Statistical inference · Lesson 9 of 13

Comparing two groups: means, proportions and spreads

Most real questions are comparisons. Did the scheme raise spending? Do two districts back the candidate equally? Is wage inequality the same in two sectors? Lesson 8 gave you the map of tests. This lesson takes the two-group branch of it apart: where the standard error of a difference comes from, why pooling works, why pairing the data can turn a failed test into a clear result, and why two intervals that overlap prove nothing.

← Statistical inference moduleLesson 9 of 13 · about 55 minutes

A question to hold on to

Eight households’ monthly spending (₹ thousand) was recorded before and after a scheme: 18, 24, 30, 21, 27, 33, 16, 25, then 22, 27, 30, 26, 33, 34, 22, 28. Run a two-sample t test and the rise is not significant. Run a paired t test on the very same sixteen numbers and it is. Which verdict is right, and why can the same data say both?

Ashanagar, inventedSolved at the end ↓

New here?What is Ashanagar?The invented district behind this page's examples: 60 households, with incomes you can inspect.

Ashanagar is an invented district. The statistics lessons use it for every worked example, so you meet the same households again as the ideas build: first as a list of numbers, then as a spread, then as a population to sample from.

It has 60 households, each with a monthly income in rupees. Unless a page says otherwise, those 60 are the whole population, and a “sample” means a few of them drawn at random. Where a page needs two variables, it uses a smaller set of 12 households with the head’s years of schooling beside income.

The figures are illustrative, made up once and fixed. They are not official statistics.

Households: 60
Mean income: ₹28,783 a month
Median income: ₹25,000 a month
Standard deviation: ₹13,366
Lowest and highest: ₹10,500 and ₹72,000

Meet the full data set →

The idea

Which design is it?

Every two-group test compares an estimate of a difference with its standard error. What separates the tests is how the data were collected and what the standard error looks like as a result. Two questions decide it: are the two sets of numbers linked unit by unit? and is the quantity a mean, a proportion or a spread?

Design	Example	Compare	Distribution under H₀
Independent means	Marks in two different classes	x̄ − ȳ	z if σ is known, otherwise t (n₁ + n₂ − 2 df pooled, or Welch)
Paired means	The same households before and after	the mean of the differences	t, n − 1 df
Two proportions	Two districts’ support for a candidate	p̂₁ − p̂₂	Standard normal
Two spreads	Wage variability in two sectors	s₁² ÷ s₂²	F, (n₁ − 1, n₂ − 1) df

The first question to ask

If one measurement in the first group has a natural partner in the second (the same household, the same student, a twin), the data are paired and the unit of analysis is the difference. Otherwise the groups are independent. Choosing wrongly changes the answer, as the puzzle at the top shows.

The foundation

The standard error of a difference

The sample means x̄ and ȳ each wobble from sample to sample, with standard errors σ₁ ÷ √n₁ and σ₂ ÷ √n₂. Their difference wobbles too, and the test statistic is the observed difference measured in units of that wobble. The size of the wobble is the one fact the whole lesson rests on: for independent samples the variances add, even though the means are subtracted.

Standard error of a difference of two independent means

SE (x ˉ - y ˉ) = \frac{σ _{1}^{2}}{n _{1}} + \frac{σ _{2}^{2}}{n _{2}}

Using the sample standard deviations s₁ and s₂ in place of σ₁ and σ₂ when the σ's are unknown.

Derive it

Why the standard error of a difference adds the variances

We want to show

For independent samples,

Var (X ˉ - Y ˉ) = \frac{σ _{1}^{2}}{n _{1}} + \frac{σ _{2}^{2}}{n _{2}}

, so the standard error is the square root of that sum, and the standard errors themselves combine like the sides of a right-angled triangle.

Each mean has its own variance.
From the standard error lesson, Var(X̄) = σ₁² ÷ n₁ and Var(Ȳ) = σ₂² ÷ n₂. The two samples are drawn separately, so X̄ and Ȳ are independent and Cov(X̄, Ȳ) = 0.
Use the rule for a combination of two variables.
From the correlation lesson, Var(aU + bV) = a²Var(U) + b²Var(V) + 2ab Cov(U, V). Put U = X̄, V = Ȳ, a = 1 and b = −1. The covariance term is zero, and b² = (−1)² = 1, so
$Var (X ˉ - Y ˉ) = Var (X ˉ) + Var (Y ˉ) = \frac{σ _{1}^{2}}{n _{1}} + \frac{σ _{2}^{2}}{n _{2}}$
The minus sign disappears. Subtracting two uncertain numbers leaves a result that is more uncertain than either one, since each mean brings its own error and neither can cancel the other for certain.
Take the square root, and see the triangle.
The standard error is the square root of the variance. In the DSE 2007 example the two classes have s = 8 and 7, with n = 40 and 50. The standard errors of the means are 1.265 and 0.99, and
$\frac{8 ^{2}}{40} + \frac{7 ^{2}}{50} = 1.6 + 0.98 = 1.606$
That is smaller than the 2.255 you would get by simply adding the two standard errors. Squares add for independent quantities, so the standard errors are the two short sides of a right-angled triangle and the standard error of the difference is the long side. The two errors are independent, so they seldom line up in the same direction at full size, and the combined error is smaller than the sum of the two.
What follows for the test.
The statistic is (x̄ − ȳ) ÷ SE, the observed gap measured in standard errors. A gap of 4 marks with SE 1.606 is 2.49 standard errors, which is why the DSE question lands between 1.96 and 2.58.

Step 1 of 4

A common slip is to add the two standard errors, 8 ÷ √40 + 7 ÷ √50, and get 2.26 where the right answer is 1.61. A second is to subtract the variances because the means are subtracted. The picture shows why neither works: independent errors combine like the legs of a right-angled triangle.

Interactive

Two independent means

Three versions of the test differ only in how the standard error is estimated. When both population standard deviations are known, use them and read the statistic against the normal curve. This is the z test of JNU 2021: with the statistic equal to x̄ − ȳ over σ√(1/n₁ + 1/n₂), a value between 1.96 and 2.58 means rejection at 5% but not at 1%. When they are unknown and the samples are large, as with the 1,000 and 1,200 mangoes of DSE 2008, z and t agree to many decimals. When they are unknown and the samples are small, you need t, and then a choice.

Pooled t. Assume the two variances are equal, combine the two sample variances into one estimate s_p², and use n₁ + n₂ − 2 degrees of freedom (JNU 2021). Best when the variances really are close.
Welch t. Make no such assumption. Use the unpooled standard error √(s₁²/n₁ + s₂²/n₂) and a fractional degrees-of-freedom formula, somewhat smaller than n₁ + n₂ − 2. Safe in both cases, and the default of most software.

The two-sample t statistic

t = \frac{x ˉ - y ˉ}{s _{p} \frac{1}{n _{1}} + \frac{1}{n _{2}}}, s_{p}^{2} = \frac{( n _{1} - 1 ) s _{1}^{2} + ( n _{2} - 1 ) s _{2}^{2}}{n _{1} + n _{2} - 2}, df = n_{1} + n_{2} - 2

Pooled version shown. Welch's replaces the denominator with √(s₁²/n₁ + s₂²/n₂).

Group 1 mean74Group 2 mean78Group 1 standard deviation8Group 2 standard deviation7Group 1 size, n₁40Group 2 size, n₂50

t curve under H₀Rejection region, 5% in allYour statistic

Difference, x̄₁ − x̄₂−4

Standard error1.582

t, on 88 df−2.53

p-value0.0133

95% interval for μ₁ − μ₂−7.14 to −0.86

What it shows

The statistic −2.53 lies in the rejection region, and the interval for the difference, −7.14 to −0.86, leaves out zero. At the 5% level we reject equal means. The two groups' own intervals still overlap, yet the difference is significant: overlapping intervals do not mean no difference. Try the three starting points: the same gap of 3.5 is not significant with eight households a side and is overwhelming with 180.

Start with the two classes. The gap is 4 marks and the pooled statistic is −2.53 on 88 degrees of freedom, with p = 0.013. Switch to Welch and the statistic is −2.49 on about 78 degrees of freedom: nearly the same, because the spreads (8 and 7) and the sizes (40 and 50) are close. Then choose “Eight and eight” and slide the sizes upward. The gap of 3.5 does not change, only the standard error does, and by 180 households a side it is overwhelming. Significance is a statement about the gap relative to its noise, never about the gap alone.

Derive it

The pooled variance and its n₁ + n₂ − 2 degrees of freedom

We want to show

If both populations have the same variance σ², the best estimate of it is

s_{p}^{2} = \frac{( n _{1} - 1 ) s _{1}^{2} + ( n _{2} - 1 ) s _{2}^{2}}{n _{1} + n _{2} - 2}

, and the statistic

\frac{x ˉ - y ˉ}{s _{p} 1/ n _{1} + 1/ n _{2}}

follows a t distribution with n₁ + n₂ − 2 degrees of freedom.

Two estimates of one number.
Suppose σ₁² = σ₂² = σ². Then s₁² and s₂² each estimate the same σ². Using both is better than using one, so we want a combined estimate. A plain average of s₁² and s₂² would treat a sample of 3 and a sample of 300 alike, which cannot be right.
Weight each by what it knows.
A sample variance is a sum of squared deviations divided by its degrees of freedom: s₁² = SS₁ ÷ (n₁ − 1). The most information-preserving combination adds the two sums of squares and divides by the total degrees of freedom:
$s_{p}^{2} = \frac{S S _{1} + S S _{2}}{( n _{1} - 1 ) + ( n _{2} - 1 )} = \frac{( n _{1} - 1 ) s _{1}^{2} + ( n _{2} - 1 ) s _{2}^{2}}{n _{1} + n _{2} - 2}$
That is a weighted average of s₁² and s₂² with weights n₁ − 1 and n₂ − 1: the bigger sample counts for more. For the two classes, s_p² = (39 × 64 + 49 × 49) ÷ 88 = 55.648, which sits between 49 and 64, closer to 49 because class 2 is larger.
Why n₁ + n₂ − 2.
SS₁ ÷ σ² is a chi-square with n₁ − 1 degrees of freedom, since one degree is spent on estimating the mean of sample 1. The same holds for SS₂ with n₂ − 1. They are independent, and independent chi-squares add by adding their degrees of freedom (the sums-of-squares result in the chi-square derivation), so
$\frac{( n _{1} + n _{2} - 2 ) s _{p}^{2}}{σ ^{2}} \sim χ_{n_{1} + n_{2} - 2}^{2}$
One degree is lost for each of the two sample means, which is the number JNU 2021 asks for.
Put it into the statistic.
Under H₀ the numerator x̄ − ȳ is normal with mean 0 and variance σ²(1/n₁ + 1/n₂). Divide by its true standard deviation and you have a standard normal Z. Replace σ by s_p and, by the definition of t (a standard normal over the square root of an independent chi-square divided by its degrees of freedom), you get t with n₁ + n₂ − 2 degrees of freedom. The standard error is s_p√(1/n₁ + 1/n₂).

Step 1 of 4

Interactive

Paired data

Take the eight households of the puzzle. Their spending before and after has a large household-to-household spread: some spend ₹16 thousand and some ₹33 thousand. That spread is not what the scheme is about, but a two-sample test has to absorb it into the standard error. The paired test removes it. For each household compute the difference, then ask whether the mean difference is zero.

The paired t statistic

t = \frac{d ˉ}{s _{d} / n}, df = n - 1, s_{d}^{2} = s_{1}^{2} + s_{2}^{2} - 2 r s_{1} s_{2}

d is after minus before for each unit. The test is a one-sample t test on the differences.

The last expression is the key. It is the variance-of-a-difference rule from the correlation lesson with the correlation r kept in. Here r = 0.94: households that spend a lot before spend a lot after. The large positive covariance is subtracted away, and the spread of the differences, 2.2, is less than half of either column’s own spread, 5.8 and 4.5.

rose or stayedfell

Analysis	Standard error	t	p-value
Independent samples, 14 df	2.595	1.35	0.1988
Paired, 7 df	0.779	4.49	0.0028

Mean rise3.5

Correlation of the columns0.94

Spread of the differences2.2

What it shows

With each household compared with itself, the rises are tightly bunched (spread 2.2), so the paired t is 4.49 and the rise is clearly real. Treated as two unrelated groups, the household-to-household differences in the level of spending swamp the rise: t is only 1.35. Pairing helps exactly when the two measurements on a unit move together.

The independent test gets t = 1.35 on 14 degrees of freedom, below the critical value 2.145, so the rise is not significant. The paired test gets t = 4.49 on 7 degrees of freedom, above 2.365, so it is. The paired test is right, because the data are paired. Use the lab’s second button to destroy the pairing, which leaves the independent test untouched and drops the paired test, since the correlation that powered it is gone.

Pause and predict

Two measurements on the same units are strongly positively correlated. Compared with treating them as independent samples, the paired test has

Watch out for

The mistakes examiners are fishing for

Running a two-sample test on paired data. It throws away the pairing and usually loses power, as the puzzle showed.
Pairing data that are not paired. If the two columns are separate households with no natural link, the order is arbitrary and a paired test is meaningless.
Getting the degrees of freedom wrong. A paired test has n − 1, where n is the number of pairs, not 2n − 2.

Proportions

Two proportions

For two sample proportions the standard error follows from the same variance-addition rule, with p(1 − p) ÷ n in place of σ² ÷ n. The new wrinkle is what to put in for p. In a test of H₀: p₁ = p₂ the null says the two populations share one proportion, so you estimate it from the combined data. In an interval there is no such assumption.

Test for two proportions

z = \frac{p ^ _{1} - p ^ _{2}}{p ˉ ( 1 - p ˉ ) ( \frac{1}{n _{1}} + \frac{1}{n _{2}} )}, p ˉ = \frac{x _{1} + x _{2}}{n _{1} + n _{2}}

p̄ is the pooled proportion: total successes over total trials.

Derive it

Why a test of two proportions pools them, and an interval does not

We want to show

Under H₀: p₁ = p₂ the standard error uses the pooled proportion

p ˉ = \frac{x _{1} + x _{2}}{n _{1} + n _{2}}

. An interval for p₁ − p₂ uses the two sample proportions separately.

The variance of a difference of proportions.
Each sample proportion has variance p(1 − p) ÷ n. The samples are independent, so by the previous derivation
$Var (p^_{1} - p^_{2}) = \frac{p _{1} ( 1 - p _{1} )}{n _{1}} + \frac{p _{2} ( 1 - p _{2} )}{n _{2}}$
The true p₁ and p₂ are unknown, so we must plug in estimates. What we plug in depends on what we are doing.
In a test, the null hypothesis tells us what to plug in.
A test asks what the world would look like if H₀ were true. H₀ says p₁ = p₂ = p, one common value. Then the variance becomes p(1 − p)(1/n₁ + 1/n₂), and the best estimate of the single p uses all the data: the total number of successes over the total number of trials,
$p ˉ = \frac{x _{1} + x _{2}}{n _{1} + n _{2}} = \frac{n _{1} p ^ _{1} + n _{2} p ^ _{2}}{n _{1} + n _{2}}$
It is a weighted average of p̂₁ and p̂₂. For District A and District B, p̄ = (168 + 96) ÷ 500 = 0.528, and the standard error is √(0.528 × 0.472 × (1/300 + 1/200)) = 0.0456.
In an interval, there is no null hypothesis.
An interval does not assume p₁ = p₂. It tries to say how large the difference is, so each proportion keeps its own variance, estimated by its own p̂(1 − p̂) ÷ n. For the same data the standard error is √(0.56 × 0.44 ÷ 300 + 0.48 × 0.52 ÷ 200) = 0.0455, and the 95% interval for p₁ − p₂ is 0.08 ± 1.96 × 0.0455, which is −0.009 to 0.169.
The two answers agree here.
z = 0.08 ÷ 0.0456 = 1.76, below 1.96, and the interval contains 0. They can disagree only in a close call, because the two standard errors, 0.0456 and 0.0455, differ slightly. The rule to remember is simple: pool for the test, do not pool for the interval.

Step 1 of 4

DSE 2007 · hard

56% of 300 voters in District A and 48% of 200 in District B favour a candidate. At 5%: (a) do the districts differ? (b) Is the candidate preferred in District A?

(a) Pool, then compare.
p̄ = 264 ÷ 500 = 0.528, SE = 0.0456, z = 0.08 ÷ 0.0456 = 1.76. That is below 1.96, so there is no significant difference (p = 0.079).
(b) A different question: one sample, one tail.
Is the true proportion in A above one half? Under H₀: p = 0.5 the standard error is √(0.25 ÷ 300) = 0.0289, so z = 0.06 ÷ 0.0289 = 2.08, above the one-sided 1.645.

Answer

No significant difference between the districts, yet the candidate is preferred in District A: option B. The two conclusions do not contradict each other. A is above one half. The eight-point gap between A and B is no bigger than noise in samples of 300 and 200 would produce, so the data do not separate the two districts.

JAM 2022 asks for the standard deviation of P̄₁ − P̄₂ when sampling is without replacement from finite populations. Each variance keeps its own n and its own finite-population correction, and the two add under a root: the same rule, with an extra factor (N − n) ÷ (N − 1) on each term.

Spreads

Two variances: the F test

To ask whether two normal populations are equally spread out, divide one sample variance by the other. Each (n − 1)s² ÷ σ² is a chi-square, so under H₀: σ₁² = σ₂² the common σ² cancels and the ratio is an F distribution with (n₁ − 1, n₂ − 1) degrees of freedom (UOH 2019). This is the reason the F distribution exists, and the chi-square, t and F derivation shows it from scratch.

The F test for two variances

F = \frac{s _{1}^{2}}{s _{2}^{2}} \sim F_{n_{1} - 1, n_{2} - 1} under H_{0} : σ_{1}^{2} = σ_{2}^{2}

Conventionally the larger sample variance goes on top, so F is at least 1, and the upper critical value for α/2 is used.

In UGC NET 2025, samples of 20 and 15 have variances 35.53 and 34.92, so F = 35.53 ÷ 34.92 = 1.02. With (19, 14) degrees of freedom the acceptance band at 5% runs from 0.378 to 2.861. An F of 1.02 is deep inside it, so there is no evidence that the variances differ.

Why does the lower limit exist if you always put the larger variance on top? You can put either on top. Swapping them turns F into 1/F and swaps the degrees of freedom, and the lower critical value for (d₁, d₂) is exactly the reciprocal of the upper critical value for (d₂, d₁): 0.378 × 2.647 = 1. Putting the larger on top just means you only need the upper limit.

Watch out for

The mistakes examiners are fishing for

Using a chi-square for two variances. A chi-square tests one variance against a number. Two variances make a ratio, and a ratio of chi-squares over their degrees of freedom is an F (UOH 2019).
Trusting the F test on non-normal data. Unlike the t test, it is quite sensitive to departures from normality. A significant F on skewed data may just reflect the skew.
Using the F test to choose between pooled and Welch. With small samples it has little power, and a non-significant result does not show the variances are equal. If in doubt, use Welch.

Intervals

Intervals for a difference, and the overlap trap

The interval for μ₁ − μ₂ is the difference plus or minus the critical value times the standard error. It carries the same verdict as the test and more: the 95% interval excludes zero exactly when the two-sided 5% test rejects. For the two classes, the difference is −4 and the 95% interval is −7.14 to −0.86, which excludes zero, matching p = 0.013.

Confidence interval for a difference of means

(x ˉ - y ˉ) \pm t_{α /2, df} \times SE (x ˉ - y ˉ)

Use the same standard error and degrees of freedom as the test.

The tempting shortcut is to build a 95% interval for each group and call the difference significant if the two do not overlap. That is a sufficient condition, not a necessary one. Two independent 95% intervals can overlap and the difference still be significant, because the standard error of the difference, √(SE₁² + SE₂²), is smaller than SE₁ + SE₂. The class example does exactly this: the two groups’ own intervals overlap, yet p = 0.013. The first plot of the lab shows both intervals.

The rule

Overlapping intervals do not mean “no difference”. Judge the difference by the interval for the difference.

When the data are not normal

With small samples from clearly skewed populations the t tests lose their footing. The replacements use ranks: the Wilcoxon rank-sum (equivalently Mann-Whitney) test for two independent samples, and the Wilcoxon signed-rank test for paired data (JNU 2021). The names are close and an examiner relies on that: the signed-rank test needs pairs, the rank-sum test does not.

Drill

Which test? A drill

Pick the test for each situation · 1 of 7

Marks of 40 students in one class and 50 in another, drawn separately. The standard deviations are 8 and 7.

Solving the opening question

The opening question, solved

Ashanagar · the puzzle

Eight households’ spending before and after a scheme gives a non-significant two-sample t and a significant paired t on the same numbers. Which verdict is right?

Are the data paired?
Yes: each “after” belongs to the same household as a “before”. So the unit of analysis is the difference, and the right test is the paired t.
Why the two-sample test fails.
It compares 27.75 with 24.25, a gap of 3.5, against the standard error of two unrelated groups, 2.595. That standard error is dominated by the fact that some households spend double what others do, which has nothing to do with the scheme. t = 1.35, p = 0.199.
Why the paired test succeeds.
The differences are 4, 3, 0, 5, 6, 1, 6, 3, with mean 3.5 and standard deviation 2.2. The standard error is 0.779, so t = 4.49 on 7 degrees of freedom, p = 0.0028. The 95% interval for the average rise is 1.66 to 5.34 thousand rupees.

Answer

The paired verdict is right: the scheme raised spending. The same data can say both because the two tests ask about different noise. The two-sample test measures the gap against household-to-household differences in the level of spending. The paired test measures each household against itself, and the correlation of 0.94 between the columns is what shrinks the noise.

Revise

The lesson on one screen

Standard error: For independent samples the variances add: SE = √(σ₁²/n₁ + σ₂²/n₂). Never add the standard errors, never subtract the variances.
Two means: z if σ is known. Otherwise t. Pooled: sₚ² = [(n₁ − 1)s₁² + (n₂ − 1)s₂²] ÷ (n₁ + n₂ − 2), df = n₁ + n₂ − 2. Welch: unpooled, fractional df.
Paired: Difference each pair, then a one-sample t on the differences with n − 1 df. Pairing helps when the two measurements are positively correlated.
Two proportions: z with the pooled p̄ in the standard error for a test. Unpooled for an interval.
Two variances: F = s₁²/s₂² on (n₁ − 1, n₂ − 1) df. Sensitive to non-normality. The lower critical value is the reciprocal of the upper with the df swapped.
Intervals: Difference ± critical value × SE. It excludes 0 exactly when the 5% two-sided test rejects. Overlapping separate intervals do not mean no difference.
Non-normal data: Wilcoxon rank-sum for independent samples, signed-rank for paired.

Practice

Check yourself

0 of 14 correct so far

Compute

A z statistic

Group 1: n = 36, mean 60, σ = 9. Group 2: n = 49, mean 56, σ = 7. Both σ are known. Find the z statistic for H₀: μ₁ = μ₂.

A pooled t

n₁ = 12, mean 52, s₁ = 10, and n₂ = 15, mean 45, s₂ = 8. Assume equal variances. Find the pooled t statistic.

A paired t

Six students sit a test before and after a revision course. The score gains are 2, −1, 3, 4, 0, 2. Find the paired t statistic.

Two proportions

In a sample of 150 men, 45 own a bank account. In 150 women, 30 do. Find the z statistic for H₀: the two proportions are equal.

An F ratio

Two normal samples give variances 40 (n = 11) and 16 (n = 9). Find the F statistic with the larger variance on top.

An interval for a difference

Using the data of problem 2 (difference 7, standard error 3.461, 25 degrees of freedom, critical value 2.060), find the upper limit of the 95% interval for μ₁ − μ₂.

Upper limit

A standard error

Two independent samples each have n = 25 and s = 10. Find the standard error of the difference of their means.

Paired, from summary figures

25 pairs have mean difference 5 and standard deviation of the differences 10. Find the paired t statistic.

Quick questions

1For independent samples, the standard error of x̄ − ȳ is

2The same households' incomes are recorded before and after a scheme. The right test of a change is

3In a test of H₀: p₁ = p₂ the standard error uses

4The pooled two-sample t statistic has how many degrees of freedom?

5Two separately built 95% confidence intervals for two group means overlap. We can conclude

6The test statistic for equality of two population variances follows

Real questions

Past papers, by level

Foundation

Definitions and one-step calculations. Most CUET and Hyderabad questions sit here.

2 questionsPractise →

Standard

Joint tables, standard errors and statement questions, up to JNU and IIT JAM level.

3 questionsPractise →

Delhi School level

This topic’s hardest questions, from the DSE entrance, are worked through in Lesson 13: Tests on means and proportions.

4 questionsLesson 13 →

All 5 past-paper questions for this lesson, easiest first: practise the full set →

These are questions from the TestStop bank. A free account answers 10 a day, each with its worked solution; All-Access opens every set with no daily limit.

← Lesson 8The tests you will be asked Lesson 10 →Chi-square tests: goodness of fit and independence

Comparing two groups: means, proportions and spreads

Which design is it?

The standard error of a difference

Two independent means

Paired data

The mistakes examiners are fishing for

Two proportions

Two variances: the F test

The mistakes examiners are fishing for

Intervals for a difference, and the overlap trap

When the data are not normal

Which test? A drill

The opening question, solved

The lesson on one screen

Check yourself

Compute

Quick questions

Past papers, by level

Foundation

Standard

Delhi School level

Read next

Comparing two groups: means, proportions and spreads

Which design is it?

The standard error of a difference

Two independent means

Paired data

The mistakes examiners are fishing for

Two proportions

Two variances: the F test

The mistakes examiners are fishing for

Intervals for a difference, and the overlap trap

When the data are not normal

Which test? A drill

The opening question, solved

The lesson on one screen

Check yourself

Compute

Quick questions

Past papers, by level

Foundation

Standard

Delhi School level

Read next