Hypothesis Testing for Normal Mean — Edexcel A-level Mathematics (Applied)

SPEC 5.3 • STATISTICAL HYPOTHESIS TESTING

Hypothesis Testing (Normal Distribution)

Testing claims about population means using z-tests when variance is known

Testing Population Means

When we have data from a normal distribution with known variance, we can test hypotheses about the population mean μ.

The Setup

Given:

X \sim N (μ, σ^{2})

where

σ^{2}

is known

We test whether μ equals a claimed value μ₀

Hypotheses

Null hypothesis:

H_{0} : μ = μ_{0}

Alternative hypothesis:

H_{1} : μ \neq = μ_{0}

(two-tailed)

H_{1} : μ > μ_{0}

(upper one-tailed)

H_{1} : μ < μ_{0}

(lower one-tailed)

Key Difference from Binomial

With normal distributions, we use a test statistic that follows a standard normal distribution, rather than calculating exact probabilities.

The Test Statistic

To conduct the test, we calculate a z-score that measures how far our sample mean is from the hypothesised mean.

Test Statistic Formula

z = \frac{x ˉ - μ _{0}}{\frac{σ}{n}}

SymbolMeaning

\overset{x}{ˉ}

Sample mean (observed value)

μ_{0}

Hypothesised population mean (from H₀)

σ

Population standard deviation (known)

n

Sample size

\frac{σ}{n}

Standard error of the mean

Critical

Under H₀, this test statistic follows

Z \sim N (0, 1)

. We compare our calculated z to critical values from the standard normal distribution.

Critical Values

The critical value is the z-score that marks the boundary of the rejection region. These values come from the standard normal distribution.

Standard Normal Critical Values

Significance LevelOne-TailedTwo-Tailed10%1.282±1.6455%1.645±1.9602.5%1.960±2.2401%2.326±2.576

Decision Rule

One-tailed (upper): Reject H₀ if z > critical value

One-tailed (lower): Reject H₀ if z < −critical value

Two-tailed: Reject H₀ if |z| > critical value

Worked Example: One-Tailed Upper01

A machine fills bottles with a mean of 500ml. The standard deviation is known to be 8ml. A sample of 25 bottles has a mean of 503.5ml. Test at the 5% level whether the machine is overfilling.

1

State hypotheses:

H_{0} : μ = 500

,

H_{1} : μ > 500

(overfilling = upper tail)

2

Calculate standard error:

SE = \frac{σ}{n} = \frac{8}{25} = \frac{8}{5} = 1.6

3

Calculate test statistic:

z = \frac{503.5 - 500}{1.6} = \frac{3.5}{1.6} = 2.1875

4

Critical value for 5% one-tailed:

z_{0.05} = 1.645

5

Compare: 2.1875 > 1.645, so reject H₀

Conclusion: There is significant evidence at the 5% level that the machine is overfilling bottles.

Worked Example: One-Tailed Lower02

The lifetime of a battery is normally distributed with σ = 12 hours. The manufacturer claims μ = 80 hours. A sample of 36 batteries has mean 76.5 hours. Test at 1% significance whether the true mean is less than claimed.

1

Hypotheses:

H_{0} : μ = 80

,

H_{1} : μ < 80

(less than = lower tail)

2

Standard error:

SE = \frac{12}{36} = \frac{12}{6} = 2

3

Test statistic:

z = \frac{76.5 - 80}{2} = \frac{- 3.5}{2} = - 1.75

4

Critical value for 1% lower-tailed:

- z_{0.01} = - 2.326

5

Compare: −1.75 > −2.326, so do not reject H₀

Conclusion: Insufficient evidence at 1% level to conclude the mean lifetime is less than 80 hours.

Two-Tailed Tests

When testing whether the mean has changed (in either direction), we use a two-tailed test.

Two-Tailed Setup

H_{0} : μ = μ_{0}

H_{1} : μ \neq = μ_{0}

The significance level α is split between both tails (α/2 in each tail).

Decision Rule

Reject H₀ if

∣ z ∣ > z_{α /2}

At 5% significance: Reject if |z| > 1.96

At 1% significance: Reject if |z| > 2.576

Exam Tip

For two-tailed tests, always use the absolute value |z| when comparing to the critical value, or equivalently check if z < −critical OR z > +critical.

Worked Example: Two-Tailed Test03

A production process has σ = 2.5g. The target weight is 50g. A sample of 40 items has mean 50.8g. At 5% significance, test whether the mean weight differs from the target.

1

Hypotheses:

H_{0} : μ = 50

,

H_{1} : μ \neq = 50

(differs = two-tailed)

2

Standard error:

SE = \frac{2.5}{40} = \frac{2.5}{6.325} = 0.3953

3

Test statistic:

z = \frac{50.8 - 50}{0.3953} = \frac{0.8}{0.3953} = 2.024

4

Critical value for 5% two-tailed:

z_{0.025} = 1.96

5

Compare: |2.024| = 2.024 > 1.96, so reject H₀

Conclusion: Significant evidence at 5% that the mean weight differs from the target of 50g.

Worked Example: Two-Tailed (Not Reject)04

IQ scores are normally distributed with σ = 15. A researcher believes the mean IQ has changed from 100. A sample of 50 people has mean 97.5. Test at 1% significance.

1

Hypotheses:

H_{0} : μ = 100

,

H_{1} : μ \neq = 100

2

Standard error:

SE = \frac{15}{50} = \frac{15}{7.071} = 2.121

3

Test statistic:

z = \frac{97.5 - 100}{2.121} = \frac{- 2.5}{2.121} = - 1.179

4

Critical value for 1% two-tailed:

z_{0.005} = 2.576

5

Compare: |−1.179| = 1.179 < 2.576, so do not reject H₀

Conclusion: Insufficient evidence at 1% level to conclude the mean IQ has changed from 100.

Critical Regions in Terms of $\overset{ˉ}{X}$

Instead of comparing z-values, we can express the critical region directly in terms of the sample mean

\overset{ˉ}{X}

.

Converting z-Critical to x̄-Critical

If the critical z-value is

z_{c}

, the corresponding critical value of

\overset{ˉ}{X}

is:

\overset{x}{ˉ}_{c} = μ_{0} + z_{c} \times \frac{σ}{n}

Critical Regions

Upper-tailed: Reject H₀ if

\overset{x}{ˉ} > \overset{x}{ˉ}_{c}

Lower-tailed: Reject H₀ if

\overset{x}{ˉ} < \overset{x}{ˉ}_{c}

Two-tailed: Reject H₀ if

\overset{x}{ˉ} < \overset{x}{ˉ}_{l o w er}

or

\overset{x}{ˉ} > \overset{x}{ˉ}_{u pp er}

Why Use This Method?

This approach is useful when asked to "find the critical region" - you give the answer in the units of the original data, making interpretation clearer in context.

05Worked Example: Finding Critical Region

A machine produces bolts with diameter D, where D is normally distributed with standard deviation 0.015 cm. The target mean diameter is 0.580 cm. After servicing, a sample of 50 bolts is taken to test if the mean has changed.

(a) Find, at the 1% level, the critical region for this test.

(b) The sample mean is 0.587 cm. Comment on this observation.

1

Hypotheses:

H_{0} : μ = 0.580

,

H_{1} : μ \neq = 0.580

(two-tailed)

2

Standard error:

SE = \frac{0.015}{50} = \frac{0.015}{7.071} = 0.002121

3

For 1% two-tailed: each tail has 0.5%, so

z_{0.005} = 2.5758

4

Lower critical value:

\overset{x}{ˉ}_{L} = 0.580 - 2.5758 \times 0.002121 = 0.580 - 0.00546 = 0.5745

5

Upper critical value:

\overset{x}{ˉ}_{U} = 0.580 + 2.5758 \times 0.002121 = 0.580 + 0.00546 = 0.5855

(a) Critical region:

\overset{ˉ}{D} \leq 0.575

or

\overset{ˉ}{D} \geq 0.585

(3 s.f.)

(b) The sample mean 0.587 cm falls inside the critical region (0.587 > 0.585), so there is sufficient evidence at 1% level that the mean diameter has changed from 0.580 cm.

P-Value Method

Instead of comparing the test statistic to critical values, we can calculate the p-value - the probability of observing a result as extreme as ours (or more extreme) if H₀ is true.

P-Value Calculation

Upper-tailed: p-value = P(Z > z)

Lower-tailed: p-value = P(Z < z)

Two-tailed: p-value = 2 × P(Z > |z|)

Decision Rule

Reject H₀ if p-value < α (significance level)

The smaller the p-value, the stronger the evidence against H₀.

Using Tables

Find P(Z < z) from standard normal tables, then:

P(Z > z) = 1 − P(Z < z)

For negative z: P(Z < z) = P(Z > |z|)

Worked Example: P-Value (Lower-Tailed)06

The mean reaction time for a task is normally distributed with μ = 0.8 seconds and σ = 0.15 seconds. After training, a sample of 20 people has mean reaction time 0.72 seconds. Using the p-value method, test at 5% whether training reduces reaction time.

1

Hypotheses:

H_{0} : μ = 0.8

,

H_{1} : μ < 0.8

(reduces = lower-tailed)

2

SE = \frac{0.15}{20} = 0.0335

,

z = \frac{0.72 - 0.8}{0.0335} = - 2.386

3

p-value = P(Z < −2.386) = P(Z > 2.386) = 1 − 0.9915 = 0.0085

4

Compare: 0.0085 < 0.05, so reject H₀

Conclusion: Significant evidence at 5% that training reduces reaction time (p = 0.0085).

Worked Example: P-Value (Two-Tailed)07

A pharmaceutical company claims a drug takes effect in 30 minutes on average, with σ = 5 minutes. A sample of 15 patients has mean 32.8 minutes. Find the p-value and test at 5% whether the mean differs from the claim.

1

Hypotheses:

H_{0} : μ = 30

,

H_{1} : μ \neq = 30

(two-tailed)

2

SE = \frac{5}{15} = 1.291

,

z = \frac{32.8 - 30}{1.291} = 2.169

3

P(Z > 2.169) = 1 − 0.9849 = 0.0151

4

p-value = 2 × 0.0151 = 0.0302

5

Compare: 0.0302 < 0.05, so reject H₀

Conclusion: Significant evidence at 5% that the mean time differs from 30 minutes (p = 0.0302).

Single Observation vs Sample Mean

Be clear about what you're testing:

Single Observation X

If testing one value:

z = \frac{x - μ _{0}}{σ}

Sample Mean X̄

If testing mean of n values:

z = \frac{x ˉ - μ _{0}}{σ / n}

Common Error

When given a sample mean, students often forget to divide σ by √n. The standard error σ/√n is SMALLER than σ, making the test more sensitive.

Why the Difference?

Sample means are less variable than individual observations. A sample mean of 503.5 from 25 bottles is much more significant than a single bottle being 503.5ml.

Worked Example: Single Observation08

Heights of adults follow N(170, 64) cm². A person claims their height of 185cm is unusually tall. Test at 5% significance (one-tailed).

1

Note: σ² = 64, so σ = 8. This is a single observation (n = 1).

2

Hypotheses:

H_{0} : μ = 170

,

H_{1} : μ > 170

(unusually tall)

3

Test statistic:

z = \frac{185 - 170}{8} = \frac{15}{8} = 1.875

4

Critical value: z₀.₀₅ = 1.645

5

Compare: 1.875 > 1.645, so reject H₀

Conclusion: There is significant evidence that 185cm is unusually tall (in the upper 5% of the population).

Link to Confidence Intervals

There is a direct connection between hypothesis tests and confidence intervals:

The Relationship

For a two-tailed test at significance level α:

Reject H₀: μ = μ₀ if and only if μ₀ lies outside the (1−α)×100% confidence interval for μ

Practical Use

If you've already calculated a 95% confidence interval for μ:

If μ₀ is outside the CI → Reject H₀ at 5% level

If μ₀ is inside the CI → Do not reject H₀ at 5% level

Example

95% CI for μ is (48.2, 52.4).

Test H₀: μ = 50 at 5% → 50 is inside CI → Do not reject

Test H₀: μ = 47 at 5% → 47 is outside CI → Reject

Worked Example: CI Method09

A sample of 30 items has x̄ = 45.2, σ = 6. (a) Find a 95% CI for μ. (b) Test H₀: μ = 43 at 5% using the CI.

1

Standard error:

SE = \frac{6}{30} = 1.095

2

95% CI:

\overset{x}{ˉ} \pm 1.96 \times SE = 45.2 \pm 1.96 \times 1.095

3

CI: 45.2 ± 2.15 = (43.05, 47.35)

4

Is μ₀ = 43 inside the CI? No, 43 < 43.05

Conclusion: Since 43 lies outside the 95% CI, we reject H₀: μ = 43 at the 5% significance level.

10

Worked Example: Full Context

A company claims delivery time is normally distributed with mean 3 days and σ = 0.5 days. After complaints, 20 deliveries averaged 3.3 days. At 5%, is there evidence delivery has slowed?

1

"Slowed" means increased time → upper-tailed test

2

H_{0} : μ = 3

,

H_{1} : μ > 3

3

SE = \frac{0.5}{20} = 0.1118

4

z = \frac{3.3 - 3}{0.1118} = \frac{0.3}{0.1118} = 2.683

5

Critical value: 1.645. Since 2.683 > 1.645, reject H₀.

Conclusion: There is significant evidence at the 5% level that delivery times have increased. The company should investigate the complaints.

Practice

Question 1: A machine produces rods with σ = 0.4mm. The target length is 20mm. A sample of 16 rods has mean 20.25mm. Test at 5% whether the mean differs from target.

H₀: μ = 20, H₁: μ ≠ 20 (two-tailed)

SE = 0.4/√16 = 0.1

z = (20.25 − 20)/0.1 = 2.5

Critical value (5%, two-tailed): 1.96

|2.5| > 1.96, so reject H₀

There is significant evidence that the mean rod length differs from 20mm.

Practice

Question 2: Exam scores are N(μ, 225). A teacher believes her class performs above average. Her 25 students score a mean of 78, compared to the population mean of 72. Test at 1% significance.

H₀: μ = 72, H₁: μ > 72 (upper-tailed)

σ = √225 = 15, SE = 15/√25 = 3

z = (78 − 72)/3 = 2

Critical value (1%, one-tailed): 2.326

2 < 2.326, so do not reject H₀

Insufficient evidence at 1% that the class performs above average.

Practice

Question 3: For the data in Practice 2, find the p-value for the test. Would the conclusion change at 5% significance?

z = 2 (from Practice 2)

p-value = P(Z > 2) = 1 − P(Z < 2) = 1 − 0.9772 = 0.0228

At 1%: 0.0228 > 0.01 → Do not reject (as before)

At 5%: 0.0228 < 0.05 → Reject H₀

Yes, at 5% we would conclude the class performs above average.

Summary

Key Formulas

ComponentFormula

Test statistic (sample mean)

z = \frac{x ˉ - μ _{0}}{σ / n}

Test statistic (single value)

z = \frac{x - μ _{0}}{σ}

Standard error

SE = σ / n

Critical Values to Know

LevelOne-TailedTwo-Tailed5%1.6451.961%2.3262.576

Exam Checklist

State hypotheses clearly with context

Use σ/√n for sample means (not just σ)

Choose correct tail based on H₁

State conclusion in context

Hypothesis Testing (Normal Distribution)

Testing Population Means

The Test Statistic

Critical Values

Two-Tailed Tests

Critical Regions in Terms of Xˉ

05Worked Example: Finding Critical Region

P-Value Method

Single Observation vs Sample Mean

Link to Confidence Intervals

Practice

Practice

Practice

Summary

Critical Regions in Terms of $\overset{ˉ}{X}$