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
25303540455055

Find the sample standard deviation s of the data: 23, 43, 45, 58, 42.

n = 5
s =
2.Descriptive statisticsHard
z −2−1012

A quantity is normally distributed with mean 50 and standard deviation 10. What proportion P of values is above 38?

μ = 50σ = 10x = 38
P =
3.ProbabilityMedium
xp(x)100.25200.2300.2400.35μ = 26.5

For the distribution shown, find the standard deviation σ of X.

x: 10, 20, 30, 40p(x): 0.25, 0.2, 0.2, 0.35
σ =
4.ProbabilityHard
n = 11, p = 0.2: P(X ≤ 1) = ?k01234567891011

11 components are tested, and each one fails with probability 0.2, independently. What is the probability P that at most one fails?

n = 11p = 0.2
P =
5.Normal distributionMedium
200 kg199.3 kgT = X₁ + … + X₈?

The masses of bags of cement are independent and normal with mean 25 kg and standard deviation 0.5 kg. Find the probability P that the total mass of 8 of them exceeds 199.3 kg. Round z to two decimals.

μ = 25 kgσ = 0.5 kgn = 8> 199.3 kg
P =
6.Normal distributionHard
557085100115130145pointsbottom 90%

The IQ scores are normally distributed with mean 100 points and standard deviation 15 points. Drag the cut-off so the shaded bottom 90% is to the left of it. Find the cut-off x.

μ = 100 pointsσ = 15 points90%
x = points
7.Confidence intervalsMedium
n = 6, sd​ = 3.70LUd̄ = −6.5E = ?90%

The blood pressure of 6 patients is measured before and after a treatment. The differences d = before − after (mmHg) have mean −6.5 and standard deviation 3.7. Find the margin of error E of the 90% confidence interval for the mean difference.

n = 6d̄ = −6.5sd = 3.790%t(5) = 2.015
E = mmHg
8.Confidence intervalsHard
μ₁ − μ₂, t(19) = 1.7290LUx̄₁ − x̄₂E = ?90%

Two fertilisers are tested. Plot group 1: 10 plots, mean yield 31.5 kg, standard deviation 9.5 kg. Plot group 2: 11 plots, mean 22.5 kg, standard deviation 8.5 kg. Assume equal variances (pooled t, df = n₁ + n₂ − 2). Find the margin of error E of the 90% confidence interval for μ₁ − μ₂.

n₁ = 10x̄₁ = 31.5s₁ = 9.5n₂ = 11x̄₂ = 22.5s₂ = 8.590%t(19) = 1.729
E = kg
9.Hypothesis tests & confidence intervalsMedium
−2.0642.064t = ?df = 24

Two teaching methods are compared. Group 1: 10 students, mean score 90, standard deviation 5. Group 2: 16 students, mean score 87.8, standard deviation 7.5. 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 = 10x̄₁ = 90s₁ = 5n₂ = 16x̄₂ = 87.8s₂ = 7.5
t =
10.Hypothesis tests & confidence intervalsHard
μ₀x̄c

A one-sided z test of H₀: μ = 500 against H₁: μ > 500 is done at α = 0.10 with a sample of 100; the standard deviation is known to be 5. In reality the mean is 500.65. Find the power 1 − β of the test in percent.

μ₀ = 500μ = 500.65σ = 5n = 100α = 0.10
1 − β = %
11.Linear regression & correlationMedium
xy274761281510221223xy510101520

The table gives 6 paired observations of x = load in kN and y = extension in mm. Find Pearson's correlation coefficient r.

n = 6
r =
12.Linear regression & correlationHard
xy34755776497211781381xy5106080ŷ = 60

The table gives 6 paired observations of x = advertising spend in k€ and y = sales in k€. Use the least-squares line ŷ = a + b·x. For which value x does the line predict ŷ = 60?

n = 6ŷ = 60
x =

3Solutions

1.Descriptive statistics

Answer: s = 12.52

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 = 211/5 = 42.2

Squared deviations

Σ(x − x̄)² = (−19.2)² + 0.8² + 2.8² + 15.8² + (−0.2)² = 626.8

Standard deviation

s = √(626.8/4) = 12.52

Answer

s = 12.52
2.Descriptive statistics

Answer: P = 0.8849

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 = −1.2

Standard normal table

Φ(−1.2) = 0.1151
P = 1 − 0.1151 = 0.8849

Answer

P = 0.8849
3.Probability

Answer: σ = 11.95

Discrete random variables: a probability distribution lists every value x a variable can take with its probability p(x); the probabilities are between 0 and 1 and add up to 1. The expected value (mean) is E(X) = Σx·p(x), the long-run average. The variance is Var(X) = Σ(x − μ)²·p(x) = E(X²) − μ², and the standard deviation σ = √Var(X). For a linear change Y = aX + b: E(Y) = a·E(X) + b, Var(Y) = a²·Var(X) and σY = |a|·σX; adding a constant does not change the spread.

Mean

μ = 10·0.25 + 20·0.2 + 30·0.2 + 40·0.35 = 26.5

Mean of the squares

E(X²) = 100·0.25 + 400·0.2 + 900·0.2 + 1600·0.35 = 845

Variance

Var(X) = 845 − 26.5² = 142.8
σ = √142.8 = 11.95

Answer

σ = 11.95
4.Probability

Answer: P = 0.3221

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.

Zero and one failure

P(X = 0) = 0.811 = 0.0859
P(X = 1) = 11·0.2·0.810 = 0.2362

Add them

P = 0.0859 + 0.2362 = 0.3221

Answer

P = 0.3221
5.Normal distribution

Answer: P = 0.6879

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 total

μT = 200 kg, σT = 0.5·√8 = 1.414 kg

Standardise

z = (199.3 − 200)/1.414 = −0.49

Probability

P = 1 − Φ(−0.49) = 0.6879

Answer

P = 0.6879
6.Normal distribution

Answer: x = 119.2 points

From an area back to z: Φ(z) is the area to the left of z. For a left-tail area A, find z with Φ(z) = A in the table. A right-tail area A means Φ(z) = 1 − A. A central area A between −z and z leaves (1 − A)/2 in each tail, so Φ(z) = (1 + A)/2. For a value x, use x = μ + zσ.

Area to the left

Φ(z) = 0.9

z from the table

z ≈ 1.28

Back to the original units

x = 100 + 1.28·15 = 119.2 points

Answer

x = 119.2 points
7.Confidence intervals

Answer: E = 3.044 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 = 5, t = 2.015

Margin of error

E = 2.015·3.7/√6 = 3.044

Answer

E = 3.044 mmHg
8.Confidence intervals

Answer: E = 6.79 kg

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.

Pooled standard deviation

sp² = (9·9.5² + 10·8.5²)/19 = 80.78
sp = 8.988

Margin of error

E = 1.729·8.988·√(1/10 + 1/11) = 6.79

Answer

E = 6.79 kg
9.Hypothesis tests & confidence intervals

Answer: t = 0.8178

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 = (9·5² + 15·7.5²)/24 = 44.53
sp = 6.673

Standard error

SE = 6.673·√(1/10 + 1/16) = 2.69

Test statistic

t = (90 − 87.8)/2.69 = 0.8178

Answer

t = 0.8178

With df = 24, the two-sided 5% critical value is 2.064; |t| is smaller, so H₀ is not rejected.

10.Hypothesis tests & confidence intervals

Answer: 1 − β = 50.72 %

Power of a test: power = 1 − β is the probability of rejecting H₀ when it is false. Find the cutoff x̄c from μ₀ and z_α, then compute the probability of landing beyond x̄c using the true mean: z = (x̄c − μ)/(σ/√n).

Critical sample mean

SE = 5/√100 = 0.5
x̄c = 500 + 1.282·0.5 = 500.6

z under the true mean

z = (500.6 − 500.65)/0.5 = −0.018

Power

power = P(Z > −0.018) = 50.72%
1 − β = 50.72%

Answer

1 − β = 50.72%
11.Linear regression & correlation

Answer: r = 0.9728

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 = 42/6 = 7
ȳ = 86/6 = 14.33

Sums of squares and products

Sxx = Σ(x − x̄)² = 70
Sxy = Σ(x − x̄)(y − ȳ) = 128
Syy = Σ(y − ȳ)² = 247.3

Correlation coefficient

r = 128/√(70·247.3) = 0.9728

Answer

r = 0.9728
12.Linear regression & correlation

Answer: x = 6.112

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 = 48/6 = 8
ȳ = 399/6 = 66.5

Sums of squares and products

Sxx = Σ(x − x̄)² = 70
Sxy = Σ(x − x̄)(y − ȳ) = 241

Regression line

b = 241/70 = 3.443
a = 66.5 − 3.443·8 = 38.96
ŷ = 38.96 + 3.443·x

Solve for x

x = (60 − 38.96)/3.443 = 6.112

Answer

x = 6.112

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