Tests whether two categorical variables are independent or associated, using observed vs expected frequencies in a contingency table.
Test statistic (given in formula booklet)
\( \chi ^2_{calc} = \sum (f_o - f_e)^2 / f_e \)
Expected frequency: \( f_{e} \) = (row total \( \times \) column total) / grand total
Degrees of freedom
\( df = (rows - 1)(columns - 1) \)
Steps: (1) State H0: variables are independent. H1: variables are not independent. (2) Calculate \( \chi ^2_{calc} ( \)GDC). (3) Find \( p- \)value. (4) If \( p \) < significance level \( \to \) reject H0.
GDC: Chi-Squared Test
TI-84 Plus CE
Enter observed data into a matrix: [2ND][MATRIX] → Edit → [A] [STAT] → TESTS \( \to \chi ^2- \)Test
Observed: [A] Expected: [B] press Calculate
Read \( \chi ^2 \) and \( p- \)value. Check [B] for expected frequencies.
TI-Nspire CX II
Open Spreadsheet, enter observed data in columns [Menu] → Statistics → Stat Tests \( \to \chi ^2 2- \)way Test
Select data range → read \( \chi ^2 \) and \( p- \)value
Casio fx-CG50
[MENU] → Statistics → TEST → CHI → 2- \)way
Enter observed matrix → Execute
Read \( \chi ^2 \) and \( p \). Press [F6] to see expected frequencies.
✗
Common error: All expected frequencies must be \( \geq 5. \) If any \( f_{e} \) < 5, combine categories or note the limitation. The IB expects you to check this.
2
Chi-Squared Goodness of Fit HL
Tests whether observed data fits an expected distribution (uniform, given proportions, Poisson, etc.).
Same test statistic, different df
\( \chi ^2_{calc} = \sum (f_o - f_e)^2 / f_e df = k - 1 - p \)
\( k = \) number of categories, \( p = \) number of estimated parameters (0 if given)
H0: The data follows the proposed distribution. H1: The data does not follow the proposed distribution.
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Degrees of freedom: For a fair die with 6 faces, \( df = 6 - 1 = 5 \). If you estimated the mean from the data for a Poisson fit, \( df = k - 1 - 1 = k - 2 \).
Spearman’s r_s measures the strength of a monotonic (not necessarily linear) relationship between two ranked variables. \( -1 \leq \) \( r_{s} \) \( \leq 1. \)
Spearman’s rank coefficient (given in formula booklet)
\( r_s = 1 - (6 \sum d^2) / (n(n^2 - 1)) \)
\( d = \) difference between ranks, \( n = \) number of data pairs
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Tied ranks: If two values are tied, assign the average of the ranks they would occupy. e.g. tied for 3rd and 4th \( \to \) both get rank 3.5.
✗
Common error: Spearman’s tests for monotonic association, not linear. It uses ranks, not raw data. Do not confuse with Pearson’s \( r \) (which measures linear correlation).
4
Poisson Distribution HL
Models the number of events in a fixed interval when events occur independently at a constant average rate \( \lambda . \)
Probability
\( P(X = x) = e^{-\lambda } \lambda ^x / x! \)
Mean & Variance
\( E(X) = \lambda Var(X) = \lambda \)
Mean = Variance is a key Poisson property
Conditions: Events occur singly, independently, at a constant mean rate, and with no upper limit.
[MENU] → Statistics → DIST → POISN Ppd for P(\( X = x), \) Pcd for P(\( X \leq x) \)
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\( P(X \geq 3): \) Use the complement: P(\( X \geq 3) = 1 - \) P(\( X \leq 2) = 1 - \) poissoncdf(\( \lambda , 2). \)
5
Confidence Intervals HL
A confidence interval gives a range of plausible values for a population parameter (usually the mean \( \mu ). \)
Confidence interval for \( \mu ( \)known \( \sigma \) or large \( n) \)
\( \bar{x} - z \times \sigma /\sqrt{}n < \mu < \bar{x} + z \times \sigma /\sqrt{}n \)
90%: \( z = 1.645 \) 95%: \( z = 1.960 \) 99%: \( z = 2.576 \)
GDC: Z-Interval
TI-84 Plus CE
[STAT] → TESTS → ZInterval
Input: Stats \( \sigma = \) known \( \bar{x} = \) sample mean \( n = \) sample size
C-Level = 0.95 → Calculate
TI-Nspire CX II
[Menu] → Statistics → Confidence Intervals → z Interval
Enter \( \sigma , \bar{x}, n, \) C-Level → read interval
Casio fx-CG50
[MENU] → Statistics → INTR → Z → 1- \)Sample
Enter C-Level, \( \sigma , \bar{x}, n \to \) Execute
✗
Common error: "95% confident" does NOT mean there is a 95% probability that \( \mu \) is in the interval. It means if we repeated the sampling, 95% of intervals would contain \( \mu . \)
“False positive” — probability = significance level \( \alpha \)
Type II Error
Failing to reject H0 when H0 is false
“False negative” — probability = \( \beta \)
Power = 1 \( - \beta = \) probability of correctly rejecting a false H0. Higher power is better.
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Memory aid: Type I = false positIve (I for “Innocent person convicted”). Type II = false negatIIve (“Guilty person goes free”).
✗
Common error: You can NEVER “accept H0”. The correct phrasing is “fail to reject H0” or “insufficient evidence to reject H0”.
✗
Common error: Decreasing \( \alpha (e.g. \) from 5% to 1%) reduces Type I errors but \( increases \) Type II errors. There is always a trade-off.
7
Exam Traps & Key Reminders
✗
State hypotheses in context. Do not write generic H0/H1. e.g. “H0: Grade and gender are independent” not just “H0: the variables are independent”.
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\( Expected frequencies \geq 5. \) State this check explicitly in \( \chi ^2 \) tests. If violated, combine adjacent categories.
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\( Compare p-value to \alpha , not \chi ^2 to \alpha . \) Write “\( p = 0.023 \) < 0.05 \( \to \) reject H0”. The IB awards marks for this comparison step.
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Conclusion in context: After rejecting or failing to reject, state the conclusion in the language of the problem. “There is sufficient evidence at the 5% level that grade and gender are not independent.”
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Default significance level: If the question does not specify, use \( \alpha = 5% \) (0.05). This is standard in the IB.
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Formula booklet: The \( \chi ^2 \) statistic formula, Spearman’s \( r_{s} \) formula, Poisson probability formula, and confidence interval formula are all given. Know when to use each one.
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