Confidence interval for a population mean

The usual situation. In almost every real experiment the population variance $\sigma^2$ is unknown — all you have is the sample. You then estimate it with the unbiased sample variance $s^2$ and the right model is the $t$-distribution, not the normal.

The unbiased estimate. From a sample $x_1, x_2, \ldots, x_n$: $s^2 = \frac{1}{n-1}\left(\sum x^2 - \frac{(\sum x)^2}{n}\right)$

The divisor is $n-1$ (NOT $n$) — this is what makes $s^2$ unbiased. Take the square root for $s$.

The confidence interval. For a sample of size $n$ from a normal population with $\bar{x}$ the sample mean: $\bar{x} \;\pm\; t_\nu\,\frac{s}{\sqrt{n}}, \qquad \nu = n-1$

• $\bar{x}$ is the centre of the interval.
• $t_\nu\,\frac{s}{\sqrt{n}}$ is the margin of error (half-width).
• $t_\nu$ is the two-tailed critical value from MF19 with $\nu = n-1$ degrees of freedom.

Reading the t-table (MF19). For a 95% interval you need the value with $2.5\%$ in the upper tail (because $2.5\%$ sits in each tail). Some common ones:

Confidence	df $\nu$	$t_\nu$
95%	9	2.262
95%	14	2.145
95%	7	2.365
95%	11	2.201
95%	19	2.093

Cambridge tip. The $t$-distribution has fatter tails than the normal, so $t_\nu > 1.96$ for every finite $\nu$. As $n \to \infty$, $t_\nu \to 1.96$ and the $t$-interval merges into the $z$-interval — which is exactly why a large sample can use $z$.

Sometimes the population standard deviation $\sigma$ is given (from long experience or a manufacturer's specification), or the sample is large enough for the Central Limit Theorem to apply. In both cases use the normal distribution.

The interval. $\bar{x} \;\pm\; z\,\frac{\sigma}{\sqrt{n}}$

where $z$ is the two-tailed standard-normal critical value:

Confidence level	$z$
90%	1.645
95%	1.960
98%	2.326
99%	2.576

Where these come from. A 95% interval leaves $2.5\%$ in each tail, so $z = \Phi^{-1}(0.975) = 1.96$. A 99% interval leaves $0.5\%$ in each tail, so $z = \Phi^{-1}(0.995) = 2.576$.

Large sample, unknown variance. If $n$ is large but $\sigma$ is unknown, you may replace $\sigma$ by $s$ and still use $z$: $\bar{x} \;\pm\; z\,\frac{s}{\sqrt{n}}$ because for large $n$, $t_\nu \approx z$. (For small $n$ this shortcut is wrong — you must use $t$.)

Cambridge tip. Decide $z$ vs $t$ FIRST. Ask: "Is $\sigma$ given, or is $n$ large?" If yes to either, use $z$. Otherwise (small sample, $\sigma$ unknown) use $t_\nu$ with $\nu = n-1$. Choosing the wrong distribution loses the method mark immediately.

The single most-tested idea. A confidence level describes the reliability of the method, not the probability for one particular interval.

Correct wording (memorise this):

"If we repeated the sampling procedure many times, about 95% of the intervals constructed this way would contain the true mean $\mu$."

WRONG wording (loses the mark):

"There is a 95% probability that $\mu$ lies in this interval."

Why is the second wrong? Once the data are collected, the interval is fixed and $\mu$ is fixed. Either $\mu$ is inside or it isn't — there is no probability left. The randomness was in which interval you happened to get.

What changes the width? The half-width (margin of error) is $\text{critical value} \times \frac{s \text{ or } \sigma}{\sqrt{n}}$.

• Higher confidence level $\Rightarrow$ WIDER. A 99% interval needs a bigger critical value ($2.576 > 1.96$), so it stretches further. More confidence costs precision.
• Larger sample size $n \Rightarrow$ NARROWER. $n$ sits under a square root, so quadrupling $n$ halves the margin. More data buys precision.
• Larger spread ($s$ or $\sigma$) $\Rightarrow$ WIDER. Noisier data give a vaguer estimate.

Cambridge tip. "Width" means the full length $2 \times$ margin. To halve the width you must quadruple $n$ (because of the $\sqrt{n}$). Examiners love this "how many times larger must the sample be?" twist.

Often you compare two populations — say a new teaching method versus the old — and want an interval for the difference $\mu_1 - \mu_2$.

Variances unknown but assumed equal — pooled t-interval. Combine the two sample variances into a single pooled estimate: $s_p^2 = \frac{(n_1-1)s_1^2 + (n_2-1)s_2^2}{n_1 + n_2 - 2}$

Then the interval is: $(\bar{x}_1 - \bar{x}_2) \;\pm\; t_\nu\,s_p\sqrt{\frac{1}{n_1} + \frac{1}{n_2}}, \qquad \nu = n_1 + n_2 - 2$

The degrees of freedom add up: $\nu = n_1 + n_2 - 2$ (you lose one for each estimated mean).

Variances known — z-interval. If $\sigma_1^2$ and $\sigma_2^2$ are known (or both samples large), use: $(\bar{x}_1 - \bar{x}_2) \;\pm\; z\,\sqrt{\frac{\sigma_1^2}{n_1} + \frac{\sigma_2^2}{n_2}}$

Interpreting a difference interval. This is the examiner's favourite follow-up:

• If the interval contains 0, the data are consistent with no difference between the means.
• If the interval is entirely positive (or entirely negative), there is evidence that one mean genuinely exceeds the other.

Cambridge tip. For the pooled $t$-interval, compute $s_p^2$ carefully — it is a weighted average of the two variances, weighted by $(n_i - 1)$. A common slip is to average $s_1^2$ and $s_2^2$ with equal weights when $n_1 \neq n_2$.