IB Mathematics AI SL Formulation of null and alternative hypotheses MAI Study Notes- New Syllabus

 Formulation of Hypotheses

Hypothesis testing allows us to assess claims about a population using sample data.

Null hypothesis: $H_0$
Alternative hypothesis: $H_1$ or $H_a$

Types of Hypotheses

One-tailed test:

$H_1: \mu >\mu_0$ (right-tailed)
$H_1: \mu < \mu_0$ (left-tailed)

Two-tailed test:

$H_1: \mu \ne \mu_0$

Example:

A researcher claims the average weight of a population is 75 kg. A student believes this claim is incorrect and takes a random sample of 40 people. The sample has:

Solution:

Null Hypothesis $H_0$: $\mu = 75$
Alternative Hypothesis $H_1$: $\mu \ne 75$
→ This is a two-tailed test, since we are checking if the mean is not equal to 75.

$z = \frac{\bar{x} – \mu_0}{\sigma / \sqrt{n}} = \frac{72.5 – 75}{6 / \sqrt{40}} = \frac{-2.5}{0.9487} \approx -2.635$

For a two-tailed test at $\alpha = 0.05$, the critical values are:

$z = \pm 1.96$

$-2.635 < -1.96$, the test statistic falls in the rejection region.

At the 5% level of significance, there is sufficient evidence to conclude that the population mean is not equal to 75 kg.

Significance Levels and p-values

Significance Level (α):

Common values: $0.05 (5\%), 0.01 (1\%), 0.10 (10\%)$

Defines the cutoff for rejecting $H_0$.

 p-value:

The probability of observing a result as extreme or more extreme than the actual sample result, assuming H₀ is true.

Interpretation:

If $p\text{-value} < \alpha$: Reject $H_0$ $\Rightarrow$ evidence supports $H_1$.
 If $p\text{-value} \geq \alpha$: Fail to reject $H_0$ $\Rightarrow$ insufficient evidence to support $H_1$.

 Calculator Use (TI or GDC):

Use $\text{normalcdf}$ or $\text{tcdf}$ functions to calculate p-values based on the test statistic.

Example:

To compare the mean weights between two populations A and B we obtain two samples:

(GDC gives that the two sample means are $\bar{x}_1 = 70.1$ and $\bar{x}_2 = 63.1$)

We will test two different claims for the population means $\mu_1, \mu_2$ with $\alpha = 0.05$:

(a) Ann claims that $\mu_1 > \mu_2$

(b) Bill claims that population means are different

Solution

(a)

We perform a one-tailed t-test:

$
\begin{aligned}
H_0&: \mu_1 = \mu_2 \\
H_1&: \mu_1 > \mu_2
\end{aligned}
$

GDC gives p-value = 0.041

Since

$
\textbf{p-value < 0.05}
$

we reject $H_0$.
That is, we accept Ann’s claim that $\mu_1 > \mu_2$.

(b)

We perform a two-tailed t-test:

$
\begin{aligned}
H_0&: \mu_1 = \mu_2 \\
H_1&: \mu_1 \ne \mu_2
\end{aligned}
$

GDC gives p-value = 0.082

Since

$
\textbf{p-value > 0.05}
$

we do not have enough evidence to reject $H_0$.
Bill is not right!

Expected vs. Observed Frequencies

Observed Frequency (O): data from the sample
Expected Frequency (E): predicted assuming $H_0$ is true

$
E = \frac{(\text{row total}) \times (\text{column total})}{\text{grand total}}
$

Example:

In a survey of 80 people, we inquired about their preferred sport

Solution:

Observed frequencies

 The expected frequencies are

 

For the first entry:

$\frac{(\text{column 1}) \times (\text{row 1})}{\text{TOTAL}} = \frac{30 \times 36}{80} = 13.5$

 Chi-Square Test for Independence

Test whether two categorical variables are independent.

Test Statistic:

$
\chi^2 = \sum \frac{(O – E)^2}{E}
$

Degrees of Freedom:

$
\text{df} = (r – 1)(c – 1)
$

Conditions:

• Expected frequencies $\geq 5$
• Random sample

EXAMPLE 1

In a survey of 80 people, we inquired about their preferred sport

Test if the favorite sport is independent of the gender. Use the significance level a=0.05.

Solution

$H_0$: gender and favorite sport are independent
$H_1$: gender and favorite sport are not independent
GDC gives

$\chi^2\text{statistic} = 7.00$
$p\text{-value} = 0.0301$
degrees of freedom = 2
Expected frequencies are automatically entered in Matrix B

Since $p\text{-value} < 0.05$ we reject $H_0$.
Thus, gender and favorite sport are not independent.

Chi-Square Goodness of Fit Test

Test if a sample distribution matches a theoretical distribution (e.g., uniform).

Test Statistic:

$
\chi^2 = \sum \frac{(O – E)^2}{E}
$

Degrees of Freedom:

$
\text{df} = k – 1
$

Where $k$ = number of categories

EXAMPLE 1
Philipp claims that the supporters of Football teams A, B, C and D are as follows

$\rm A$	$\rm B$	$\rm C$	$\rm D$
$30\%$	$30\%$	$26\%$	$14\%$

In a sample of 40 people we found

$\rm A$	$\rm B$	$\rm C$	$\rm D$
$11$	$13$	$10$	$6$

We test Phillip’s claim for $\alpha = 0.05$

Solution

Goodness of fit Chi-squared test with

$H_0$: data follow the given distribution
$H_1$: data do not follow the given distribution

$0.30$

$0.30$

$0.26$

$0.14$

observed frequencies in List 1,
probabilities in List 2
expected frequencies in List 3. (multiply List 3 by 40)

List 1	List 2	List 3
$11$	$0.30$	$12$
$13$	$0.30$	$12$
$10$	$0.26$	$10.4$
$6$	$0.14$	$5.6$

Use $\text{GDC: Statistics – TEST – CHI – GOF}$
Observed: List 1
Expected: List 3
$d.f.= k-1 = 3$

GDC gives

$\chi^2\text{statistic} = 0.210$
$p\text{-value} = 0.976$

Since $p\text{-value} > 0.05$, we do not have enough evidence to reject $H_0$. We may accept that Philipp’s claim about the distribution of the people is true.

 t-Tests and Comparison of Means

When population standard deviation is unknown → use t-distribution

One-sample t-test:

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

Where:

$\bar{x}$ = sample mean
$\mu$ = population mean
$s$ = sample standard deviation
$n$ = sample size
$\text{df} = n – 1$

Two-sample t-test (unpaired):

$
t = \frac{\bar{x}_1 – \bar{x}_2}{\sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}}}
$

Where:

$\bar{x}_1, \bar{x}_2$ = sample means
$s_1, s_2$ = sample standard deviations
$n_1, n_2$ = sample sizes

Degrees of freedom: approximately

$
\text{df} = \min(n_1 – 1, n_2 – 1)
$

Assumptions:

Independent samples
Normal distribution (or large sample size)
Equal variances not required (Welch’s test used)

 One-tailed vs. Two-tailed Tests

One-tailed Test:

Tests for effect in a single direction:

$
H_1: \mu \mu_0 \quad \text{or} \quad H_1: \mu < \mu_0
$

Two-tailed Test:

Tests for difference in either direction:

$
H_1: \mu \ne \mu_0
$

Critical Region:

One-tailed: entire $\alpha$ in one tail
Two-tailed: split $\alpha$ into both tails
(e.g., 0.025 in each if $\alpha = 0.05$)

Example:

To compare the mean weights between two populations A and B we obtain two samples:

(GDC gives that the two sample means are $\bar{x}_1 = 70.1$ and $\bar{x}_2 = 63.1$)

We will test two different claims for the population means $\mu_1, \mu_2$ with $\alpha = 0.05$:

(a) Ann claims that $\mu_1 > \mu_2$

(b) Bill claims that population means are different

Solution

(a)

We perform a one-tailed t-test:

$
\begin{aligned}
H_0&: \mu_1 = \mu_2 \\
H_1&: \mu_1 > \mu_2
\end{aligned}
$

GDC gives p-value = 0.041

Since

$
\textbf{p-value < 0.05}
$

we reject $H_0$.
That is, we accept Ann’s claim that $\mu_1 > \mu_2$.

(b)

We perform a two-tailed t-test:

$
\begin{aligned}
H_0&: \mu_1 = \mu_2 \\
H_1&: \mu_1 \ne \mu_2
\end{aligned}
$

GDC gives p-value = 0.082

Since

$
\textbf{p-value > 0.05}
$

we do not have enough evidence to reject $H_0$.
Bill is not right!