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
2025303540455055

Find the sample standard deviation s of the data: 40, 22, 16, 40, 55, 17, 38.

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

Exam scores are normally distributed with mean 60 and standard deviation 10. Find the z-score z of a score of 75.

μ = 60σ = 10x = 75
z =
3.ProbabilityMedium
xp(x)100.2200.3300.15400.35μ = 26.5

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

x: 10, 20, 30, 40p(x): 0.2, 0.3, 0.15, 0.35
σ =
4.ProbabilityHard
xp(x)100.35200.15300.15400.35μ

X has the distribution shown, and Y = 3X + 10. Find the standard deviation σY of Y.

x: 10, 20, 30, 40p(x): 0.35, 0.15, 0.15, 0.35Y = 3X + 10
σY =
5.Normal distributionMedium
247.75252.75P = ?

The bag weights are normally distributed with mean 250 g and standard deviation 5 g. What fraction lies between 247.75 and 252.75 g? Give P as a decimal.

μ = 250σ = 5x1 = 247.75x2 = 252.75
P =
6.Normal distributionHard
0.03 mm0D − d?

Hole diameters D are normal with mean 25.03 mm and standard deviation 0.015 mm. Shaft diameters d are independent and normal with mean 25 mm and standard deviation 0.01 mm. A randomly chosen shaft fits when D − d > 0. Find the probability P that it fits. Round z to two decimals.

hole: μ = 25.03 mm, σ = 0.015 mmshaft: μ = 25 mm, σ = 0.01 mm
P =
7.Confidence intervalsMedium
μ₁ − μ₂0LU = ?x̄₁ − x̄₂E90%

The battery life of two phone models is compared. Model 1: 50 phones, mean 41.5 h, known standard deviation 15 h. Model 2: 50 phones, mean 43 h, known standard deviation 4 h. Find the upper limit of the 90% confidence interval for μ₁ − μ₂.

n₁ = 50x̄₁ = 41.5 hσ₁ = 15 hn₂ = 50x̄₂ = 43 hσ₂ = 4 h90%
upper limit = h
8.Confidence intervalsHard
n = 21σ = ?Us = 8.395%

The diameters of 21 machined shafts have a sample standard deviation of 8.3 µm. Assuming the diameters are normally distributed, find the lower limit σ of the 95% confidence interval for the population standard deviation. With df = 20, 2.5% of the χ² distribution lies below 9.591 and 2.5% lies above 34.17.

n = 21s = 8.3 µmχ² = 9.591, 34.17
σ = µm
9.Hypothesis tests & confidence intervalsMedium
t −2.01102.011

A sample of 49 students has a mean score of 98.97 and a sample standard deviation of 8. Find the test statistic t for H₀: μ = 100.

μ₀ = 100x̄ = 98.97s = 8n = 49
t =
10.Hypothesis tests & confidence intervalsHard
μ₀x̄c

A one-sided z test of H₀: μ = 80 against H₁: μ > 80 is done at α = 0.05 with a sample of 25; the standard deviation is known to be 7.5. In reality the mean is 82.4. Find the power 1 − β of the test in percent.

μ₀ = 80μ = 82.4σ = 7.5n = 25α = 0.05
1 − β = %
11.Linear regression & correlationMedium
xy330440537649746856xy468304050

The table gives 6 paired observations of x = advertising spend in k€ and y = sales in k€. Find Pearson's correlation coefficient r.

n = 6
r =
12.Linear regression & correlationHard
xy128225322424520xy242025x = 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, 29) is made. Find its residual e = y − ŷ.

n = 5(2.5, 29)
e =

3Solutions

1.Descriptive statistics

Answer: s = 14.56

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 = 228/7 = 32.57

Squared deviations

Σ(x − x̄)² = 7.429² + (−10.57)² + (−16.57)² + 7.429² + 22.43² + (−15.57)² + 5.429² = 1272

Standard deviation

s = √(1272/6) = 14.56

Answer

s = 14.56
2.Descriptive statistics

Answer: z = 1.5

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 = (75 − 60)/10 = 1.5

Answer

z = 1.5
3.Probability

Answer: σ = 11.52

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 = 10·0.2 + 20·0.3 + 30·0.15 + 40·0.35 = 26.5

Mean of the squares

E(X²) = 100·0.2 + 400·0.3 + 900·0.15 + 1600·0.35 = 835

Variance

Var(X) = 835 − 26.5² = 132.8
σ = √132.8 = 11.52

Answer

σ = 11.52
4.Probability

Answer: σY = 38.54

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 = 25, Var(X) = 165, σX = 12.85

Linear change

σY = |3|·12.85 = 38.54

Answer

σY = 38.54

The mean does move: E(Y) = 3·25 + 10 = 85.

5.Normal distribution

Answer: P = 0.3825

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 = (247.75 − 250)/5 = −0.45
z₂ = (252.75 − 250)/5 = 0.55

Area between

P = Φ(0.55) − Φ(−0.45) = 0.7088 − 0.3264 = 0.3825

Answer

P = 0.3825
6.Normal distribution

Answer: P = 0.9515

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 = 25.03 − 25 = 0.03 mm, σ = 0.01803 mm

Standardise

z = (0 − 0.03)/0.01803 = −1.66

Probability

P = 1 − Φ(−1.66) = 0.9515

Answer

P = 0.9515
7.Confidence intervals

Answer: upper limit = 2.112 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 = 41.5 − 43 = −1.5

Standard error

SE = √(15²/50 + 4²/50) = 2.195

Margin of error

E = 1.645·2.195 = 3.612

Interval

−1.5 ± 3.612 → [−5.112, 2.112]

Answer

upper limit = 2.112 h

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

8.Confidence intervals

Answer: σ = 6.35 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 = 20·8.3² = 1378

Limits for σ²

1378/34.17 = 40.32
1378/9.591 = 143.7

Limit for σ

σ = √40.32 = 6.35 µm

Answer

σ = 6.35 µm

95% interval for σ: [6.35, 11.99] µm.

9.Hypothesis tests & confidence intervals

Answer: t = −0.9013

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 = 8/√49 = 8/7 = 1.143

Test statistic

t = (98.97 − 100)/1.143 = −0.9013

Answer

t = −0.9013
t0.025; df = 48 = 2.011

|t| < 2.011, so H₀ is not rejected at the 5% level (two-sided).

10.Hypothesis tests & confidence intervals

Answer: 1 − β = 48.21 %

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 = 7.5/√25 = 1.5
x̄c = 80 + 1.645·1.5 = 82.47

z under the true mean

z = (82.47 − 82.4)/1.5 = 0.045

Power

power = P(Z > 0.045) = 48.21%
1 − β = 48.21%

Answer

1 − β = 48.21%
11.Linear regression & correlation

Answer: r = 0.9244

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 = 33/6 = 5.5
ȳ = 258/6 = 43

Sums of squares and products

Sxx = Σ(x − x̄)² = 17.5
Sxy = Σ(x − x̄)(y − ȳ) = 80
Syy = Σ(y − ȳ)² = 428

Correlation coefficient

r = 80/√(17.5·428) = 0.9244

Answer

r = 0.9244
12.Linear regression & correlation

Answer: e = 4.35

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 = 119/5 = 23.8

Sums of squares and products

Sxx = Σ(x − x̄)² = 10
Sxy = Σ(x − x̄)(y − ȳ) = −17

Regression line

b = −17/10 = −1.7
a = 23.8 − (−1.7)·3 = 28.9
ŷ = 28.9 − 1.7·x

Predicted value

ŷ = 28.9 + (−1.7)·2.5 = 24.65

Residual

e = 29 − 24.65 = 4.35

Answer

e = 4.35

The point lies above the line.

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