also have expressed the critical value as a z score. Because the sample size is fairly large, a z score analysis produces a similar result - a critical value equal to 2.58.

- Compute margin of error (ME): ME = critical value * standard error = 2.61 * 0.82 = 2.1

- Find standard deviation or standard error. Since we do not
know the standard deviation of the population, we cannot compute the
standard deviation of the sample mean; instead, we compute the standard
error (SE). Because the sample size is much smaller than the
population size, we can use the "approximate" formula for the
standard error.
- Specify the confidence interval. The range of the confidence
interval is defined by the
*sample statistic*__+__*margin of error*. And the uncertainty is denoted by the confidence level.

Therefore, the 99% confidence interval is 112.9 to 117.1. That is, we are 99%
confident that the true population mean is in the range
defined by 115 __+__ 2.1.