Normal hypothesis tests for a population mean

A Normal hypothesis test for a mean uses an observation or sample mean to assess a claim about an unknown population mean $\mu$. The calculation asks how unusual the observed result would be if the null hypothesis were true.

The key model for a random sample of size $n$ is

$$X\sim\operatorname N(\mu,\sigma^2) \quad\Longrightarrow\quad \bar X\sim\operatorname N\left(\mu,\frac{\sigma^2}{n}\right).$$

Therefore the standard deviation of $\bar X$, called its standard error, is

$$\boxed{\operatorname{sd}(\bar X)=\frac{\sigma}{\sqrt n}}.$$

This lesson concerns A level questions in which the population variance $\sigma^2$ is known. It does not cover the $t$ test used when $\sigma^2$ must be estimated.

Prerequisites

You should be able to:

• standardise a Normal random variable using $Z=(X-\mu)/\sigma$;
• use inverse Normal probabilities and calculator distribution functions;
• distinguish a population mean $\mu$ from a sample mean $\bar x$;
• write null and alternative hypotheses and interpret a significance level.

Review the Normal distribution and hypothesis testing language where needed.

From a population to a test statistic

Suppose the null hypothesis is $H_0:\mu=\mu_0$. Under $H_0$,

$$\bar X\sim\operatorname N\left(\mu_0,\frac{\sigma^2}{n}\right),$$

so an observed sample mean $\bar x$ can be standardised as

$$\boxed{z=\frac{\bar x-\mu_0}{\sigma/\sqrt n}}.$$

The sign and magnitude of $z$ both matter:

• $z>0$ means that $\bar x$ is above the null mean;
• $z<0$ means that $\bar x$ is below the null mean;
• $|z|$ measures the distance from the null mean in standard errors.

For one observation, $n=1$, so the denominator is simply $\sigma$.

Variance is not standard deviation

If $X\sim\operatorname N(50,16)$, then $\sigma^2=16$ and $\sigma=4$. For a sample of size $25$, the standard error is $4/\sqrt{25}=0.8$, not $16/25$ and not $4/25$.

Worked example 1: form the null model

Packet masses are modelled by $X\sim\operatorname N(200,6^2)$ grams. A random sample of $36$ packets is taken. Under the claim that the population mean remains $200$ g,

$$\bar X\sim\operatorname N\left(200,\frac{6^2}{36}\right) =\operatorname N(200,1).$$

Thus $\operatorname{sd}(\bar X)=1$ g. Sample means vary much less than individual packet masses.

If the sample mean is $197.8$ g, then

$$z=\frac{197.8-200}{1}=-2.2.$$

The sample mean is $2.2$ standard errors below the claimed mean.

Choose the alternative hypothesis first

Let $\mu$ be the population mean and let $\mu_0$ be the value claimed under $H_0$.

Investigation	Hypotheses	Relevant evidence
Has the mean increased?	$H_0:\mu=\mu_0$, $H_1:\mu>\mu_0$	upper tail
Has the mean decreased?	$H_0:\mu=\mu_0$, $H_1:\mu<\mu_0$	lower tail
Has the mean changed?	$H_0:\mu=\mu_0$, $H_1:\mu\neq\mu_0$	both tails

Hypotheses concern the population parameter $\mu$, not the observed statistic $\bar x$. Decide whether the test has one tail or two from the question, before inspecting the data.

Self-check 1

A filling machine is set to deliver a mean of $500$ ml. An engineer investigates whether it is overfilling. A sample has mean $\bar x=503.1$ ml. Write the hypotheses.

Answer

Let $\mu$ be the population mean fill volume. Since overfilling means a larger mean,

$$H_0:\mu=500, \qquad H_1:\mu>500.$$

Writing a hypothesis about $\bar x$ is incorrect because $503.1$ ml is already observed.

Method 1: use a p value

The p value is the probability, assuming $H_0$, of obtaining the observed result or one more extreme in the direction specified by $H_1$.

For an observed standardised value $z$:

$$\begin{array}{c|c} H_1 & \text{p value} \\ \hline \mu>\mu_0 & P(Z\geq z) \\ \mu<\mu_0 & P(Z\leq z) \\ \mu\neq\mu_0 & 2P(Z\geq |z|) \end{array}$$

Then

$$\boxed{\text{reject }H_0\text{ if the p value}\leq\alpha.}$$

For a two tailed test, doubling the smaller tail probability counts equally extreme results on the opposite side of $\mu_0$. This works because the Normal distribution is symmetric.

Worked example 2: lower tailed test

A manufacturer states that the lifetime of a component is Normally distributed with standard deviation $120$ hours and mean $1500$ hours. A random sample of $36$ components has mean lifetime $1464$ hours. Test at the $5\%$ level whether the population mean lifetime has decreased.

Let $\mu$ be the population mean lifetime. The hypotheses are

$$H_0:\mu=1500, \qquad H_1:\mu<1500.$$

Under $H_0$,

$$\bar X\sim\operatorname N\left(1500,\frac{120^2}{36}\right),$$

so the standard error is

$$\frac{120}{\sqrt{36}}=20.$$

The observed test statistic is

$$z=\frac{1464-1500}{20}=-1.8.$$

This is a lower tailed test, so

$$p=P(Z\leq-1.8)\approx0.0359.$$

Since $0.0359<0.05$, reject $H_0$. There is sufficient evidence at the $5\%$ significance level to suggest that the population mean lifetime has decreased.

Worked example 3: upper tailed test with no rejection

The mass of produce in a container is modelled as Normal with known standard deviation $8$ g. Historically its mean is $250$ g. After a process adjustment, a random sample of $16$ containers has mean mass $252.6$ g. Test at the $5\%$ level whether the mean has increased.

$$H_0:\mu=250, \qquad H_1:\mu>250.$$

Under $H_0$, the standard error is

$$\frac{8}{\sqrt{16}}=2.$$

Therefore

$$z=\frac{252.6-250}{2}=1.3$$

and the upper tail probability is

$$p=P(Z\geq1.3)\approx0.0968.$$

Since $0.0968>0.05$, do not reject $H_0$. There is insufficient evidence at the $5\%$ level to suggest that the population mean mass has increased.

The sample mean is above $250$ g, but the difference is not large relative to ordinary sampling variation.

Two tailed tests

A two tailed alternative $H_1:\mu\neq\mu_0$ treats a result equally far below or above $\mu_0$ as evidence against $H_0$. At overall significance level $\alpha$, each tail contains probability $\alpha/2$.

Worked example 4: calculate and double one tail

A machine produces rods whose lengths are Normally distributed with known standard deviation $0.8$ mm. It is set to a mean of $50$ mm. A random sample of $25$ rods has mean $50.34$ mm. Test at the $5\%$ level whether the mean length has changed.

$$H_0:\mu=50, \qquad H_1:\mu\neq50.$$

The standard error is

$$\frac{0.8}{\sqrt{25}}=0.16,$$

giving

$$z=\frac{50.34-50}{0.16}=2.125.$$

The upper tail probability is approximately

$$P(Z\geq2.125)=0.0168.$$

Hence the two tailed p value is

$$p=2(0.0168)=0.0336.$$

Since $0.0336<0.05$, reject $H_0$. There is sufficient evidence at the $5\%$ level to suggest that the population mean rod length has changed.

Do not forget to double

The sample happens to lie above $50$ mm, but the planned alternative was $\mu\neq50$. Using only $0.0168$ would silently turn the test into a one tailed test.

Self-check 2

A two tailed test gives $z=-1.72$. Find the p value and state the decision at the $5\%$ level.

Answer

By symmetry,

$$p=2P(Z\leq-1.72)\approx2(0.0427)=0.0854.$$

Since $0.0854>0.05$, do not reject $H_0$.

Method 2: use critical values

A critical value separates results that lead to rejection from those that do not. Common standard Normal boundaries are:

Test	Significance level	Reject $H_0$ when
upper tailed	$5\%$	$z\geq1.645$
lower tailed	$5\%$	$z\leq-1.645$
two tailed	$5\%$	$
upper tailed	$1\%$	$z\geq2.326$
lower tailed	$1\%$	$z\leq-2.326$
two tailed	$1\%$	$

You may instead convert a $z$ boundary into a critical sample mean. For an upper tailed test,

$$\bar x_{\mathrm c}=\mu_0+z_{1-\alpha}\frac{\sigma}{\sqrt n}.$$

For a lower tailed test, use the negative boundary. For a two tailed test, use $z_{1-\alpha/2}$ in both directions.

Because the Normal model is continuous, whether a single boundary point is written with $<$ or $\leq$ does not alter its probability. The usual decision convention includes equality in the critical region.

Worked example 5: critical sample mean

Weekly demand is Normally distributed with known standard deviation $24$ units. A supplier tests

$$H_0:\mu=400 \qquad\text{against}\qquad H_1:\mu>400$$

using the mean of $64$ weeks at the $5\%$ level. Find the critical value of $\bar X$.

Under $H_0$, the standard error is

$$\frac{24}{\sqrt{64}}=3.$$

The upper $5\%$ standard Normal boundary is $1.645$, so

$$\bar x_{\mathrm c}=400+1.645(3)=404.935.$$

Thus the critical region is approximately

$$\boxed{\bar X\geq404.94}.$$

A sample mean of $405.2$ lies in this region, so $H_0$ would be rejected. A sample mean of $404.8$ does not.

Worked example 6: two critical boundaries

The fill mass of a product is Normal with known standard deviation $5$ g. A sample of $100$ items is used to test $H_0:\mu=80$ against $H_1:\mu\neq80$ at the $1\%$ level.

The standard error is

$$\frac{5}{\sqrt{100}}=0.5.$$

For a $1\%$ two tailed test, the boundary is $z=\pm2.576$. Therefore

$$80\pm2.576(0.5)=80\pm1.288.$$

The critical regions are

$$\boxed{\bar X\leq78.712\quad\text{or}\quad\bar X\geq81.288}.$$

Notice that $0.5\%$ is placed in each tail, giving $1\%$ in total.

Testing a single observation

Some questions give one observation from a Normal population rather than a sample mean. Then use $n=1$:

$$z=\frac{x-\mu_0}{\sigma}.$$

Worked example 7: one observation

Under a stated model, journey time $X$ is Normal with standard deviation $4$ minutes and mean $30$ minutes. One randomly selected journey takes $38.5$ minutes. Test at the $5\%$ level whether journey times are longer than the model claims.

$$H_0:\mu=30, \qquad H_1:\mu>30.$$

For one observation,

$$z=\frac{38.5-30}{4}=2.125.$$

Thus

$$p=P(Z\geq2.125)\approx0.0168<0.05.$$

Reject $H_0$. There is sufficient evidence at the $5\%$ level to suggest that the population mean journey time is greater than $30$ minutes.

Do not divide $4$ by another square root: the observation is not a sample mean based on an unstated sample size.

A reliable exam method

1. Define $\mu$ in context.
2. Write $H_0$ and $H_1$, choosing the tail from the question.
3. Under $H_0$, write the distribution of $X$ or $\bar X$.
4. Calculate the correct standard deviation, using $\sigma/\sqrt n$ for a sample mean.
5. Find a p value or critical region.
6. Compare with the significance level and state reject or do not reject $H_0$.
7. Give a contextual conclusion using the direction in $H_1$.

Worked example 8: full test from start to finish

The breaking strength of a cable is Normally distributed with standard deviation $18$ N. The manufacturer claims a mean strength of $320$ N. A regulator suspects the true mean is lower. A random sample of $49$ cables has mean strength $315.5$ N. Test the regulator’s suspicion at the $1\%$ level.

Let $\mu$ be the population mean breaking strength in newtons.

$$H_0:\mu=320, \qquad H_1:\mu<320.$$

Under $H_0$,

$$\bar X\sim\operatorname N\left(320,\frac{18^2}{49}\right),$$

with standard error

$$\frac{18}{7}\approx2.5714.$$

Hence

$$z=\frac{315.5-320}{18/7}=-1.75.$$

The lower tail p value is

$$P(Z\leq-1.75)\approx0.0401.$$

Since $0.0401>0.01$, do not reject $H_0$. There is insufficient evidence at the $1\%$ significance level to suggest that the population mean breaking strength is below $320$ N.

The result would be significant at $5\%$, but the question demands the stricter $1\%$ standard.

Common misconceptions

• Using $\sigma$ instead of $\sigma/\sqrt n$. A sample mean has less variation than an individual value.
• Using variance in the denominator. Standardisation always divides by a standard deviation.
• Writing hypotheses about $\bar x$. The hypotheses concern the unknown population mean $\mu$.
• Choosing the tail from the data. The wording of the investigation determines $H_1$.
• Forgetting the second tail. For $H_1:\mu\neq\mu_0$, double the smaller one tail probability.
• Saying that $H_0$ is proved. A non-significant result means only that the evidence was insufficient to reject $H_0$.
• Ignoring context. Finish with a statement about the population mean, not only a calculator probability.

Final self-check

A process produces values that are Normally distributed with known variance $144$. A random sample of $36$ values has mean $74.5$. Test $H_0:\mu=78$ against $H_1:\mu<78$ at the $5\%$ level.

Answer

Since $\sigma=\sqrt{144}=12$, the standard error is

$$\frac{12}{\sqrt{36}}=2.$$

Therefore

$$z=\frac{74.5-78}{2}=-1.75$$

and

$$p=P(Z\leq-1.75)\approx0.0401.$$

Since $0.0401<0.05$, reject $H_0$. There is sufficient evidence at the $5\%$ level to suggest that the population mean is less than $78$.

Using the critical value method gives the same decision because $-1.75<-1.645$.

Next steps

• Compare exact discrete testing in binomial hypothesis tests.
• Learn how a sample association is tested in hypothesis tests for correlation.
• Revisit sampling to understand why random and independent observations matter.
• Practise selecting an appropriate model in choosing a distribution.