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
3035404550

Find the sample standard deviation s of the data: 54, 36, 47, 27, 35.

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

A quantity is normally distributed with mean 70 and standard deviation 20. What proportion P of values is above 54?

μ = 70σ = 20x = 54
P =
3.ProbabilityMedium
xp(x)00.1510.3520.1530.35μ = 1.7

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

x: 0, 1, 2, 3p(x): 0.15, 0.35, 0.15, 0.35
σ =
4.ProbabilityHard
xp(x)100.35200.1300.3400.25μ

X has the distribution shown, and Y = 2X + 1. Find the standard deviation σY of Y.

x: 10, 20, 30, 40p(x): 0.35, 0.1, 0.3, 0.25Y = 2X + 1
σY =
5.Normal distributionMedium
440560P = ?

The battery lifetimes are normally distributed with mean 500 h and standard deviation 40 h. What fraction lies between 440 and 560 h? Give P as a decimal.

μ = 500σ = 40x1 = 440x2 = 560
P =
6.Normal distributionHard
235240245250255260265gbottom 70%

The bag weights are normally distributed with mean 250 g and standard deviation 5 g. Drag the cut-off so the shaded bottom 70% is to the left of it. Find the cut-off x.

μ = 250 gσ = 5 g70%
x = g
7.Confidence intervalsMedium
μ₁ − μ₂0LUx̄₁ − x̄₂E = ?90%

The battery life of two phone models is compared. Model 1: 60 phones, mean 91 h, known standard deviation 5 h. Model 2: 60 phones, mean 103 h, known standard deviation 11 h. Find the margin of error E of the 90% confidence interval for μ₁ − μ₂.

n₁ = 60x̄₁ = 91 hσ₁ = 5 hn₂ = 60x̄₂ = 103 hσ₂ = 11 h90%
E = h
8.Confidence intervalsHard
μ₁ − μ₂, t(19) = 1.7290LU = ?x̄₁ − x̄₂E90%

Two fertilisers are tested. Plot group 1: 8 plots, mean yield 85.5 kg, standard deviation 8 kg. Plot group 2: 13 plots, mean 82 kg, standard deviation 7 kg. Assume equal variances (pooled t, df = n₁ + n₂ − 2). Find the upper limit of the 90% confidence interval for μ₁ − μ₂.

n₁ = 8x̄₁ = 85.5s₁ = 8n₂ = 13x̄₂ = 82s₂ = 790%t(19) = 1.729
upper limit = kg
9.Hypothesis tests & confidence intervalsMedium
−2.1792.179t = ?df = 12

13 employees take a typing test before and after a training course. For each person the difference d = after − before (words per minute) is recorded. The mean difference is 10.49 and the standard deviation of the differences is 9.7. Find the paired test statistic t for H₀: μd = 0.

n = 13d̄ = 10.49sd = 9.7
t =
10.Hypothesis tests & confidence intervalsHard
−1.961.96z = ?

In a trial, 30 of 100 patients on treatment A recover, and 20 of 80 patients on treatment B. Using the pooled proportion, find the test statistic z for H₀: p₁ = p₂.

x₁ = 30n₁ = 100x₂ = 20n₂ = 80
z =
11.Linear regression & correlationMedium
xy144617502070237426922998xy152025306080100

The table gives 6 paired observations of x = temperature in °C and y = ice creams sold. Their sums are given. Find Pearson's correlation coefficient r from these sums.

n = 6Σx = 129Σy = 430Σx² = 2931Σy² = 33060Σxy = 9830
r =
12.Linear regression & correlationHard
xy127224325422519xy242025x = 3.5

The table gives 5 paired observations of x = car age in years and y = price in k€. Use the least-squares line ŷ = a + b·x. A new observation (3.5, 25) is made. Find its residual e = y − ŷ.

n = 5(3.5, 25)
e =

3Solutions

1.Descriptive statistics

Answer: s = 10.66

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 = 199/5 = 39.8

Squared deviations

Σ(x − x̄)² = 14.2² + (−3.8)² + 7.2² + (−12.8)² + (−4.8)² = 454.8

Standard deviation

s = √(454.8/4) = 10.66

Answer

s = 10.66
2.Descriptive statistics

Answer: P = 0.7881

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 = −0.8

Standard normal table

Φ(−0.8) = 0.2119
P = 1 − 0.2119 = 0.7881

Answer

P = 0.7881
3.Probability

Answer: σ = 1.1

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 = 0·0.15 + 1·0.35 + 2·0.15 + 3·0.35 = 1.7

Mean of the squares

E(X²) = 0·0.15 + 1·0.35 + 4·0.15 + 9·0.35 = 4.1

Variance

Var(X) = 4.1 − 1.7² = 1.21
σ = √1.21 = 1.1

Answer

σ = 1.1
4.Probability

Answer: σY = 24.06

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.

Spread of X

μ = 24.5, Var(X) = 144.8, σX = 12.03

Linear change

σY = |2|·12.03 = 24.06

Answer

σY = 24.06

The mean does move: E(Y) = 2·24.5 + 1 = 50.

5.Normal distribution

Answer: P = 0.8664

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 = (440 − 500)/40 = −1.5
z₂ = (560 − 500)/40 = 1.5

Area between

P = Φ(1.5) − Φ(−1.5) = 0.9332 − 0.06681 = 0.8664

Answer

P = 0.8664
6.Normal distribution

Answer: x = 252.6 g

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) = 0.7

z from the table

z ≈ 0.52

Back to the original units

x = 250 + 0.52·5 = 252.6 g

Answer

x = 252.6 g
7.Confidence intervals

Answer: E = 2.566 h

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.

Point estimate

x̄₁ − x̄₂ = 91 − 103 = −12

Standard error

SE = √(5²/60 + 11²/60) = 1.56

Margin of error

E = 1.645·1.56 = 2.566

Answer

E = 2.566 h

The interval does not contain 0, so the means differ significantly.

8.Confidence intervals

Answer: upper limit = 9.237 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 = (7·8² + 12·7²)/19 = 54.53
sp = 7.384

Margin of error

E = 1.729·7.384·√(1/8 + 1/13) = 5.737

Interval

3.5 ± 5.737 → [−2.237, 9.237]

Answer

upper limit = 9.237 kg
9.Hypothesis tests & confidence intervals

Answer: t = 3.899

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).

Standard error of the mean difference

SE = 9.7/√13 = 2.69

Test statistic

t = 10.49/2.69 = 3.899

Answer

t = 3.899

With df = 12, the two-sided 5% critical value is 2.179; |t| is larger, so H₀ is rejected.

10.Hypothesis tests & confidence intervals

Answer: z = 0.7442

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.

Sample proportions

p̂₁ = 30/100 = 0.3
p̂₂ = 20/80 = 0.25
p̂ = 50/180 = 0.2778

Standard error under H₀

SE = √(0.2778·0.7222·(1/100 + 1/80)) = 0.06719

Test statistic

z = (0.3 − 0.25)/0.06719 = 0.7442

Answer

z = 0.7442

|z| < 1.96, so H₀ is not rejected at the 5% level (two-sided).

11.Linear regression & correlation

Answer: r = 0.9842

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 = 9830 − 129·430/6 = 585
Sxx = 2931 − 129²/6 = 157.5
Syy = 33060 − 430²/6 = 2243

Correlation coefficient

r = 585/√(157.5·2243) = 0.9842

Answer

r = 0.9842
12.Linear regression & correlation

Answer: e = 2.5

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 = 15/5 = 3
ȳ = 117/5 = 23.4

Sums of squares and products

Sxx = Σ(x − x̄)² = 10
Sxy = Σ(x − x̄)(y − ȳ) = −18

Regression line

b = −18/10 = −1.8
a = 23.4 − (−1.8)·3 = 28.8
ŷ = 28.8 − 1.8·x

Predicted value

ŷ = 28.8 + (−1.8)·3.5 = 22.5

Residual

e = 25 − 22.5 = 2.5

Answer

e = 2.5

The point lies above the line.

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