Every computation of the measurement block, worked by hand: grouped mean, median, mode and quartiles; QD, mean deviation, SD, combined SD and CV; Pearson r and Spearman ρ; z, T, stanines and percentile ranks; Spearman–Brown, KR-20, KR-21, Rulon, Flanagan and SEM; p, D and the distractor table; correction for guessing and the CRT score; blueprint weightage; the SE of the mean and χ².
GROUPED DATA: MEAN, MEDIAN, MODE, QUARTILES
1.1 Reading a frequency table
| Term |
Meaning |
Example |
| Exclusive series |
20–30, 30–40 …: the upper limit belongs to the next class (ch03) |
A score of 30 goes into 30–40 |
| Inclusive series |
20–29, 30–39 …: both printed limits belong to the class |
A score of 29 is in 20–29 |
| Exact limits |
An inclusive class widened by half the gap between classes |
20–29 runs from 19.5 to 29.5 |
| Class width h |
Exact upper limit − exact lower limit |
29.5 − 19.5 = 10, not 29 − 20 = 9 |
| Mid-point X |
(Lower limit + upper limit) ÷ 2 |
20–29: 24.5; 20–30: 25 |
| Cumulative frequency cf |
Running total of f from the lowest class upward |
The basis of the median, quartiles and percentiles |
ch03's rule: an inclusive series must be converted to exact limits before medians, quartiles and percentiles are computed. The mid-point comes out the same from printed or exact limits.
Worked table W — marks of 40 pupils (exclusive series, h = 10)
Marks f Mid-point X cf
20–30 2 25 2
30–40 10 35 12
40–50 10 45 22
50–60 16 55 38
60–70 2 65 40
N = 40
1.2 The mean
Direct method: M = ΣfX / N
Step-deviation method: d′ = (X − A) / h A = an assumed mean (any mid-point)
M = A + (Σfd′ / N) × h
Worked example 1.2 — table W
Direct: ΣfX = 2 × 25 + 10 × 35 + 10 × 45 + 16 × 55 + 2 × 65
= 50 + 350 + 450 + 880 + 130 = 1860
M = 1860 / 40 = 46.5
Step deviation: A = 45, h = 10
d′ = −2, −1, 0, +1, +2
fd′ = −4, −10, 0, +16, +4 Σfd′ = 6
M = 45 + (6 / 40) × 10 = 45 + 1.5 = 46.5 (the same answer)
The choice of A never changes M. Raise A by one class width and every d′ falls by 1, so Σfd′ falls by N, and (Σfd′/N) × h falls by exactly h: the two changes cancel.
Trap: three slips account for most wrong options: forgetting the final "× h"; dividing Σfd′ by the number of classes instead of N; and multiplying each f by a class limit instead of the mid-point. In an inclusive series the mid-point of 20–29 is 24.5, not 25.
1.3 The median, quartiles and percentiles
Median = L + [(N/2 − cf) / f] × h
Q₁ = L + [(N/4 − cf) / f] × h
Q₃ = L + [(3N/4 − cf) / f] × h
Pₖ = L + [(kN/100 − cf) / f] × h (Q₁ = P₂₅, Median = P₅₀, Q₃ = P₇₅)
L = exact lower limit of the class that holds the position
cf = cumulative frequency BELOW that class
f = frequency OF that class
h = class width
Find the position (N/2, N/4, 3N/4 or kN/100), run down the cf column to the first class whose cf reaches it, read L, the cf below, f and h, and substitute.
Worked example 1.3 — table W
Median: N/2 = 20. The cf runs 2, 12, 22 …, so the 20th score lies in 40–50.
L = 40, cf below = 12, f = 10
Median = 40 + [(20 − 12) / 10] × 10 = 40 + 8 = 48
Q₁: N/4 = 10 → class 30–40 (cf below 2, f = 10)
Q₁ = 30 + [(10 − 2) / 10] × 10 = 38
Q₃: 3N/4 = 30 → class 50–60 (cf below 22, f = 16)
Q₃ = 50 + [(30 − 22) / 16] × 10 = 50 + 5 = 55
P₁₀: 10N/100 = 4 → class 30–40 (cf below 2, f = 10)
P₁₀ = 30 + [(4 − 2) / 10] × 10 = 32
In an inclusive series only L changes: the class 40–49 starts at 39.5, and h is still 10.
Trap: using the cf of the median class (22 here) instead of the cf below it (12). The formula then gives 40 + [(20 − 22)/10] × 10 = 38, a value below the median class itself, which is how to catch the slip.
Trap: the position is N/2, not (N + 1)/2. (N + 1)/2 locates the middle score of raw scores arranged in order; the grouped formula uses N/2, N/4 and 3N/4.
1.4 The mode
Interpolation formula (grouped data):
Mode = L + [(f₁ − f₀) / (2f₁ − f₀ − f₂)] × h
L = exact lower limit of the modal class (the class with the largest f)
f₁ = frequency of the modal class
f₀ = frequency of the class before it f₂ = frequency of the class after it
Empirical relationship (ch03): Mode = 3 Median − 2 Mean
Worked example 1.4 — table W
Modal class 50–60: f₁ = 16, f₀ = 10, f₂ = 2
Mode = 50 + [(16 − 10) / (32 − 10 − 2)] × 10 = 50 + (6 / 20) × 10 = 53
Empirical: 3 × 48 − 2 × 46.5 = 144 − 93 = 51
The two methods need not agree: the empirical relationship is only an approximation, for a moderately skewed distribution, so a question has to say which method it wants. In table W, Mean 46.5 < Median 48 < Mode 53: the mean lies on the side of the tail, the negative-skew order of ch03.
Trap: the numerator is f₁ − f₀ (the class before), not f₁ − f₂. The mid-point of the modal class (55 here) is only a crude mode, not the interpolated one.
