Mathematics 9709 · Probability & Statistics 2

Hypothesis test for a mean

Is my sample mean surprising enough to doubt the claimed $\mu$ ? Stand the claim up as $H_0$ , measure how many standard errors $(\sigma/\sqrt n)$ your $\bar x$ sits from it — that's $z$ — and reject only if it lands past the critical cut-off.

z = (x̄ − μ₀) ÷ standard error

The standard error $\sigma/\sqrt n$ is how much a sample mean wobbles: averaging $n$ values tames the spread by $\sqrt n$ . State $H_0{:}\ \mu=\mu_0$ and a direction for $H_1$ , then read off the critical $z$ : 1.645 (one-tailed 5%), 1.960 (two-tailed 5%), 2.326 (one-tailed 1%). Use $\sigma$ if the variance is known, else $s$ from a large sample.

z = (x̄ − μ₀) ÷ SE, SE = σ/√n. Compare with the critical z; hypotheses are about μ, and the conclusion is non-definite and in context.

Worked example

Start simple: variance known, one-tailed

Cucumbers are $N(310,22^2)$ . A new feed is claimed to increase mass; a sample of $n=40$ gives $\bar x=316$ . Test at 5%.

$H_0{:}\ \mu=310$ , $H_1{:}\ \mu>310$ — one-tailed (looking only for an increase).
$z=\dfrac{316-310}{22/\sqrt{40}}=1.725$ ; critical value $1.645$ .
$1.725>1.645$ , so $\bar x$ is in the critical region.

Answer

z=1.725>1.645\ \Rightarrow\ \text{reject }H_0

There is evidence the feed increases mass.

Two-tailed, raw data, with assumptions

A two-tailed test asks whether $\mu$ differs either way, so the 5% is split and the cut-off is $\pm1.96$ . With raw data you first compute $\bar x$ , and you must state the assumptions.

Worked example

Two-tailed test from raw data

Candle burn times are normal with $\sigma=5.7$ ; the maker claims a mean of 250 min. Ten candles give 245, 247, 236, 255, 250, 239, 241, 252, 251, 243. Test the claim at 5%, stating any assumptions.

Assumptions: the sample is random (representative). Normality and $\sigma$ are given, so no further assumption is needed.
$\bar x=\tfrac{2459}{10}=245.9$ ; $H_0{:}\ \mu=250$ , $H_1{:}\ \mu\neq250$ , critical $\pm1.96$ .
$z=\dfrac{245.9-250}{5.7/\sqrt{10}}=-2.275$ .
$-2.275<-1.96$ , so it lands in the lower critical region.

Answer

z=-2.275<-1.96\ \Rightarrow\ \text{reject }H_0

There is evidence the mean burn time is not 250 minutes.

Variance unknown: a large sample lets you use s

If $\sigma$ is unknown but the sample is large ( $n\ge30$ ), estimate it: find $s^2$ first, then use $s/\sqrt n$ as the standard error.

Worked example

Variance unknown, borderline reject

42 students estimate “one minute”; $\sum x=2471,\ \sum x^2=146801$ (seconds). Test at 10% whether the mean is below 60.

$\bar x=\tfrac{2471}{42}=58.83$ ; $H_0{:}\ \mu=60$ , $H_1{:}\ \mu<60$ , critical $-1.282$ .
$s^2=\tfrac{1}{41}\big(146801-\tfrac{2471^2}{42}\big)=34.73$ .
$z=\dfrac{58.83-60}{\sqrt{34.73/42}}=-1.283$ ; just past $-1.282$ .

Answer

-1.283<-1.282\Rightarrow\text{(just) reject }H_0

When $z$ sits this close to the cut-off, advise that more data be collected — the verdict is fragile.

Variance unknown + large sample → estimate σ with s, SE = s/√n, then z exactly as before.

Your turn— tap to reveal the worked answer (borderline accept)

Pesto is $N(380,6.4)$ g; check at 5% whether the mean has reduced. Ten jars give $\bar x=378.7$ .

$H_1{:}\ \mu<380$ , $z=\dfrac{378.7-380}{\sqrt{6.4/10}}=-1.625$ vs $-1.645$ .

Answer

-1.625>-1.645\ \Rightarrow\ \text{do not reject }H_0

There is no evidence of a reduction.

Examiner note

Hypotheses about

\mu

, not

\bar x

; the right critical

z

for the tail; the unbiased

s^2

; a non-definite contextual conclusion (“evidence that …”, not “the mean is …”). (9709/62/O/N/24 Q7) (9709/62/M/J/22 Q2)

Common mistake

Stating hypotheses about

\bar x

instead of the parameter

\mu

; the wrong critical

z

(1.96 for a one-tailed 5% test); using

n

not

n-1

for

s^2

; forgetting to state the assumptions for raw-data tests; and a conclusion that isn't in context.

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Hypothesis test for a mean

z = (x̄ − μ₀) ÷ standard error

z = (x̄ − μ₀) ÷ SE, SE = σ/√n. Compare with the critical z; hypotheses are about μ, and the conclusion is non-definite and in context.

Worked example

Start simple: variance known, one-tailed

Cucumbers are $N(310,22^2)$ . A new feed is claimed to increase mass; a sample of $n=40$ gives $\bar x=316$ . Test at 5%.

$H_0{:}\ \mu=310$ , $H_1{:}\ \mu>310$ — one-tailed (looking only for an increase).
$z=\dfrac{316-310}{22/\sqrt{40}}=1.725$ ; critical value $1.645$ .
$1.725>1.645$ , so $\bar x$ is in the critical region.

Answer

z=1.725>1.645\ \Rightarrow\ \text{reject }H_0

There is evidence the feed increases mass.

Two-tailed, raw data, with assumptions

A two-tailed test asks whether $\mu$ differs either way, so the 5% is split and the cut-off is $\pm1.96$ . With raw data you first compute $\bar x$ , and you must state the assumptions.

Worked example

Two-tailed test from raw data

Candle burn times are normal with $\sigma=5.7$ ; the maker claims a mean of 250 min. Ten candles give 245, 247, 236, 255, 250, 239, 241, 252, 251, 243. Test the claim at 5%, stating any assumptions.

Assumptions: the sample is random (representative). Normality and $\sigma$ are given, so no further assumption is needed.
$\bar x=\tfrac{2459}{10}=245.9$ ; $H_0{:}\ \mu=250$ , $H_1{:}\ \mu\neq250$ , critical $\pm1.96$ .
$z=\dfrac{245.9-250}{5.7/\sqrt{10}}=-2.275$ .
$-2.275<-1.96$ , so it lands in the lower critical region.

Answer

z=-2.275<-1.96\ \Rightarrow\ \text{reject }H_0

There is evidence the mean burn time is not 250 minutes.

Variance unknown: a large sample lets you use s

If $\sigma$ is unknown but the sample is large ( $n\ge30$ ), estimate it: find $s^2$ first, then use $s/\sqrt n$ as the standard error.

Worked example

Variance unknown, borderline reject

42 students estimate “one minute”; $\sum x=2471,\ \sum x^2=146801$ (seconds). Test at 10% whether the mean is below 60.

$\bar x=\tfrac{2471}{42}=58.83$ ; $H_0{:}\ \mu=60$ , $H_1{:}\ \mu<60$ , critical $-1.282$ .
$s^2=\tfrac{1}{41}\big(146801-\tfrac{2471^2}{42}\big)=34.73$ .
$z=\dfrac{58.83-60}{\sqrt{34.73/42}}=-1.283$ ; just past $-1.282$ .

Answer

-1.283<-1.282\Rightarrow\text{(just) reject }H_0

When $z$ sits this close to the cut-off, advise that more data be collected — the verdict is fragile.

Variance unknown + large sample → estimate σ with s, SE = s/√n, then z exactly as before.

Your turn— tap to reveal the worked answer (borderline accept)

Pesto is $N(380,6.4)$ g; check at 5% whether the mean has reduced. Ten jars give $\bar x=378.7$ .

$H_1{:}\ \mu<380$ , $z=\dfrac{378.7-380}{\sqrt{6.4/10}}=-1.625$ vs $-1.645$ .

Answer

-1.625>-1.645\ \Rightarrow\ \text{do not reject }H_0

There is no evidence of a reduction.

Examiner note

Hypotheses about

\mu

, not

\bar x

; the right critical

z

for the tail; the unbiased

s^2

; a non-definite contextual conclusion (“evidence that …”, not “the mean is …”). (9709/62/O/N/24 Q7) (9709/62/M/J/22 Q2)

Common mistake

Stating hypotheses about

\bar x

instead of the parameter

\mu

; the wrong critical

z

(1.96 for a one-tailed 5% test); using

n

not

n-1

for

s^2

; forgetting to state the assumptions for raw-data tests; and a conclusion that isn't in context.