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
152025303540

Find the sample standard deviation s of the data: 18, 28, 43, 19, 33, 13, 33.

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

A quantity is normally distributed with mean 70 and standard deviation 5. What proportion P of values is above 73?

μ = 70σ = 5x = 73
P =
3.ProbabilityMedium
xp(x)00.310.3520.35μ = 1.05

For the distribution shown, find the variance Var(X) of X.

x: 0, 1, 2p(x): 0.3, 0.35, 0.35
Var(X) =
4.ProbabilityHard
€p(x)−100.83400.151900.02

A game costs €10 to play. You win €50 with probability 0.15 and €200 with probability 0.02; otherwise you win nothing. The stake is not returned. Find the expected net gain E(G) per game.

stake = €10€50: p = 0.15€200: p = 0.02
E(G) = €
5.Normal distributionMedium
−3−2−10123zshaded area to the right of z = 0.0359

Drag z so that the shaded area under the standard normal curve, to the right of z, is 0.0359. Find z to two decimals.

A = 0.0359
z =
6.Normal distributionHard
235240245250255260265gbottom 80%

The bag weights are normally distributed with mean 250 g and standard deviation 5 g. Drag the cut-off so the shaded bottom 80% is to the left of it. Find the cut-off x.

μ = 250 gσ = 5 g80%
x = g
7.Confidence intervalsMedium
μ₁ − μ₂0LUx̄₁ − x̄₂E = ?95%

The battery life of two phone models is compared. Model 1: 100 phones, mean 134.5 h, known standard deviation 14 h. Model 2: 80 phones, mean 145.5 h, known standard deviation 4 h. Find the margin of error E of the 95% confidence interval for μ₁ − μ₂.

n₁ = 100x̄₁ = 134.5 hσ₁ = 14 hn₂ = 80x̄₂ = 145.5 hσ₂ = 4 h95%
E = h
8.Confidence intervalsHard
p ≈ 0.3, n = ?p̂ − Ep̂ + Ep̂E = 0.0190%

A pilot study suggests that about 0.3 of products are defective. How many products n must be inspected so that a 90% confidence interval for the proportion has a margin of error of at most 0.01?

p ≈ 0.3E = 0.0190%
n =
9.Hypothesis tests & confidence intervalsMedium
t −1.99801.998

A sample of 64 students has a mean score of 501.3 and a sample standard deviation of 15. Find the test statistic t for H₀: μ = 500.

μ₀ = 500x̄ = 501.3s = 15n = 64
t =
10.Hypothesis tests & confidence intervalsHard
μ₀x̄c

A one-sided z test of H₀: μ = 50 against H₁: μ > 50 is done at α = 0.01 with a sample of 100; the standard deviation is known to be 20. In reality the mean is 57.2. Find the power 1 − β of the test in percent.

μ₀ = 50μ = 57.2σ = 20n = 100α = 0.01
1 − β = %
11.Linear regression & correlationMedium
xy35455876296511801377xy510607080

The table gives 6 paired observations of x = hours studied and y = exam score. Find Pearson's correlation coefficient r.

n = 6
r =
12.Linear regression & correlationHard
xy127225327422523619xy2462025x = 5.5

The table gives 6 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 (5.5, 19) is made. Find its residual e = y − ŷ.

n = 6(5.5, 19)
e =

3Solutions

1.Descriptive statistics

Answer: s = 10.56

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 = 187/7 = 26.71

Squared deviations

Σ(x − x̄)² = (−8.714)² + 1.286² + 16.29² + (−7.714)² + 6.286² + (−13.71)² + 6.286² = 669.4

Standard deviation

s = √(669.4/6) = 10.56

Answer

s = 10.56
2.Descriptive statistics

Answer: P = 0.2743

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 = 0.6

Standard normal table

Φ(0.6) = 0.7257
P = 1 − 0.7257 = 0.2743

Answer

P = 0.2743
3.Probability

Answer: Var(X) = 0.6475

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 = 0·0.3 + 1·0.35 + 2·0.35 = 1.05

Mean of the squares

E(X²) = 0·0.3 + 1·0.35 + 4·0.35 = 1.75

Variance

Var(X) = 1.75 − 1.05² = 0.6475

Answer

Var(X) = 0.6475
4.Probability

Answer: E(G) = 1.5 €

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.

Net outcomes

G: 40 (p = 0.15), 190 (p = 0.02), −10 (p = 0.83)

Expected value

E(G) = 40·0.15 + 190·0.02 − 10·0.83 = 1.5 €

Answer

E(G) = 1.5 €

A positive value means you win money on average.

5.Normal distribution

Answer: z = 1.8

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 of z

Φ(z) = 1 − 0.0359 = 0.9641

Read the z-table backwards

Find 0.9641 inside the table and read off z.

z = 1.80

Answer

z = 1.80
6.Normal distribution

Answer: x = 254.2 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) = 0.8

z from the table

z ≈ 0.84

Back to the original units

x = 250 + 0.84·5 = 254.2 g

Answer

x = 254.2 g
7.Confidence intervals

Answer: E = 2.881 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 = 134.5 − 145.5 = −11

Standard error

SE = √(14²/100 + 4²/80) = 1.47

Margin of error

E = 1.96·1.47 = 2.881

Answer

E = 2.881 h

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

8.Confidence intervals

Answer: n = 5683

Confidence interval for a proportion: p̂ ± z·√(p̂(1 − p̂)/n), where p̂ = x/n. To reach a margin E you need n ≥ z²·p(1 − p)/E², rounded up; without a prior estimate use p = 0.5, which gives the largest n.

Solve for n

n = 1.645²·0.3·0.7/0.01² = 5683

Round up

n = 5683

Answer

n = 5683
9.Hypothesis tests & confidence intervals

Answer: t = 0.6933

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

SE = 15/√64 = 15/8 = 1.875

Test statistic

t = (501.3 − 500)/1.875 = 0.6933

Answer

t = 0.6933
t0.025; df = 63 = 1.998

|t| < 1.998, so H₀ is not rejected at the 5% level (two-sided).

10.Hypothesis tests & confidence intervals

Answer: 1 − β = 89.87 %

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 = 20/√100 = 2
x̄c = 50 + 2.326·2 = 54.65

z under the true mean

z = (54.65 − 57.2)/2 = −1.274

Power

power = P(Z > −1.274) = 89.87%
1 − β = 89.87%

Answer

1 − β = 89.87%
11.Linear regression & correlation

Answer: r = 0.9446

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 = 396/6 = 66

Sums of squares and products

Sxx = Σ(x − x̄)² = 70
Sxy = Σ(x − x̄)(y − ȳ) = 184
Syy = Σ(y − ȳ)² = 542

Correlation coefficient

r = 184/√(70·542) = 0.9446

Answer

r = 0.9446
12.Linear regression & correlation

Answer: e = −1.919

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 = 21/6 = 3.5
ȳ = 143/6 = 23.83

Sums of squares and products

Sxx = Σ(x − x̄)² = 17.5
Sxy = Σ(x − x̄)(y − ȳ) = −25.5

Regression line

b = −25.5/17.5 = −1.457
a = 23.83 − (−1.457)·3.5 = 28.93
ŷ = 28.93 − 1.457·x

Predicted value

ŷ = 28.93 + (−1.457)·5.5 = 20.92

Residual

e = 19 − 20.92 = −1.919

Answer

e = −1.919

The point lies below the line.

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