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: 57, 18, 55, 26, 32, 54, 50.

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

A quantity is normally distributed with mean 170 and standard deviation 12. What proportion P of values is below 146?

μ = 170σ = 12x = 146
P =
3.ProbabilityMedium
xp(x)00.3520.2540.2560.15μ = 2.4

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

x: 0, 2, 4, 6p(x): 0.35, 0.25, 0.25, 0.15
Var(X) =
4.ProbabilityHard
xp(x)00.3510.320.35μ

X has the distribution shown, and Y = 10X + 20. Find the standard deviation σY of Y.

x: 0, 1, 2p(x): 0.35, 0.3, 0.35Y = 10X + 20
σY =
5.Normal distributionMedium
−3−2−10123zshaded area to the right of z = 0.8023

Drag z so that the shaded area under the standard normal curve, to the right of z, is 0.8023. Find z to two decimals.

A = 0.8023
z =
6.Normal distributionHard
μ = 250x = ?99%

The bag weights are normally distributed with mean 250 g and standard deviation 5 g. Find the value x below which 99% of them lie (the 99th percentile).

μ = 250σ = 5P = 99%
x = g
7.Confidence intervalsMedium
x̄E₁ = 7.499%, n = 50x̄E₂ = ?99%, n = 200

A 99% confidence interval for a mean, based on 50 observations and a known σ, has a margin of error of 7.4. What is the margin of error E₂ if the same study uses 200 observations instead?

99%n₁ = 50E₁ = 7.4n₂ = 200
E₂ =
8.Confidence intervalsHard
n = 10Lσ = ?s = 9.495%

The diameters of 10 machined shafts have a sample standard deviation of 9.4 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 = 9.4 µmχ² = 2.7, 19.023
σ = µm
9.Hypothesis tests & confidence intervalsMedium
−2.0642.064t = ?df = 24

Two teaching methods are compared. Group 1: 12 students, mean score 58.5, standard deviation 10. Group 2: 14 students, mean score 63.7, standard deviation 2.5. 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 = 12x̄₁ = 58.5s₁ = 10n₂ = 14x̄₂ = 63.7s₂ = 2.5
t =
10.Hypothesis tests & confidence intervalsHard
H₀H₁μ₀x̄cμ₀ + 5α = 0.01βσ = 12, power 90%, n = ?

A one-sided z test at α = 0.01 must detect a shift of the mean of 5 units with a power of 90%. The standard deviation is 12. Find the minimum sample size n.

α = 0.01power = 90%δ = 5σ = 12zα = 2.326zβ = 1.282
n =
11.Linear regression & correlationMedium
xy248348454555659764xy24650556065

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
xy16351839204322542451xy204050x = 17

The table gives 5 paired observations of x = temperature in °C and y = ice creams sold. Use the least-squares line ŷ = a + b·x. A new observation (17, 41) is made. Find its residual e = y − ŷ.

n = 5(17, 41)
e =

3Solutions

1.Descriptive statistics

Answer: s = 15.99

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 = 292/7 = 41.71

Squared deviations

Σ(x − x̄)² = 15.29² + (−23.71)² + 13.29² + (−15.71)² + (−9.714)² + 12.29² + 8.286² = 1533

Standard deviation

s = √(1533/6) = 15.99

Answer

s = 15.99
2.Descriptive statistics

Answer: P = 0.02275

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) = 0.02275
P = 0.02275

Answer

P = 0.02275
3.Probability

Answer: Var(X) = 4.64

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 = 0·0.35 + 2·0.25 + 4·0.25 + 6·0.15 = 2.4

Mean of the squares

E(X²) = 0·0.35 + 4·0.25 + 16·0.25 + 36·0.15 = 10.4

Variance

Var(X) = 10.4 − 2.4² = 4.64

Answer

Var(X) = 4.64
4.Probability

Answer: σY = 8.367

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 = 1, Var(X) = 0.7, σX = 0.8367

Linear change

σY = |10|·0.8367 = 8.367

Answer

σY = 8.367

The mean does move: E(Y) = 10·1 + 20 = 30.

5.Normal distribution

Answer: z = −0.85

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 − 0.8023 = 0.1977

Read the z-table backwards

Find 0.1977 inside the table and read off z.

z = −0.85

Answer

z = −0.85
6.Normal distribution

Answer: x = 261.6 g

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 for the percentile

Φ(z) = 0.99 → z = 2.326

Back to x

x = 250 + 2.326·5 = 261.63 g

Answer

x = 261.63 g
7.Confidence intervals

Answer: E₂ = 3.7

Confidence interval for a mean: x̄ ± margin of error. With a known population σ the margin is z·σ/√n; with only the sample standard deviation s, use t with n − 1 degrees of freedom: t·s/√n. Critical values: z = 1.645 (90%), 1.96 (95%), 2.576 (99%). To get a margin E or smaller, you need n ≥ (z·σ/E)², rounded up.

Ratio of margins

E₂/E₁ = (2.576/2.576)·√(50/200) = 0.5

New margin

E₂ = 7.4·0.5 = 3.7

Answer

E₂ = 3.7
8.Confidence intervals

Answer: σ = 17.16 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 = 9·9.4² = 795.2

Limits for σ²

795.2/19.023 = 41.8
795.2/2.7 = 294.5

Limit for σ

σ = √294.5 = 17.16 µm

Answer

σ = 17.16 µm

95% interval for σ: [6.466, 17.16] µm.

9.Hypothesis tests & confidence intervals

Answer: t = −1.884

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 = (11·10² + 13·2.5²)/24 = 49.22
sp = 7.016

Standard error

SE = 7.016·√(1/12 + 1/14) = 2.76

Test statistic

t = (58.5 − 63.7)/2.76 = −1.884

Answer

t = −1.884

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: n = 75

Sample size for a test: to detect a true shift δ with power 1 − β in a one-sided z test at level α, the standard error must satisfy δ = (z_α + z_β)·σ/√n, so n = ((z_α + z_β)·σ/δ)², rounded up.

Solve for n

n = ((2.326 + 1.282)·12/5)² = 74.98

Round up

n = 75

Answer

n = 75
11.Linear regression & correlation

Answer: r = 0.9749

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 = 27/6 = 4.5
ȳ = 328/6 = 54.67

Sums of squares and products

Sxx = Σ(x − x̄)² = 17.5
Sxy = Σ(x − x̄)(y − ȳ) = 57
Syy = Σ(y − ȳ)² = 195.3

Correlation coefficient

r = 57/√(17.5·195.3) = 0.9749

Answer

r = 0.9749
12.Linear regression & correlation

Answer: e = 3.65

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 = 100/5 = 20
ȳ = 222/5 = 44.4

Sums of squares and products

Sxx = Σ(x − x̄)² = 40
Sxy = Σ(x − x̄)(y − ȳ) = 94

Regression line

b = 94/40 = 2.35
a = 44.4 − 2.35·20 = −2.6
ŷ = −2.6 + 2.35·x

Predicted value

ŷ = −2.6 + 2.35·17 = 37.35

Residual

e = 41 − 37.35 = 3.65

Answer

e = 3.65

The point lies above the line.

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