IB Mathematics AI AHL Test for proportion MAI Study Notes - New Syllabus

IB Mathematics AI AHL Test for proportion MAI Study Notes

LEARNING OBJECTIVE

Critical values and critical regions.

Key Concepts:

Critical values and critical regions.
Test for population mean for normal/Poisson/Binomial distribution.
Hypothesis testing
Type I and II errors

MAI HL and SL Notes – All topics

Critical Values

Boundaries that define the critical (rejection) region in a distribution.

Depend on:

Significance level $(α)$
One-tailed or two-tailed test
Standard Normal Distribution $(z):$

Two-tailed test

$α = 0.05 →$ Critical values: $±1.96$

One-tailed test

$α = 0.05 →$ Critical value: $±1.645$

Critical Region

Range of test statistic values that lead to rejecting the null hypothesis (H₀).

Determined by:

Critical values

Area of critical region $= α$

Cases of $H_1$

$μ<μ_0$
The red area $X<r$

$μ>μ-0$
The red area $X>r$

$μ\neq μ_0$
The red area $X<r$ or $X>s$

We will call the remaining region non-critical.

Example

For a sample of $ n = 40 $, we know $ \bar{x} = 23 $. We also know that the population standard deviation is $ \sigma = 2.8 $.

For the test:

$ H_0: \mu = 24 $
$ H_1: \mu \ne 24 $
$ \alpha = 0.05 $

Find the critical region and the non-critical region.

▶️Answer/Explanation

Solution:

(a) Critical and Non-Critical Region:

Sampling distribution under $ H_0 $: $ N(24, \left(\frac{2.8}{\sqrt{40}}\right)^2) $
Standard deviation: $ \frac{2.8}{\sqrt{40}} \approx 0.4425 $
Using inverse normal (invN) for 2-tailed test at $ \alpha = 0.05 $:
Critical Region: $ (-\infty, 23.1) \cup (24.9, +\infty) $
Non-Critical Region: $ [23.1, 24.9] $

Z-Test (Normal Distribution)

Use when $σ$ is known
Applicable regardless of sample size

Formula:

$
z = \frac{\bar{x} – \mu_0}{\sigma / \sqrt{n}}
$

Example

For a sample of $ n = 40 $ data (normally distributed), we know that:
$ \bar{x} = 23 $, $ s_{n-1} = 3 $
The population standard deviation is $ \sigma = 2.8 $.

There is a CLAIM that $ \mu = 24 $.

Conduct a two-tail test for this claim with $ \alpha = 0.05 $
Conduct a one-tail test for this claim (against $ \mu < 24 $)

Use the significance level $ \alpha = 0.05 $

▶️Answer/Explanation

Solution

Since we know $ \sigma $, we use the Z-test (note: $ s_{n-1} = 3 $ was not necessary).

Use GDC: Statistics → TEST → Z – 1 SAMPLE → Data: Variable

Part	Hypotheses	p-value	Decision	Conclusion
(a) Two-tail test	$ H_0: \mu = 24 $ $ H_1: \mu \ne 24 $	0.0239	Reject $ H_0 $	$ \mu \ne 24 $
(b) One-tail test	$ H_0: \mu = 24 $ $ H_1: \mu < 24 $	0.0119	Reject $ H_0 $	$ \mu < 24 $

T-Test (t-Distribution)

Use when $σ$ is unknown
Works for any sample size, but especially small samples $(n ≤ 30)$

Formula:

$
t = \frac{\bar{x} – \mu_0}{s / \sqrt{n}}
$

Note:

Use z-test when σ is known, $t-$test when $σ$ is unknown.

Example

For a sample of $ n = 20 $ data, we know that:
$ \bar{x} = 23 $, $ s_{n-1} = 3 $

There is a CLAIM that $ \mu = 24 $.

Conduct a two-tail test for this claim with $ \alpha = 0.05 $
Conduct a one-tail test for this claim (against $ \mu < 24 $)

Use the significance level $ \alpha = 0.05 $

▶️Answer/Explanation

Solution

Since we don’t know $ \sigma $, we use the t-test.

Use GDC: Statistics → TEST → t – 1 SAMPLE → Data: Variable

Part	Hypotheses	p-value	Decision	Conclusion
(a) Two-tail test	$ H_0: \mu = 24 $ $ H_1: \mu \ne 24 $	0.152	Do not reject $ H_0 $	Not enough evidence to say $ \mu \ne 24 $
(b) One-tail test	$ H_0: \mu = 24 $ $ H_1: \mu < 24 $	0.0762	Do not reject $ H_0 $	Not enough evidence to say $ \mu < 24 $

Unpaired Samples

Independent groups
Example: Two different classes

Paired Samples

Dependent observations (e.g., before/after)
Analyze the difference
Example: Weight before and after treatment → Paired t-test

Approach	Design	Type of Data	Appropriate Test
Unpaired Samples	Two different classes of students are randomly assigned: • Class 1 uses Method A • Class 2 uses Method B After teaching, both take the same math test.	Independent groups – different students in each group.	Use an unpaired (independent) t-test Because the groups are unrelated and sampled independently.
Paired Samples	The same group of students: • First learns with Method A and takes a test • Then learns with Method B and takes another test	Dependent data – two scores per student (A vs B)	Use a paired t-test Because we compare the same individuals under both conditions.

Test for Proportion (Binomial Distribution)

Use when outcomes are binary (success/failure)

Test Statistic:

$
z = \frac{\hat{p} – p_0}{\sqrt{\frac{p_0(1 – p_0)}{n}}}
$

$\hat{p}$: sample proportion
$p_0$: hypothesized proportion
$n$: sample size

Example:

Claim: 80% satisfaction → In a sample of 100, only 75 satisfied → Test with binomial distribution

Example

In a sample of 200 people, there are 50 smokers. That is:
$ n = 200 $
$ x = 50 $
Observed proportion: $ \hat{p} = \frac{50}{200} = 0.25 $

For the test:

$ H_0: p = 0.30 $
$ H_1: p < 0.30 $ with $ \alpha = 0.05 $

Find the critical region and the non-critical region.

▶️Answer/Explanation

Solution

We use the distribution $ B(200, 0.30) $
For the critical region, we find when $ P(X \le r) < 0.05 $.
By trial and error, we find that $ r = 48 $, since $ P(X \le 48) = 0.035 $

Critical Region :$ X \le 48 $

Non-Critical Region : $ X \ge 49 $

Test for Mean Using Poisson Distribution

Use for count data (e.g., defects)

Test Statistic:

$
z = \frac{\bar{x} – \lambda_0}{\sqrt{\lambda_0 / n}}
$

$\bar{x}$: sample mean
$\lambda_0$: hypothesized mean
$n$: number of observations

Example:

Old rate = 2/day, after 10 days observe 15 → Use Poisson test to evaluate change

Testing Correlation Coefficient

Hypothesis:

$ρ = 0$ (no linear correlation)

Test Statistic:

$
t = \frac{r \sqrt{n – 2}}{\sqrt{1 – r^2}} \quad \text{with } (n – 2) \text{ degrees of freedom}
$

Purpose:

Test if there is a significant linear relationship between two variables

Example

The number of accidents per day in a certain area follows a Poisson distribution.
We record the number of accidents in 5 different days:

Day 1	Day 2	Day 3	Day 4	Day 5
7	8	8	6	10

The statistic is $ \sum X_i = 39 $

For the test:
$ H_0: \lambda = 35 $
$ H_1: \lambda > 35 $, with $ \alpha = 0.10 $

Find the critical region and the non-critical region.

▶️Answer/Explanation

Solution

We use the Poisson distribution $ \text{Po}(35) $
For the critical region, we find when $ P(X \ge r) < 0.10 $
By trial and error, $ r = 44 $ with $ P(X \ge 44) = 0.079 $

Critical Region : $ X \ge 44 $

Non-Critical Region : $ X \le 43 $

Type I

Rejecting $H_0$ when it’s true $α$ (Significance level)

Type II

Failing to reject H₀ when it’s false $β $

Note:

Type I → False positive
Type II → False negative

Probability of Type II Error:

$
\beta = P\left(Z < \frac{z_{\text{crit}} – (\mu – \mu_0)}{\sigma / \sqrt{n}}\right)
$

Example:

For a sample of $ n = 40 $, we know $ \bar{x} = 23 $. We also know that the population standard deviation is $ \sigma = 2.8 $.

For the test:

$ H_0: \mu = 24 $
$ H_1: \mu \ne 24 $
$ \alpha = 0.05 $

State the TYPE I error.
It was finally found that $ \mu = 22.9 $. Find the TYPE II error.

▶️Answer/Explanation

Solution:

(a) Type I Error:

This is the probability of rejecting $ H_0 $ when $ H_0 $ is true. Thus,
Type I Error = $ \alpha = 0.05 $

(b) Type II Error:

Suppose the true mean is $ \mu = 22.9 $
We now use $ N(22.9, \left(\frac{2.8}{\sqrt{40}}\right)^2) $
Compute $ P(23.1 < X < 24.9) $ using normal CDF (Ncd)
Type II Error = 0.326

Example:

Context: Heights of 5^th graders around the world follow $ X \sim N(142, 5^2) $.

You believe students at your school are taller, so you collect a sample of $ n = 32 $ students.

After conducting a hypothesis test at a 5% significance level, the true mean is later revealed to be $ \mu = 146 $. Calculate $ P(\text{Type II Error}) $.

▶️ Solution/Explanation

Hypotheses

$ H_0: \mu = 142 \quad \text{(normal population)} \\ H_1: \mu > 142 \quad \text{(taller than normal)} $

Sampling Distribution under $ H_0 $

The sample mean $ \bar{X} \sim N\left(142, \frac{25}{32} \right) $, since $ \sigma^2 = 5^2 = 25 $.

Standard error: $ \text{SE} = \sqrt{\frac{25}{32}} \approx 0.990 $

Step 3: Critical Value (C.V.) for 5% significance level (1-tailed test)

We find the value of $ \bar{x} $ such that: $ P(\bar{X} \leq \text{C.V.}) = 0.95 \quad \text{(because upper 5% is rejection region)} $ Using: $ \text{C.V.} = \text{invNorm}(0.95, 142, 0.990) \approx 143.74 $

Type II Error

Type II error is the probability of failing to reject $ H_0 $ when $ \mu = 146 $ is true:

Under $ \mu = 146 $, the distribution of $ \bar{X} \sim N(146, 0.990) $.

So, $ P(\text{Type II Error}) = P(\bar{X} \leq 143.74 \mid \mu = 146) = \text{normalcdf}(-\infty, 143.74, 146, 0.990) \approx 0.0166 $

Conclusion:

There is a 1.66% chance of failing to detect that the students are taller than normal. This is the probability of a Type II error for a sample of size 32.

IB only requires one-tailed tests for discrete distributions (binomial, Poisson). Define critical region to maximize Type I error without exceeding $α$

Assumptions for Hypothesis Tests

Before performing a hypothesis test, certain assumptions must be satisfied to ensure the validity of the results.

Z-Test Assumptions

The population is normally distributed, or the sample size is large enough (typically $n ≥ 30$) for the Central Limit Theorem to apply.
The population standard deviation ($σ$) is known.
The data points are independent (no dependence between observations).

T-Test Assumptions

The population is normally distributed — this is especially important when the sample size is small (typically $n ≤ 30$).
The population standard deviation (σ) is unknown, and is estimated using the sample standard deviation (s).
The data points are independent.

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IB Mathematics AI AHL Test for proportion MAI Study Notes - New Syllabus

IB Mathematics AI AHL Test for proportion MAI Study Notes

CRITICAL VALUES AND REGIONS

USE OF NORMAL AND T-DISTRIBUTION

PAIRED VS. UNPAIRED SAMPLES

TEST FOR PROPORTION

TYPE I AND TYPE II ERRORS

DISCRETE DISTRIBUTIONS AND ONE-TAILED TESTS

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