What if the sample size is small and the population is heavily skewed?

If $n < 30$ and the population distribution is heavily skewed or contains extreme outliers, the t-distribution is not strictly valid. Non-parametric methods or bootstrapping should be used instead.

Does a 95% confidence interval mean there is a 95% probability the true mean is in my specific interval?

No. The 95% confidence level means that if you were to take many random samples and construct an interval for each, approximately 95% of those constructed intervals would contain the true population mean. The specific interval calculated either contains the true mean or it does not (probability 1 or 0).

When should I use a z-test for a mean in practice?

Almost never. In real engineering and scientific applications, the true population standard deviation $\sigma$ is rarely known if the population mean is also unknown. The t-interval is the standard for practical application.

How to construct a confidence interval for a mean

Q: What if the sample size is small and the population is heavily skewed?

If $n < 30$ and the population distribution is heavily skewed or contains extreme outliers, the t-distribution is not strictly valid. Non-parametric methods or bootstrapping should be used instead.

Q: Does a 95% confidence interval mean there is a 95% probability the true mean is in my specific interval?

No. The 95% confidence level means that if you were to take many random samples and construct an interval for each, approximately 95% of those constructed intervals would contain the true population mean. The specific interval calculated either contains the true mean or it does not (probability 1 or 0).

Q: When should I use a z-test for a mean in practice?

Almost never. In real engineering and scientific applications, the true population standard deviation $\sigma$ is rarely known if the population mean is also unknown. The t-interval is the standard for practical application.

A confidence interval for a population mean provides a range of values that is likely to contain the true population mean. It applies when you have a simple random sample and either the population is normally distributed or the sample size is large enough ( $n \ge 30$ ) for the Central Limit Theorem to apply.

The setup

Identify the sample mean $\bar{x}$ , the sample size $n$ , and the desired confidence level $C$ . Determine whether the population standard deviation $\sigma$ is known. If $\sigma$ is known, you will use the standard normal distribution ( $z$ ). If $\sigma$ is unknown, use the sample standard deviation $s$ and the Student's t-distribution ( $t$ ) with degrees of freedom $df = n - 1$ .

The steps

Find the critical value. For a known $\sigma$ , find $z^*$ corresponding to the middle $C$ area of the standard normal distribution. For an unknown $\sigma$ , find $t^*$ for the middle $C$ area with $df = n - 1$ .
Calculate the Standard Error (SE). $SE = \frac{\sigma}{\sqrt{n}}$ or $SE = \frac{s}{\sqrt{n}}$ .
Calculate the Margin of Error (ME). $ME = ext{critical value} imes SE$ .
Construct the interval. Subtract and add the ME to the sample mean: $(\bar{x} - ME, \bar{x} + ME)$ .

Checking the result

Verify that the sample mean $\bar{x}$ is exactly in the center of the calculated interval. The distance from the lower bound to $\bar{x}$ must equal the distance from $\bar{x}$ to the upper bound. Increasing the confidence level or decreasing the sample size should result in a wider interval.

Common errors

A frequent error is using a $z$ -score instead of a $t$ -score when the population standard deviation $\sigma$ is unknown. Always use the $t$ -distribution if you computed the standard deviation from the sample data. Another error is dividing by $n$ instead of $\sqrt{n}$ when calculating the Standard Error.

Worked example

A random sample of 36 apples yields a sample mean weight of 150 grams and a sample standard deviation of 12 grams. Construct a 95% confidence interval for the true mean weight of the apples.

Identify parameters: $\bar{x} = 150$ , $s = 12$ , $n = 36$ , $C = 0.95$ .
Since $\sigma$ is unknown, use the t-distribution. $df = n - 1 = 35$ .
Find the critical value $t^*$ for 95% confidence and $df = 35$ . Using a t-table, $t^* \approx 2.030$ .
Calculate the Standard Error: $SE = \frac{s}{\sqrt{n}} = \frac{12}{\sqrt{36}} = \frac{12}{6} = 2$ .
Calculate the Margin of Error: $ME = t^* imes SE = 2.030 imes 2 = 4.06$ .
Construct the interval: $(\bar{x} - ME, \bar{x} + ME) = (150 - 4.06, 150 + 4.06) = (145.94, 154.06)$ .

The 95% confidence interval is $(145.94, 154.06)$ grams.

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References: OpenStax Introductory Statistics, Chapter 8: Confidence Intervals · Mathematical Statistics with Applications by Wackerly, Mendenhall, and Scheaffer