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
1520253035404550

Find the sample standard deviation s of the data: 32, 24, 11, 53, 22, 41.

n = 6
s =
2.Descriptive statisticsHard
z −2−1012

A quantity is normally distributed with mean 170 and standard deviation 15. What proportion P of values is below 200?

μ = 170σ = 15x = 200
P =
3.ProbabilityMedium
xp(x)00.0510.320.2530.2540.15μ = 2.15

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

x: 0, 1, 2, 3, 4p(x): 0.05, 0.3, 0.25, 0.25, 0.15
Var(X) =
4.ProbabilityHard
€p(x)−50.65100.3950.05

A game costs €5 to play. You win €15 with probability 0.3 and €100 with probability 0.05; otherwise you win nothing. The stake is not returned. Find the expected net gain E(G) per game.

stake = €5€15: p = 0.3€100: p = 0.05
E(G) = €
5.Normal distributionMedium
22.442.8P = ?

The delivery times are normally distributed with mean 32 min and standard deviation 6 min. What fraction lies between 22.4 and 42.8 min? Give P as a decimal.

μ = 32σ = 6x1 = 22.4x2 = 42.8
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
μ₁ − μ₂0LU = ?x̄₁ − x̄₂E95%

The battery life of two phone models is compared. Model 1: 60 phones, mean 145.5 h, known standard deviation 15 h. Model 2: 40 phones, mean 147 h, known standard deviation 20 h. Find the upper limit of the 95% confidence interval for μ₁ − μ₂.

n₁ = 60x̄₁ = 145.5 hσ₁ = 15 hn₂ = 40x̄₂ = 147 hσ₂ = 20 h95%
upper limit = h
8.Confidence intervalsHard
n = 10Lσ = ?s = 9.195%

The diameters of 10 machined shafts have a sample standard deviation of 9.1 µm. Assuming the diameters are normally distributed, find the upper limit σ of the 95% confidence interval for the population standard deviation. With df = 9, 2.5% of the χ² distribution lies below 2.7 and 2.5% lies above 19.023.

n = 10s = 9.1 µmχ² = 2.7, 19.023
σ = µm
9.Hypothesis tests & confidence intervalsMedium
t −2.01102.011

A sample of 49 students has a mean score of 194.7 and a sample standard deviation of 12. Find the test statistic t for H₀: μ = 200.

μ₀ = 200x̄ = 194.7s = 12n = 49
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 64; the standard deviation is known to be 24. In reality the mean is 39.5. Find the power 1 − β of the test in percent.

μ₀ = 50μ = 39.5σ = 24n = 64α = 0.01
1 − β = %
11.Linear regression & correlationMedium
xy2424446448501059xy5105060

The table gives 5 paired observations of x = advertising spend in k€ and y = sales in k€. Find Pearson's correlation coefficient r.

n = 5
r =
12.Linear regression & correlationHard
xy125219321417514xy24152025x = 2.5

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 (2.5, 25) is made. Find its residual e = y − ŷ.

n = 5(2.5, 25)
e =

3Solutions

1.Descriptive statistics

Answer: s = 14.92

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 = 183/6 = 30.5

Squared deviations

Σ(x − x̄)² = 1.5² + (−6.5)² + (−19.5)² + 22.5² + (−8.5)² + 10.5² = 1114

Standard deviation

s = √(1114/5) = 14.92

Answer

s = 14.92
2.Descriptive statistics

Answer: P = 0.9772

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 = 2

Standard normal table

Φ(2) = 0.9772
P = 0.9772

Answer

P = 0.9772
3.Probability

Answer: Var(X) = 1.327

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 = 0·0.05 + 1·0.3 + 2·0.25 + 3·0.25 + 4·0.15 = 2.15

Mean of the squares

E(X²) = 0·0.05 + 1·0.3 + 4·0.25 + 9·0.25 + 16·0.15 = 5.95

Variance

Var(X) = 5.95 − 2.15² = 1.327

Answer

Var(X) = 1.327
4.Probability

Answer: E(G) = 4.5 €

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: 10 (p = 0.3), 95 (p = 0.05), −5 (p = 0.65)

Expected value

E(G) = 10·0.3 + 95·0.05 − 5·0.65 = 4.5 €

Answer

E(G) = 4.5 €

A positive value means you win money on average.

5.Normal distribution

Answer: P = 0.9093

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 = (22.4 − 32)/6 = −1.6
z₂ = (42.8 − 32)/6 = 1.8

Area between

P = Φ(1.8) − Φ(−1.6) = 0.9641 − 0.0548 = 0.9093

Answer

P = 0.9093
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 = 5.768 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 = 145.5 − 147 = −1.5

Standard error

SE = √(15²/60 + 20²/40) = 3.708

Margin of error

E = 1.96·3.708 = 7.268

Interval

−1.5 ± 7.268 → [−8.768, 5.768]

Answer

upper limit = 5.768 h

The interval contains 0, so the data do not show a significant difference.

8.Confidence intervals

Answer: σ = 16.61 µm

Interval for a variance: (n − 1)s²/σ² has a χ² distribution with n − 1 degrees of freedom. The 95% interval for σ² runs from (n − 1)s² divided by the upper χ² value (2.5% above it) to (n − 1)s² divided by the lower χ² value (2.5% below it); the interval for σ is the square root of both limits. It is not symmetric around s.

Sum of squares

(n − 1)s² = 9·9.1² = 745.3

Limits for σ²

745.3/19.023 = 39.18
745.3/2.7 = 276

Limit for σ

σ = √276 = 16.61 µm

Answer

σ = 16.61 µm

95% interval for σ: [6.259, 16.61] µm.

9.Hypothesis tests & confidence intervals

Answer: t = −3.092

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 = 12/√49 = 12/7 = 1.714

Test statistic

t = (194.7 − 200)/1.714 = −3.092

Answer

t = −3.092
t0.025; df = 48 = 2.011

|t| > 2.011, so H₀ is rejected at the 5% level (two-sided).

10.Hypothesis tests & confidence intervals

Answer: 1 − β = 87.98 %

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 = 24/√64 = 3
x̄c = 50 − 2.326·3 = 43.02

z under the true mean

z = (43.02 − 39.5)/3 = 1.174

Power

power = P(Z < 1.174) = 87.98%
1 − β = 87.98%

Answer

1 − β = 87.98%
11.Linear regression & correlation

Answer: r = 0.911

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 = 239/5 = 47.8

Sums of squares and products

Sxx = Σ(x − x̄)² = 40
Sxy = Σ(x − x̄)(y − ȳ) = 80
Syy = Σ(y − ȳ)² = 192.8

Correlation coefficient

r = 80/√(40·192.8) = 0.911

Answer

r = 0.911
12.Linear regression & correlation

Answer: e = 4.6

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 = 15/5 = 3
ȳ = 96/5 = 19.2

Sums of squares and products

Sxx = Σ(x − x̄)² = 10
Sxy = Σ(x − x̄)(y − ȳ) = −24

Regression line

b = −24/10 = −2.4
a = 19.2 − (−2.4)·3 = 26.4
ŷ = 26.4 − 2.4·x

Predicted value

ŷ = 26.4 + (−2.4)·2.5 = 20.4

Residual

e = 25 − 20.4 = 4.6

Answer

e = 4.6

The point lies above the line.

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