A large sample of n = 100 has x = 30 and known σ = 5. Find a 95\% confidence interval for μ. [2 marks]

ScotlandStatisticsSyllabus dot point

How do you estimate a population mean from a sample and express how confident you are?

Calculate point estimates of a population mean and variance, construct and interpret confidence intervals for a population mean using the normal and Student's t-distributions, and construct a confidence interval for a population proportion.

A focused answer to the SQA Advanced Higher Statistics estimation content: point estimates of the population mean and variance, confidence intervals for a mean using the normal distribution and Student's t-distribution, the role of degrees of freedom, and confidence intervals for a population proportion.

Generated by Claude Opus 4.813 min answerUpdated 2026-06-16

Reviewed by: AI editorial process; not yet individually human-reviewed

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Quick answer

A point estimate of the population mean $\mu$ is the sample mean $\bar{x}$ ; of the population variance $\sigma^2$ it is the sample variance $s^2$ (using the $n-1$ divisor). A confidence interval for $\mu$ is $\bar{x} \pm (\text{critical value}) \times (\text{standard error})$ . When $\sigma$ is known (or $n$ is large) use $z$ and standard error $\dfrac{\sigma}{\sqrt{n}}$ ; when $\sigma$ is unknown and estimated by $s$ with small $n$ , use Student's t-distribution with $n-1$ degrees of freedom and standard error $\dfrac{s}{\sqrt{n}}$ . A confidence interval for a proportion is $\hat{p} \pm z\sqrt{\dfrac{\hat{p}(1-\hat{p})}{n}}$ .

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What this dot point is asking
Point estimation
Confidence intervals for a mean
Student's t-distribution
Confidence intervals for a proportion
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What this dot point is asking

Estimation turns a sample into a statement about the population. The SQA wants you to give a single best estimate (a point estimate) of a population mean or variance, and then a confidence interval that says how precise that estimate is, choosing the normal distribution or Student's t-distribution appropriately and interpreting the result correctly.

Point estimation

Before an interval, you need a single best estimate of each unknown parameter.

Confidence intervals for a mean

A confidence interval is a range, centred on the point estimate, that captures the true mean with a stated long-run reliability.

The structure is always estimate plus or minus margin of error, and the margin of error is the critical value multiplied by the standard error.

Construct a 95% confidence interval with known sigma

A confidence interval is built from three pieces: the point estimate, the critical value, and the standard error. Because $\sigma$ is known here, we use the standard normal critical value $z = 1.96$ rather than a t-value, and the standard error uses $\sigma$ directly.

Step 1: Compute the standard error

The standard error measures how much the sample mean typically varies from the population mean. It is $\sigma$ divided by the square root of $n$ .

\text{SE} = \dfrac{\sigma}{\sqrt{n}} = \dfrac{10}{\sqrt{25}} = \dfrac{10}{5} = 2

Step 2: Choose the critical value

For a 95% confidence interval with known $\sigma$ (or a large sample), we use the standard normal distribution. The critical value that cuts off 2.5% in each tail is $z = 1.96$ .

Step 3: Assemble the interval as estimate plus or minus margin of error

The margin of error is the critical value multiplied by the standard error. Subtract it and add it to the point estimate to form the interval.

80 \pm 1.96 \times 2 = 80 \pm 3.92

Final answer: the 95% confidence interval for $\mu$ is $(76.08,\ 83.92)$ .

Student's t-distribution

When the population standard deviation is unknown and is estimated by the sample $s$ , the extra uncertainty means the normal distribution is too narrow; Student's t-distribution corrects this.

Confidence intervals for a proportion

For a population proportion, the same plus-or-minus structure applies, using the standard error of $\hat{p}$ .

Try this

Q1. A large sample of $n = 100$ has $\bar{x} = 30$ and known $\sigma = 5$ . Find a $95\%$ confidence interval for $\mu$ . [2 marks]

Cue. Standard error $= \dfrac{5}{\sqrt{100}} = 0.5$ , so $30 \pm 1.96 \times 0.5 = 30 \pm 0.98 = (29.02, 30.98)$ .

Q2. State the degrees of freedom for a t-interval from a sample of size $20$ . [1 mark]

Cue. $n - 1 = 19$ degrees of freedom.

Exam-style practice questions

Practice questions written in the style of SQA exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

AH style: t-interval4 marksA sample of

n = 16

has mean

\bar{x} = 25.0

and sample standard deviation

s = 4.0

. Construct a

95\%

confidence interval for the population mean. (Use

t_{0.025, 15} = 2.131

Show worked answer →

Since $\sigma$ is unknown and $n$ is small, use the t-distribution with $n - 1 = 15$ degrees of freedom (1 mark).

Standard error: $\dfrac{s}{\sqrt{n}} = \dfrac{4.0}{\sqrt{16}} = \dfrac{4.0}{4} = 1.0$ (1 mark).

Margin of error: $t \times \dfrac{s}{\sqrt{n}} = 2.131 \times 1.0 = 2.131$ (1 mark).

Interval: $25.0 \pm 2.131 = (22.87, 27.13)$ (1 mark). Markers reward choosing the t-distribution, the standard error, the margin of error and the final interval.

AH style: interpretation3 marksA

95\%

confidence interval for a mean is

(48.2, 51.8)

. Explain what '95% confidence' means, and state how the interval would change for a 99% level.

Show worked answer →

A $95\%$ confidence level means that if the sampling and interval procedure were repeated many times, about $95\%$ of the intervals produced would contain the true population mean (1 mark). It is the method, not this one interval, that is right $95\%$ of the time; the true mean is fixed and either is or is not in $(48.2, 51.8)$ (1 mark).

A $99\%$ interval would be wider, because a higher confidence level uses a larger critical value, increasing the margin of error to be more certain of capturing the mean (1 mark). Markers reward the long-run-frequency interpretation and the point that higher confidence widens the interval.

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Sources & how we know this

SQA Advanced Higher Statistics Course Specification (C803 77) — SQA (2023)
SQA Advanced Higher Statistics Data Booklet — SQA (2019)