Confidence Intervals for the Mean

A sample mean of $182$ g is a useful estimate of a population mean, but on its own it hides how uncertain it is. A confidence interval gives a range of plausible values instead, such as ” $176$ g to $188$ g, with $95\%$ confidence”. In this topic you’ll build these intervals for the mean of a normal population, both when the population standard deviation is known and when you have to estimate it, and learn to interpret them correctly. It’s the same idea as the margin of error in a poll, made precise.

Key ideas

Point estimates and interval estimates

The sample mean $\bar{x}$ is a point estimate of the population mean $\mu$ (see linear combinations of random variables). A confidence interval surrounds it with a margin that reflects sampling variability:

\bar{x} - \text{margin} \quad \text{to} \quad \bar{x} + \text{margin}

The margin depends on how spread out the population is, how big the sample is, and how confident you want to be.

When σ is known: use the normal distribution

If the population is normal with known standard deviation $\sigma$ , then $\bar{X} \sim N\left(\mu, \dfrac{\sigma^2}{n}\right)$ (see the central limit theorem). A confidence interval for $\mu$ is

\bar{x} - z\frac{\sigma}{\sqrt{n}} \ \lt\ \mu \ \lt\ \bar{x} + z\frac{\sigma}{\sqrt{n}}

where $z$ is chosen so that the middle area under the standard normal curve equals the confidence level:

Confidence level	$90\%$	$95\%$	$99\%$
$z$	$1.645$	$1.960$	$2.576$

For $95\%$ , you need $2.5\%$ in each tail, so $z$ is the inverse normal of $0.975$ , which is $1.95996\ldots$ On a GDC, the z-interval function (for example ZInterval) does all of this from $\sigma$ , $\bar{x}$ , $n$ , and the level, or straight from a list of data.

When σ is unknown: use the t-distribution

Usually you don’t know $\sigma$ , so you estimate it with $s_{n-1}$ , the sample standard deviation. Using an estimate adds extra uncertainty, so instead of $z$ you use a value from the t-distribution with $n - 1$ degrees of freedom:

\bar{x} - t\frac{s_{n-1}}{\sqrt{n}} \ \lt\ \mu \ \lt\ \bar{x} + t\frac{s_{n-1}}{\sqrt{n}}

The t-distribution looks like the standard normal but with fatter tails, so $t$ is a bit bigger than $z$ and the interval is a bit wider. In practice you use your GDC’s t-interval function (for example TInterval), entering the data or $\bar{x}$ , $s_{n-1}$ (labelled $Sx$ ), and $n$ .

The IB rule is simple: σ known → normal (z); σ unknown → t, regardless of the sample size. Both methods assume the population is normal.

What the confidence level means

A $95\%$ confidence interval comes from a method that captures the true mean in $95\%$ of samples. If you took many samples and built an interval from each, about $95\%$ of the intervals would contain $\mu$ and about $5\%$ would miss it.

Twenty

95\%

confidence intervals from twenty different samples. The true mean

\mu

is fixed; the intervals move. Here

19

of the

20

capture

\mu

Once you’ve calculated a particular interval, like $(176.1, 187.9)$ , $\mu$ is either in it or not; you just don’t know which. So the correct interpretation is:

“We are $95\%$ confident that the population mean is between $176.1$ g and $187.9$ g.”

meaning the method you used works $95\%$ of the time. It does not mean that $95\%$ of the individual values lie in the interval, and it’s not about the sample mean (which is definitely in the middle of the interval).

What affects the width

The width of the interval is $2 \times z\dfrac{\sigma}{\sqrt{n}}$ (or the same with $t$ and $s_{n-1}$ ).

Higher confidence → wider interval. To be more confident of catching $\mu$ , you need a bigger net: $z$ goes from $1.645$ to $1.960$ to $2.576$ .
Bigger sample → narrower interval. The width is proportional to $\dfrac{1}{\sqrt{n}}$ , so multiplying $n$ by $4$ halves the width.
More variable population → wider interval.

Using an interval to judge a claim

If a claimed value of $\mu$ lies outside a confidence interval, the data give evidence against the claim. If it lies inside, the claim is plausible (but not proven). This connects confidence intervals to hypothesis testing.

Worked examples

Example 1: σ known

The masses of apples from an orchard are normally distributed with standard deviation $15$ g. A random sample of $25$ apples has mean mass $182$ g. Find a $95\%$ confidence interval for the mean mass of all the apples.

Solution. The standard deviation of $\bar{X}$ is $\dfrac{15}{\sqrt{25}} = 3$ g.

\begin{aligned} \text{margin} &= 1.95996\ldots \times 3 = 5.8798\ldots \\ \text{interval} &= 182 \pm 5.8798\ldots \end{aligned}

The $95\%$ confidence interval is $176.1 \lt \mu \lt 187.9$ g (to 1 d.p.), or $(176, 188)$ to 3 s.f. A GDC z-interval with $\sigma = 15$ , $\bar{x} = 182$ , $n = 25$ gives the same result.

Example 2: σ unknown, from data

A random sample of $10$ students recorded how many hours they slept on a school night:

7.2,\ 6.5,\ 8.1,\ 7.8,\ 6.9,\ 7.4,\ 8.4,\ 6.2,\ 7.0,\ 7.5

Assuming sleep times are normally distributed, find a $95\%$ confidence interval for the mean sleep time of all students at the school.

Solution. $\sigma$ is unknown, so use the t-distribution with $10 - 1 = 9$ degrees of freedom.

From one-variable statistics: $\bar{x} = 7.3$ and $s_{n-1} = 0.68799\ldots$

GDC t-interval (data, $95\%$ ): the interval is $6.81 \lt \mu \lt 7.79$ hours (3 s.f.).

By formula, as a check: $t = 2.262\ldots$ (the t-value with $9$ degrees of freedom that leaves $2.5\%$ in the upper tail), so

7.3 \pm 2.262\ldots \times \frac{0.68799\ldots}{\sqrt{10}} = 7.3 \pm 0.4922\ldots

which matches. We are $95\%$ confident that the mean sleep time of all students at the school is between $6.81$ and $7.79$ hours.

Example 3: Checking a claim

A company claims its bags of rice have a mean mass of $500$ g. A consumer group weighs a random sample and calculates a $99\%$ confidence interval for the mean of $(493.2, 498.6)$ g.

(a) Interpret the interval.
(b) Does the interval support the company’s claim?
(c) Would a $95\%$ interval from the same sample be wider or narrower?

Solution.

(a) We are $99\%$ confident that the mean mass of all the company’s bags of rice is between $493.2$ g and $498.6$ g.

(b) No. The claimed value $500$ g lies above the whole interval, so the data suggest the true mean is less than $500$ g.

(c) Narrower: a lower confidence level uses a smaller $z$ (or $t$ ) value. Since $500$ is outside the $99\%$ interval, it’s also outside the narrower $95\%$ interval.

Example 4: Choosing the sample size

A researcher wants to estimate the mean commute time of workers in a city. Commute times are normally distributed with standard deviation $12$ minutes. How large a sample is needed for a $95\%$ confidence interval to have a width of at most $5$ minutes?

Solution. The width is $2 \times 1.95996\ldots \times \dfrac{12}{\sqrt{n}}$ .

\begin{aligned} 2 \times 1.95996\ldots \times \frac{12}{\sqrt{n}} &\le 5 \\ \sqrt{n} &\ge \frac{2 \times 1.95996\ldots \times 12}{5} = 9.4078\ldots \\ n &\ge 88.507\ldots \end{aligned}

The sample must contain at least $89$ workers. (Always round up: $88$ would give an interval slightly wider than $5$ minutes.)

Common mistakes

Using z when σ is unknown. If you’ve calculated the standard deviation from the sample, use the t-distribution with $n - 1$ degrees of freedom, even for a large sample.

Forgetting to divide by √n. The margin uses the standard deviation of $\bar{X}$ , which is $\dfrac{\sigma}{\sqrt{n}}$ , not $\sigma$ .

Using sₙ instead of sₙ₋₁. For a t-interval, use the unbiased estimate $s_{n-1}$ ( $Sx$ on the GDC), not $\sigma x$ .

Saying “there is a 95% probability that μ is in this interval”. Once the interval is calculated, $\mu$ is either in it or not. Say “we are $95\%$ confident”, and know that it describes the long-run success rate of the method.

Thinking the interval contains 95% of the data. A confidence interval estimates the mean. Individual values are much more spread out than the interval.

Rounding the sample size down. If the calculation gives $n \ge 88.5$ , you need $n = 89$ .

Practice

1. (Warm-up) A normal population has $\sigma = 5$ . A random sample of $16$ values has mean $64.0$ . Find a $95\%$ confidence interval for $\mu$ .

Solution

64.0 \pm 1.95996\ldots \times \frac{5}{\sqrt{16}} = 64.0 \pm 2.4499\ldots

The interval is $61.6 \lt \mu \lt 66.4$ (3 s.f.).

2. (Warm-up) A confidence interval for a mean is $(12.4, 15.0)$ . Find the sample mean and the margin.

Solution

The sample mean is the midpoint: $\dfrac{12.4 + 15.0}{2} = 13.7$ . The margin is half the width: $\dfrac{15.0 - 12.4}{2} = 1.3$ .

3. (Warm-up) A $90\%$ confidence interval for the mean height of Grade 12 students is $(168.2, 173.6)$ cm. Which statements are correct?

(a) $90\%$ of Grade 12 students are between $168.2$ cm and $173.6$ cm tall.
(b) We are $90\%$ confident that the mean height of all Grade 12 students is between $168.2$ cm and $173.6$ cm.
(c) If many samples were taken, about $90\%$ of the intervals made this way would contain the population mean.
(d) The sample mean is between $168.2$ cm and $173.6$ cm with probability $0.9$ .

Solution

(b) and (c) are correct. (a) is wrong because the interval is about the mean, not individual heights. (d) is wrong because the sample mean is known exactly: it’s $170.9$ cm, the centre of the interval.

4. (Core) A random sample of $20$ bags of chips from a normal population has mean mass $48.3$ g and $s_{n-1} = 6.2$ g. Find a $95\%$ confidence interval for the population mean.

Solution

$\sigma$ is unknown, so use $t$ with $19$ degrees of freedom: $t = 2.0930\ldots$

48.3 \pm 2.0930\ldots \times \frac{6.2}{\sqrt{20}} = 48.3 \pm 2.9016\ldots

The interval is $45.4 \lt \mu \lt 51.2$ g (3 s.f.). A GDC t-interval from the summary statistics gives the same.

5. (Core) The lifetime of a type of phone battery is normally distributed with standard deviation $0.8$ years. A sample of $40$ batteries has mean lifetime $3.42$ years. Find a $90\%$ confidence interval for the mean lifetime.

Solution

$\sigma$ is known, so use $z = 1.6448\ldots$ (inverse normal of $0.95$ ).

3.42 \pm 1.6448\ldots \times \frac{0.8}{\sqrt{40}} = 3.42 \pm 0.20806\ldots

The interval is $3.21 \lt \mu \lt 3.63$ years (3 s.f.).

6. (Core) The times, in seconds, for $8$ randomly chosen swimmers to complete a $50$ m length were:

31.5,\ 29.8,\ 33.2,\ 30.4,\ 32.7,\ 28.9,\ 31.1,\ 30.6

Assuming the times are normally distributed, find a $99\%$ confidence interval for the mean time, and interpret it.

Solution

$\sigma$ is unknown, so use a t-interval with $7$ degrees of freedom. From the GDC: $\bar{x} = 31.025$ , $s_{n-1} = 1.4320\ldots$

GDC t-interval ( $99\%$ ): $29.3 \lt \mu \lt 32.8$ seconds (3 s.f.).

(Check: $t = 3.4994\ldots$ and $31.025 \pm 3.4994\ldots \times \dfrac{1.4320\ldots}{\sqrt{8}} = 31.025 \pm 1.7718\ldots$ .)

We are $99\%$ confident that the mean time for all such swimmers is between $29.3$ and $32.8$ seconds.

7. (Core) A normal population has $\sigma = 10$ .

(a) Find the width of a $95\%$ confidence interval for $\mu$ from a sample of $25$ .
(b) Find the width if the sample size is $100$ instead. How does it compare with (a)?
(c) Find the width of a $99\%$ interval from a sample of $25$ .

Solution

(a) $2 \times 1.95996\ldots \times \dfrac{10}{\sqrt{25}} = 7.84$ (3 s.f.)

(b) $2 \times 1.95996\ldots \times \dfrac{10}{\sqrt{100}} = 3.92$ (3 s.f.). Four times the sample size gives half the width.

8. (Challenge) A $95\%$ confidence interval for the mean of a normal population with known $\sigma$ is $(71.2, 76.8)$ . Find the $99\%$ confidence interval from the same sample.

Solution

The centre is $\bar{x} = 74$ and the margin is $2.8$ . So

1.95996\ldots \times \frac{\sigma}{\sqrt{n}} = 2.8 \quad\Rightarrow\quad \frac{\sigma}{\sqrt{n}} = 1.42859\ldots

For $99\%$ : margin $= 2.5758\ldots \times 1.42859\ldots = 3.6798\ldots$

The $99\%$ interval is $70.3 \lt \mu \lt 77.7$ (3 s.f.).

9. (Challenge) A machine fills bottles with a volume that is normally distributed with standard deviation $0.15$ litres. Quality control wants a $99\%$ confidence interval for the mean volume with a width of no more than $0.05$ litres. Find the smallest sample size needed.

Solution

\begin{aligned} 2 \times 2.5758\ldots \times \frac{0.15}{\sqrt{n}} &\le 0.05 \\ \sqrt{n} &\ge \frac{2 \times 2.5758\ldots \times 0.15}{0.05} = 15.454\ldots \\ n &\ge 238.85\ldots \end{aligned}

The smallest sample size is $239$ bottles.