A gym survey finds 90% of people exercise daily. What went wrong?

Population, sample and types of data

• The population is the whole group you want to know about. A sample is the part you collect data from.
• In a random sample every member of the population has an equal chance of being chosen.
• A sample that over- or under-represents part of the population is biased.
• Discrete data are counted (goals, siblings). Continuous data are measured and can take any value in a range (time, mass).
• Check reliability: deal with missing values and recording errors first.

Sampling techniques

• Non-random methods are quicker but more likely to be biased.

Cumulative frequency, box plots and outliers

• Classes are inequalities with no gaps ($10\le t<20$10 ≤ t < 20). A histogram uses equal class widths, frequency upwards.
• Cumulative frequency: plot running totals at upper class ends. Read the median at 2n​n/2, Q1​Q₁ at 4n​n/4, Q3​Q₃ at 43n​3n/4, the $p$pth percentile at $p\%$p% of $n$n. IQR=Q3​−Q1​IQR = Q₃ − Q₁.
• An outlier is over $1.5\times \text{IQR}$1.5 × IQR from the nearest quartile: mark it with a cross. A symmetric box plot suggests possible normality.
• Compare box plots by median (average) and IQR or range (spread).

Give everyone a £100 pay rise. What happens to the standard deviation?

Mean, median and mode

• Mean: add the values and divide by how many there are.

• Median: the middle value when the data are in order, at position 2n+1​(n+1)/2.
• Mode: the most frequent value.
• In IB SL questions the data set is treated as the whole population unless the question says otherwise.

Grouped data

• When data are grouped you don't know the exact values, so use each class's mid-interval value as $x$x.
• Estimated mean =∑f∑fx​= (Σ fx)/(Σ f) with $x$x the mid-interval values. It is an estimate, not exact.
• The modal class is the class with the highest frequency (equal class widths only).

Measuring spread

• Range = largest − smallest. IQR =Q3​−Q1​= Q₃ − Q₁, the spread of the middle half; outliers do not affect it.
• Standard deviation $\sigma$σ measures the typical distance of the values from the mean. Variance =σ2= σ².
• Find $\sigma$σ with your GDC's one-variable statistics. Treat the data as the population and use σx​σ_x, not the sample value sx​s_x.
• Find quartiles of discrete data on your GDC too. Methods differ, so a hand answer may not match exactly.

Changing every value

• Add or subtract a constant $c$c: the mean, median and quartiles move by $c$c. The standard deviation, IQR and range do not change, because every gap stays the same.
• Multiply by a positive constant $k$k: the mean and the standard deviation are both multiplied by $k$k. The variance is multiplied by k2k².
• Do multiplication and addition in the order the question gives them.

Ice-cream sales and drownings rise together. Does one cause the other?

Scatter diagrams and correlation

• A scatter diagram shows bivariate data. Describe the pattern as positive, negative or no correlation, and as strong or weak.
• Pearson's product-moment correlation coefficient $r$r measures linear correlation: $-1\le r\le 1$−1 ≤ r ≤ 1. Find it with your GDC.
• $r$r near 1 or $-1$−1: strong linear correlation. $r$r near 0: no linear correlation (there may still be a curved relationship).
• Correlation does not imply causation. A third variable may drive both.

The regression line of $y$y on $x$x

• Your GDC gives the regression line of $y$y on $x$x, $y=ax+b$y = ax + b. It always passes through (xˉ,yˉ​)(x̄, ȳ).
• $a$a is the gradient: the change in $y$y for each increase of 1 in $x$x. $b$b is the value of $y$y when $x=0$x = 0, which may have no meaning in context.
• Use it to predict $y$y from a given $x$x inside the data range. Predicting outside it is extrapolation and may be unreliable.
• Don't rearrange a $y$y on $x$x line to predict $x$x from $y$y.

Spearman's rank correlation

• Spearman's rank correlation coefficient rs​r_s: rank each variable separately, then find Pearson's $r$r for the ranks. Use your GDC.
• Tied values share the average of the ranks they occupy: two values in 2nd and 3rd place both get 2.5.
• rs​=1r_s = 1 or $-1$−1 for any monotonic relationship (always increasing or always decreasing), even a curved one. Pearson's $r$r only measures linearity.
• rs​r_s is less sensitive to outliers than $r$r.

In a class of 30, how likely is it that two share a birthday?

Sample spaces and events

• A trial has outcomes; the set of all outcomes is the sample space $U$U; an event is a set of outcomes.
• For equally likely outcomes:

• The complement A′A' means "not $A$A": P(A′)=1−P(A)P(A') = 1 − P(A).
• Relative frequency (successes ÷ trials) estimates a probability from an experiment.
• Expected number of occurrences in $n$n trials $=n\times \mathrm{P}(A)$= n × P(A).

Combined and conditional probability

• "$A$A or $B$B" includes both, so subtract the overlap once:

• Mutually exclusive events cannot happen together: $\mathrm{P}(A\cap B)=0$P(A ∩ B) = 0.
• Conditional probability ("$A$A given $B$B") restricts you to $B$B: P(A∣B)=P(B)P(A∩B)​P(A | B) = (P(A ∩ B))/(P(B)).
• Independent events don't affect each other: $\mathrm{P}(A\cap B)=\mathrm{P}(A)\mathrm{P}(B)$P(A ∩ B) = P(A) P(B).

Tree diagrams and "without replacement"

• On a tree diagram, multiply along the branches for "this and then that", and add the end results for "either path".
• Each set of branches from one point adds to 1.
• Without replacement, the second-stage probabilities change: there is one fewer item, and one fewer of the kind you took.
• For "at least one", it is often quicker to use $1-\mathrm{P}(\text{none})$1 − P(none).

A £1 dice game pays out £5 if you roll a six. Is the game fair?

Discrete random variables

• A discrete random variable $X$X takes separate values, each with a probability. Write a value as $x$x and its probability as $P(X=x)$P(X=x).
• The probability distribution lists every value with its probability, in a table or by a formula, e.g. P(X=x)=181​(4+x)P(X=x) = 1/18(4+x) for $x\in \{1,2,3\}$x ∈ {1, 2, 3}.
• The probabilities are all between 0 and 1 and sum to 1: 185​+186​+187​=15/18 + 6/18 + 7/18 = 1. Use this to find an unknown probability.

Expected value and fair games

• The expected value (mean) is the long-run average value of $X$X:

• $\mathrm{E}(X)$E(X) need not be a value $X$X can take. The expected number of occurrences in $n$n trials is $np$np: 128 students each absent with probability 0.1 gives 12.8 absent on average.
• In a game, let $X$X be the player's gain (winnings minus the cost to play). The game is fair when $\mathrm{E}(X)=0$E(X) = 0.

The binomial distribution

• Use $X\sim \mathrm{B}(n,p)$X ~ B(n, p) when there is a fixed number $n$n of trials, each with two outcomes (success or failure), the trials are independent, and the probability of success $p$p is constant. $X$X counts the successes.
• P(X=r)=(rn​)pr(1−p)n−rP(X = r) = C(n, r)p^r(1−p)^(n−r). Find probabilities with your GDC: binomial pdf gives $P(X=r)$P(X = r), binomial cdf gives $P(X\le r)$P(X ≤ r).
• Mean $\mathrm{E}(X)=np$E(X) = np and variance $\mathrm{Var}(X)=np(1-p)$Var(X) = np(1−p).

Heights average 170 cm. How rare is someone over 2 metres tall?

The normal curve

• X∼N(μ,σ2)X ~ N(μ, σ²) has a bell-shaped curve, symmetric about the mean $\mu$μ, so mean, median and mode are equal. The second parameter is the variance σ2σ².
• Probabilities are areas under the curve; the total area is 1. Many measurements (lengths, masses, times) are close to normal.
• About 68% of values lie within $\mu \pm \sigma$μ ± σ, 95% within $\mu \pm 2\sigma$μ ± 2σ and 99.7% within $\mu \pm 3\sigma$μ ± 3σ.

Probabilities and inverse normal on your GDC

• Sketch first: draw the curve, mark $\mu$μ and shade the area you want.
• Normal cdf needs a lower bound, an upper bound, $\mu$μ and $\sigma$σ. With no lower limit enter a huge negative number such as −1099−10⁹⁹; with no upper limit, 109910⁹⁹.
• Inverse normal takes an area and returns the value $x$x with that area to its left (unless you set the tail). For the top 5%, enter 0.95.
• $P(X=a)=0$P(X = a) = 0 for a continuous variable, so $P(X<a)=P(X\le a)$P(X < a) = P(X ≤ a).

Are left-handers more likely to play tennis? How could you test it?

Hypotheses, significance and $p$p-values

• The null hypothesis H0​H₀ is the "no effect" claim: no association, or equal means. The alternative hypothesis H1​H₁ is what you are looking for evidence of.
• Fix the significance level (1%, 5% or 10%) before testing.
• The $p$p-value is the probability of a result at least as extreme as yours, assuming H0​H₀ is true.
• If $p$p-value $<$< significance level, reject H0​H₀. Otherwise there is insufficient evidence to reject H0​H₀. Conclude in context; a test never proves H0​H₀.

The χ2chi² test for independence

• Data in a contingency table. H0​H₀: the two variables are independent. H1​H₁: they are not independent.
• Expected frequency =grand totalrow total×column total​= (row total × column total)/(grand total). Each should be greater than 5.

• Degrees of freedom $=(r-1)(c-1)$= (r − 1)(c − 1) for $r$r rows and $c$c columns of data. Your GDC gives χcalc2​chi²_calc and the $p$p-value.
• Reject H0​H₀ if $p$p-value $<$< significance level, or χcalc2​>chi²_calc > the critical value.

The χ2chi² goodness of fit test

• Tests whether data fit a claimed distribution (equal chances, or given proportions). H0​H₀: the data fit the distribution. H1​H₁: they do not.
• Expected frequency $=$= total $\times$× claimed probability. Use the same χcalc2​chi²_calc formula.
• Degrees of freedom $=n-1$= n − 1 for $n$n categories. Only upper-tail tests are used: a large χcalc2​chi²_calc means a poor fit.
• If you estimate a parameter of the distribution from the data (such as a mean), subtract one more degree of freedom for each parameter estimated.

The $t$t-test for two means

• Compares the means of two populations using two independent samples. The variables must be normally distributed; assume equal variances and use the pooled two-sample $t$t-test.
• H0​H₀: μ1​=μ2​μ₁ = μ₂. Two-tailed: H1​H₁: μ1​=μ2​μ₁ ≠ μ₂. One-tailed: H1​H₁: μ1​>μ2​μ₁ > μ₂ (or $<$<), when the claim has a direction.
• Choose the tail from the claim before looking at the data. Your GDC gives $t$t and the $p$p-value; compare $p$p with the significance level.