Mini-Lesson

Hypothesis testing

2 missions. Guess, learn, then check. Your place is saved.

Mission 1A coin lands heads 15 times in 20 throws. Is it biased?
Mission 2Bags should average 500 g. Thirty bags average 497 g. Suspicious?

The specification points and sources are listed under the lesson: see Sources.

Mission 1

A coin lands heads 15 times in 20 throws. Is it biased?

Make a guess: A coin lands heads 15 times in 20 throws. Is it biased?

Hypothesis testing · binomial tests

The language of a hypothesis test

The null hypothesis $H_{0}$ H₀ states a value of the population parameter, e.g. $H_{0} : p = 0.5$ H₀colon p = 0.5. The alternative hypothesis $H_{1}$ H₁ says how it may differ: $p > 0.5$ p > 0.5 or $p < 0.5$ p < 0.5 (1-tail), or $p \neq = 0.5$ p ≠ 0.5 (2-tail).
Define $p$ p in context, e.g. " $p$ p is the probability that the coin lands heads".
The test statistic is the sample count, $X \sim B (n, p)$ X ~ B(n, p) assuming $H_{0}$ H₀ is true.
The significance level is the cut-off probability for "too unlikely if $H_{0}$ H₀ is true", e.g. 5%.

Quick check

?A dice should show a six with probability

\frac{1}{6}

. Jo suspects her dice shows a six less often than this, and rolls it 40 times. Which hypotheses should she test?

Worked example Hypothesis testing · binomial tests

A 1-tail test using a p-value

Question

A sweet maker says 30% of its sweets are red. Asha thinks the proportion is higher. A random bag of 25 sweets has 12 red. Test Asha's claim at the 5% level.

1
State hypotheses: $H_{0} : p = 0.3$ H₀colon p = 0.3, $H_{1} : p > 0.3$ H₁colon p > 0.3, where $p$ p is the proportion of red sweets
2
Model under $H_{0}$ H₀: $X \sim B (25, 0.3)$ X ~ B(25, 0.3), where $X$ X is the number of red sweets
3
Find the probability of 12 or more: $P (X \geq 12) = 1 - P (X \leq 11) = 1 - 0.9558 = 0.0442$ P(X ≥ 12) = 1 − P(X ≤ 11) = 1 − 0.9558 = 0.0442
4
Compare with the level: $0.0442 < 0.05$ 0.0442 < 0.05, so the result is significant: reject $H_{0}$ H₀

Answer

There is sufficient evidence at the 5% level to reject

H_{0}

. It is likely that more than 30% of the sweets are red.

⚠️ Watch out

the p-value is

P (X \geq 12)

, not

P (X = 12) = 0.0268

. You need the result or anything more extreme.

Hypothesis testing · binomial tests

Critical regions and the actual significance level

The critical region is the set of values that lead you to reject $H_{0}$ H₀; a critical value is its boundary value (a 2-tail test has two). The rest is the acceptance region.
For a 2-tail test at 5%, each tail has probability at most 2.5%. Compare $x$ x with $n p$ np, the expected number of successes, to see which tail it is in.
For discrete counts the actual significance level, $P (X in critical region)$ P(X in critical region), is usually below the stated level.
The significance level is the probability of incorrectly rejecting $H_{0}$ H₀.

Worked example Hypothesis testing · binomial tests

A 2-tail critical region

Question

X \sim B (30, 0.4)

. Find the critical region for a test of

H_{0} : p = 0.4

against

H_{1} : p \neq = 0.4

at the 5% level, and its actual significance level.

1
Each tail may have at most 0.025
2
Lower tail, using cumulative probabilities: $P (X \leq 6) = 0.0172$ P(X ≤ 6) = 0.0172 but $P (X \leq 7) = 0.0435$ P(X ≤ 7) = 0.0435, so $X \leq 6$ X ≤ 6
3
Upper tail: $P (X \geq 18) = 1 - P (X \leq 17) = 0.0212$ P(X ≥ 18) = 1 − P(X ≤ 17) = 0.0212 but $P (X \geq 17) = 0.0481$ P(X ≥ 17) = 0.0481, so $X \geq 18$ X ≥ 18
4
Add the two tails: $0.0172 + 0.0212 = 0.0384$ 0.0172 + 0.0212 = 0.0384

Answer

critical region

X \leq 6

or

X \geq 18

; actual significance level 0.0384 (3.84%)

⚠️ Watch out

P (X \geq 18)

is

1 - P (X \leq 17)

. Using

1 - P (X \leq 18)

gives the probability of 19 or more.

Calculate

?A website says 35% of visitors buy something. After a redesign, the manager believes the proportion has increased. She will test

H_{0} : p = 0.35

against

H_{1} : p > 0.35

at the 5% level using a random sample of 30 visitors. Find the critical value.

Calculate

?For the test in the last question, find the actual significance level. Give your answer to 3 significant figures.

Sort it

?Is each conclusion well written?

Tap a statement, then tap the group it belongs in. Then press Check.

Quick check

?A test is carried out at the 5% significance level. What does the 5% represent?

Mission check

?A coin lands heads 15 times in 20 throws. Test at the 5% level whether it is biased.

Mission 2

Bags should average 500 g. Thirty bags average 497 g. Suspicious?

Make a guess: Bags should average 500 g. Thirty bags average 497 g. Suspicious?

Hypothesis testing · correlation and the normal mean

The sample mean as a random variable (A-level only)

Each random sample of size $n$ n gives a different mean, so the sample mean $X ˉ$ X̄ is a random variable.
If $X \sim N (μ, σ^{2})$ X ~ N(μ, σ²), then

X ˉ \sim N (μ, \frac{σ ^{2}}{n})

Its standard deviation is $\frac{σ}{n}$ σ/(√n): averages vary much less than single values.
To test a mean, state $H_{0} : μ = μ_{0}$ H₀colon μ = μ₀ and $H_{1}$ H₁ in terms of $μ$ μ, then standardise:

Z = \frac{X ˉ - μ _{0}}{σ / n} \sim N (0, 1)

Quick check

?

X \sim N (80, 1 5^{2})

. A random sample of 25 values of

X

is taken. What is the distribution of the sample mean

X ˉ

?

Worked example Hypothesis testing · correlation and the normal mean

A test for a normal mean

Question

Journey times on a route are normally distributed with mean 42 minutes and standard deviation 6 minutes. After roadworks, a random sample of 16 journeys has mean 44.1 minutes. Assuming the standard deviation is unchanged, test at the 5% level whether the mean journey time has increased.

1
State hypotheses: $H_{0} : μ = 42$ H₀colon μ = 42, $H_{1} : μ > 42$ H₁colon μ > 42, where $μ$ μ is the mean journey time in minutes
2
Under $H_{0}$ H₀: $X ˉ \sim N (42, \frac{6 ^{2}}{16})$ X̄ ~ N(42, 6²/16), so the standard deviation of $X ˉ$ X̄ is $\frac{6}{4} = 1.5$ 6/4 = 1.5
3
Find the p-value: $P (X ˉ \geq 44.1) = P (Z \geq 1.4) = 0.0808$ P(X̄ ≥ 44.1) = P(Z ≥ 1.4) = 0.0808
4
Compare: $0.0808 > 0.05$ 0.0808 > 0.05, so the result is not significant: do not reject $H_{0}$ H₀

Answer

There is insufficient evidence at the 5% level that the mean journey time has increased.

⚠️ Watch out

using

σ = 6

instead of

\frac{6}{16}

gives

z = 0.35

. The test is about the mean of 16 journeys, not one journey.

Calculate

?The volumes of drinks dispensed by a machine are normally distributed with standard deviation 12 ml. To test

H_{0} : μ = 250

against

H_{1} : μ \neq = 250

at the 5% level, a random sample of 36 drinks is used. Find the upper critical value for

x ˉ

. Give your answer to 2 decimal places.

ml

Calculate

?Bags of compost are filled with masses that are normally distributed with standard deviation 0.5 kg. A supplier claims the mean is 20 kg. A random sample of 10 bags has mean 19.7 kg. Find the p-value for a test of

H_{0} : μ = 20

against

H_{1} : μ < 20

. Give your answer to 3 significant figures.

Hypothesis testing · correlation and the normal mean

Testing for correlation (A-level only)

The product moment correlation coefficient (PMCC) $r$ r measures how close points lie to a straight line: $- 1 \leq r \leq 1$ −1 ≤ r ≤ 1; $r = \pm 1$ r = ± 1 means all on a line.
$r$ r is the sample value; $ρ$ ρ the population value. Test $H_{0} : ρ = 0$ H₀colon ρ = 0 against $H_{1} : ρ > 0$ H₁colon ρ > 0, $ρ < 0$ ρ < 0 or $ρ \neq = 0$ ρ ≠ 0.
Compare $∣ r ∣$ |r| with the tabled critical value for $n$ n. Reject $H_{0}$ H₀ if $∣ r ∣ \geq$ |r| ≥ critical value (sign matching $H_{1}$ H₁), or if the p-value is at most the level.
The data are assumed to be from a bivariate normal distribution.

Worked example Hypothesis testing · correlation and the normal mean

A 1-tail correlation test

Question

On 12 randomly chosen days, a garden records hours of sunshine and number of visitors. The PMCC is

r = 0.53

. The critical value for

n = 12

at the 5% level (1-tail) is 0.4973. Test at the 5% level whether there is positive correlation.

1
State hypotheses: $H_{0} : ρ = 0$ H₀colon ρ = 0, $H_{1} : ρ > 0$ H₁colon ρ > 0, where $ρ$ ρ is the population PMCC between sunshine and visitors
2
Compare: $0.53 > 0.4973$ 0.53 > 0.4973, so $r$ r is in the critical region
3
Reject $H_{0}$ H₀

Answer

There is evidence at the 5% level of positive correlation between hours of sunshine and number of visitors.

⚠️ Watch out

correlation is not causation. The conclusion is about a linear relationship, not about sunshine causing visits.

Quick check

?For a sample of 20 pairs,

r = - 0.41

. Tables give critical values for

n = 20

of 0.3783 (5% 1-tail, 10% 2-tail) and 0.4438 (2.5% 1-tail, 5% 2-tail). Test

H_{0} : ρ = 0

against

H_{1} : ρ \neq = 0

at the 5% level.

Sort it

?Which hypothesis test fits each question?

Tap a statement, then tap the group it belongs in. Then press Check.

Mission check

?Bag masses have standard deviation 8 g. A sample of 30 bags averages 497 g. Is the factory's 500 g mean suspicious at the 5% level?

Recap

Big ideas

State $H_{0}$ H₀ and $H_{1}$ H₁ about the population parameter $p$ p (defined in context); the test statistic $X \sim B (n, p)$ X ~ B(n, p) assumes $H_{0}$ H₀.
Reject $H_{0}$ H₀ if the p-value is at most the significance level, or if $x$ x is in the critical region. For a 2-tail test, halve the level for each tail.
Conclude in context and without certainty: "there is evidence at the 5% level that …". The significance level is the probability of incorrectly rejecting $H_{0}$ H₀.
If $X \sim N (μ, σ^{2})$ X ~ N(μ, σ²) then $X ˉ \sim N (μ, \frac{σ ^{2}}{n})$ X̄ ~ N(μ, σ²/n); test a mean with $Z = \frac{X ˉ - μ _{0}}{σ / n}$ Z = (X̄ − μ₀)/(σ/√n).
Test correlation with $H_{0} : ρ = 0$ H₀colon ρ = 0; compare $∣ r ∣$ |r| with the tabled critical value for $n$ n (or use the p-value), choosing the 1-tail or 2-tail column.
Conclude in context, without certainty, and remember that correlation is not causation.

🏆

Mini-lesson complete!

⭐⭐⭐

You've worked through Hypothesis testing for OCR A-level Mathematics A (H240). 🎉

Mission 1A coin lands heads 15 times in 20 throws. Is it biased?
Mission 2Bags should average 500 g. Thirty bags average 497 g. Suspicious?

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Specification points and sources for this lesson are listed at the bottom of the page.