Introductory Statistics: cram pack

Formula sheets, the key method steps for every topic and 12 practice problems, with full solutions at the back.

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1Formulas and methods

Descriptive statistics

Mean
x̄ = Σx/n
Sample standard deviation
s = √(Σ(x − x̄)²/(n − 1))
Median
middle value of the sorted data
z-score
z = (x − μ)/σ
Normal distribution
68% within ±1σ, 95% within ±2σ

Probability

Complement
P(not A) = 1 − P(A)
Addition rule
P(A ∪ B) = P(A) + P(B) − P(A ∩ B)
Independent events
P(A ∩ B) = P(A)·P(B)
Conditional probability
P(A|B) = P(A ∩ B)/P(B)
Binomial
P(X = k) = C(n,k)·pk(1 − p)n−k

Normal distribution

z-score
z = (x − μ)/σ
Below x
P(X < x) = Φ(z)
Above x
P(X > x) = 1 − Φ(z)
Between
Φ(z2) − Φ(z1)
Percentile
x = μ + z·σ

Method

  1. Convert the value to a z-score: z = (x − μ)/σ.
  2. Look up Φ(z), the area to the left of z.
  3. For the area to the right use 1 − Φ(z); for an interval subtract the two left areas.
  4. For a percentile, work backwards: find z for the area, then x = μ + z·σ.

Confidence intervals

σ known
x̄ ± z·σ/√n
σ unknown
x̄ ± t·s/√n, df = n − 1
Critical z
90%: 1.645, 95%: 1.96, 99%: 2.576
Sample size
n = (z·σ/E)², rounded up

Method

  1. Decide between z (population σ known) and t (only the sample s, with n − 1 degrees of freedom).
  2. Find the critical value for the confidence level.
  3. Margin of error: E = critical value × standard deviation/√n.
  4. Interval: x̄ ± E.

Hypothesis tests & confidence intervals

z-test for a mean
z = (x̄ − μ0)/(σ/√n)
t-test for a mean
t = (x̄ − μ0)/(s/√n), df = n − 1
Two-sided at 5%
reject H0 if |z| > 1.96
Confidence interval
x̄ ± z·σ/√n

Method

  1. State H₀ and H₁ and decide whether the test is one- or two-sided.
  2. Calculate the test statistic: (x̄ − μ₀)/(σ/√n), or with s and t.
  3. Compare with the critical value, or find the p-value.
  4. Reject H₀ if the statistic is beyond the critical value (p below α), and state the conclusion in words.

Linear regression & correlation

Sums of squares
Sxx = Σ(x − x̄)² = Σx² − (Σx)²/n, Sxy = Σxy − ΣxΣy/n
Slope
b = Sxy/Sxx
Intercept
a = ȳ − b·x̄
Correlation
r = Sxy/√(SxxSyy)
Residual
e = y − ŷ

2Practice problems

1.Descriptive statisticsMedium
15202530354045505560

Find the sample standard deviation s of the data: 55, 55, 34, 21, 43, 60, 14.

n = 7
s =
2.Descriptive statisticsHard
z −2−1012

A quantity is normally distributed with mean 50 and standard deviation 12. What proportion P of values is above 62?

μ = 50σ = 12x = 62
P =
3.ProbabilityMedium
P(at least one six) = ??1?2?3?4?5?6?7?8P(six) = 1/6 on each roll

A fair die is rolled 8 times. What is the probability P of getting at least one six?

n = 8
P =
4.ProbabilityHard
n = 11, p = 0.4: P(X = 4) = ?k01234567891011

A test has 11 multiple-choice questions. A student who guesses gets each one right with probability 0.4, independently. What is the probability P of exactly 4 correct answers?

n = 11p = 0.4k = 4
P =
5.Normal distributionMedium
2941.3P = ?

The delivery times are normally distributed with mean 32 min and standard deviation 6 min. What fraction lies between 29 and 41.3 min? Give P as a decimal.

μ = 32σ = 6x1 = 29x2 = 41.3
P =
6.Normal distributionHard
10 cm0X − Y?

The heights X of men and Y of women in a group are independent and normal: X has mean 180 cm and standard deviation 10 cm, Y has mean 170 cm and standard deviation 12 cm. Find the probability P that X is greater than Y. Round z to two decimals.

μX = 180 cm, σX = 10 cmμY = 170 cm, σY = 12 cm
P =
7.Confidence intervalsMedium
n = 15, sd​ = 5.20LU = ?d̄ = 4.5E99%

The blood pressure of 15 patients is measured before and after a treatment. The differences d = before − after (mmHg) have mean 4.5 and standard deviation 5.2. Find the upper limit of the 99% confidence interval for the mean difference.

n = 15d̄ = 4.5sd = 5.299%t(14) = 2.977
upper limit = mmHg
8.Confidence intervalsHard
σ = 38, n = ?x̄ − Ex̄ + Ex̄E = 499%

The population standard deviation is 38. How large a sample n do you need so that a 99% confidence interval for the mean has a margin of error of at most 4?

σ = 38E = 499%
n =
9.Hypothesis tests & confidence intervalsMedium
−2.0932.093t = ?df = 19

Two teaching methods are compared. Group 1: 7 students, mean score 63, standard deviation 4. Group 2: 14 students, mean score 49.1, standard deviation 11. Assume equal population variances and use the pooled two-sample t test. Find the test statistic t for H₀: μ₁ = μ₂, taking the difference as x̄₁ − x̄₂ (group 1 minus group 2).

n₁ = 7x̄₁ = 63s₁ = 4n₂ = 14x̄₂ = 49.1s₂ = 11
t =
10.Hypothesis tests & confidence intervalsHard
z −1.64501.645

A sample of 36 measurements has mean 72.3. The population standard deviation is 8. Find the lower limit L of the 90% confidence interval for the mean.

x̄ = 72.3σ = 8n = 3690%
L =
11.Linear regression & correlationMedium
xy136245346454552660xy246405060

The table gives 6 paired observations of x = hours studied and y = exam score. Their sums are given. Find Pearson's correlation coefficient r from these sums.

n = 6Σx = 21Σy = 293Σx² = 91Σy² = 14657Σxy = 1100
r =
12.Linear regression & correlationHard
xy284136178231029xy510102030x = 5

The table gives 5 paired observations of x = load in kN and y = extension in mm. Use the least-squares line ŷ = a + b·x. A new observation (5, 18) is made. Find its residual e = y − ŷ.

n = 5(5, 18)
e =

3Solutions

1.Descriptive statistics

Answer: s = 17.94

Descriptive statistics: the mean is the sum divided by n; the median is the middle value of the sorted data (the average of the two middle values when n is even). The sample standard deviation is s = √(Σ(x − x̄)²/(n − 1)). A z-score z = (x − μ)/σ says how many standard deviations a value lies from the mean.

Mean

x̄ = 282/7 = 40.29

Squared deviations

Σ(x − x̄)² = 14.71² + 14.71² + (−6.286)² + (−19.29)² + 2.714² + 19.71² + (−26.29)² = 1931

Standard deviation

s = √(1931/6) = 17.94

Answer

s = 17.94
2.Descriptive statistics

Answer: P = 0.1587

Descriptive statistics: the mean is the sum divided by n; the median is the middle value of the sorted data (the average of the two middle values when n is even). The sample standard deviation is s = √(Σ(x − x̄)²/(n − 1)). A z-score z = (x − μ)/σ says how many standard deviations a value lies from the mean.

Standardise

z = 1

Standard normal table

Φ(1) = 0.8413
P = 1 − 0.8413 = 0.1587

Answer

P = 0.1587
3.Probability

Answer: P = 0.7674

Probability: with equally likely outcomes, P(A) = favourable/total. For independent events multiply: P(A and B) = P(A)·P(B). The complement often helps: P(at least one) = 1 − P(none). Binomial: in n independent trials with success probability p, P(X = k) = C(n, k)·pᵏ·(1 − p)ⁿ⁻ᵏ.

Complement

P(no six) = (5/6)8 = 0.2326
P = 1 − 0.2326 = 0.7674

Answer

P = 0.7674
4.Probability

Answer: P = 0.2365

Probability: with equally likely outcomes, P(A) = favourable/total. For independent events multiply: P(A and B) = P(A)·P(B). The complement often helps: P(at least one) = 1 − P(none). Binomial: in n independent trials with success probability p, P(X = k) = C(n, k)·pᵏ·(1 − p)ⁿ⁻ᵏ.

Binomial formula

P(X = 4) = 330·0.44·0.67 = 330·0.0256·0.02799 = 0.2365

Answer

P = 0.2365
5.Normal distribution

Answer: P = 0.6309

Normal distribution: convert a value x to a z-score, z = (x − μ)/σ, which counts standard deviations from the mean. A z-table (or Φ(z)) gives the area to the left of z, which is P(X < x). The area to the right is 1 − Φ(z), and the area between two values is the difference of their two areas.

z-scores

z₁ = (29 − 32)/6 = −0.5
z₂ = (41.3 − 32)/6 = 1.55

Area between

P = Φ(1.55) − Φ(−0.5) = 0.9394 − 0.3085 = 0.6309

Answer

P = 0.6309
6.Normal distribution

Answer: P = 0.7389

Sums and differences of normal variables: if X and Y are independent and normal, X + Y and X − Y are normal too. Means add or subtract: E(X ± Y) = μX ± μY. Variances always add, also for a difference: Var(X ± Y) = σX² + σY², so σ = √(σX² + σY²). Standard deviations never add directly. The total of n independent items with the same distribution has mean n·μ and standard deviation σ·√n. Then standardise as usual: z = (x − mean)/σ and read Φ(z).

Distribution of the difference

μ = 180 − 170 = 10 cm, σ = 15.62 cm

Standardise

z = (0 − 10)/15.62 = −0.64

Probability

P = 1 − Φ(−0.64) = 0.7389

Answer

P = 0.7389
7.Confidence intervals

Answer: upper limit = 8.497 mmHg

Interval for a difference: estimate ± critical value × standard error. For two independent means with known σ: SE = √(σ₁²/n₁ + σ₂²/n₂) and z. With equal unknown variances: pooled s_p and t with n₁ + n₂ − 2 degrees of freedom. For paired data, work with the differences: d̄ ± t·s_d/√n.

Critical value

df = 14, t = 2.977

Margin of error

E = 2.977·5.2/√15 = 3.997

Interval

4.5 ± 3.997 → [0.503, 8.497]

Answer

upper limit = 8.497 mmHg
8.Confidence intervals

Answer: n = 599

Confidence interval for a mean: x̄ ± margin of error. With a known population σ the margin is z·σ/√n; with only the sample standard deviation s, use t with n − 1 degrees of freedom: t·s/√n. Critical values: z = 1.645 (90%), 1.96 (95%), 2.576 (99%). To get a margin E or smaller, you need n ≥ (z·σ/E)², rounded up.

Solve for n

n = (2.576·38/4)² = 598.9

Round up

A sample size is a whole number, and rounding down would make the margin too large.

n = 599

Answer

n = 599
9.Hypothesis tests & confidence intervals

Answer: t = 3.204

Hypothesis test: the test statistic = (estimate − value under H₀)/(standard error under H₀). The p-value is the probability, assuming H₀ is true, of a statistic at least as extreme as the one observed: the area in one tail for a one-sided H₁, twice the tail area for a two-sided H₁. Reject H₀ when p ≤ α.

Pooled variance

sp² = (6·4² + 13·11²)/19 = 87.84
sp = 9.372

Standard error

SE = 9.372·√(1/7 + 1/14) = 4.339

Test statistic

t = (63 − 49.1)/4.339 = 3.204

Answer

t = 3.204

With df = 19, the two-sided 5% critical value is 2.093; |t| is larger, so H₀ is rejected.

10.Hypothesis tests & confidence intervals

Answer: L = 70.11

Testing a mean: the test statistic measures how many standard errors the sample mean lies from the value claimed in H₀: z = (x̄ − μ₀)/(σ/√n) when σ is known, t = (x̄ − μ₀)/(s/√n) when it is estimated from the sample. A confidence interval is x̄ ± (critical value)·(standard error).

Margin of error

E = 1.645·8/√36 = 2.193

Interval limit

L = 72.3 − 2.193 = 70.11

Answer

L = 70.11
11.Linear regression & correlation

Answer: r = 0.9535

Least-squares regression: with x̄ and ȳ the means, Sxx = Σ(x − x̄)², Syy = Σ(y − ȳ)² and Sxy = Σ(x − x̄)(y − ȳ), the line ŷ = a + b·x has slope b = Sxy/Sxx and intercept a = ȳ − b·x̄, so it always passes through (x̄, ȳ). Pearson's correlation coefficient r = Sxy/√(Sxx·Syy) lies between −1 and 1; r² is the fraction of the variation in y explained by the line.

Centred sums

Sxy = 1100 − 21·293/6 = 74.5
Sxx = 91 − 21²/6 = 17.5
Syy = 14657 − 293²/6 = 348.8

Correlation coefficient

r = 74.5/√(17.5·348.8) = 0.9535

Answer

r = 0.9535
12.Linear regression & correlation

Answer: e = 2.6

Least-squares regression: with x̄ and ȳ the means, Sxx = Σ(x − x̄)², Syy = Σ(y − ȳ)² and Sxy = Σ(x − x̄)(y − ȳ), the line ŷ = a + b·x has slope b = Sxy/Sxx and intercept a = ȳ − b·x̄, so it always passes through (x̄, ȳ). Pearson's correlation coefficient r = Sxy/√(Sxx·Syy) lies between −1 and 1; r² is the fraction of the variation in y explained by the line.

Means

x̄ = 30/5 = 6
ȳ = 90/5 = 18

Sums of squares and products

Sxx = Σ(x − x̄)² = 40
Sxy = Σ(x − x̄)(y − ȳ) = 104

Regression line

b = 104/40 = 2.6
a = 18 − 2.6·6 = 2.4
ŷ = 2.4 + 2.6·x

Predicted value

ŷ = 2.4 + 2.6·5 = 15.4

Residual

e = 18 − 15.4 = 2.6

Answer

e = 2.6

The point lies above the line.

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