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: 54, 25, 34, 13, 33, 32, 28.

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

Exam scores are normally distributed with mean 100 and standard deviation 15. Find the z-score z of a score of 85.

μ = 100σ = 15x = 85
z =
3.ProbabilityMedium
xp(x)10.220.3530.240.25μ = 2.5

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

x: 1, 2, 3, 4p(x): 0.2, 0.35, 0.2, 0.25
Var(X) =
4.ProbabilityHard
€p(x)−30.88120.1570.02

A game costs €3 to play. You win €15 with probability 0.1 and €60 with probability 0.02; otherwise you win nothing. The stake is not returned. Find the expected net gain E(G) per game.

stake = €3€15: p = 0.1€60: p = 0.02
E(G) = €
5.Normal distributionMedium
−3−2−10123zshaded area to the right of z = 0.1685

Drag z so that the shaded area under the standard normal curve, to the right of z, is 0.1685. Find z to two decimals.

A = 0.1685
z =
6.Normal distributionHard
35455565758595pointstop 80%

The exam scores are normally distributed with mean 65 points and standard deviation 10 points. Drag the cut-off so the shaded top 80% is to the right of it. Find the cut-off x.

μ = 65 pointsσ = 10 points80%
x = points
7.Confidence intervalsMedium
n = 20, sd​ = 2.80LU = ?d̄ = −5.2E99%

The blood pressure of 20 patients is measured before and after a treatment. The differences d = before − after (mmHg) have mean −5.2 and standard deviation 2.8. Find the upper limit of the 99% confidence interval for the mean difference.

n = 20d̄ = −5.2sd = 2.899%t(19) = 2.861
upper limit = mmHg
8.Confidence intervalsHard
n = 30Lσ = ?s = 9.995%

The diameters of 30 machined shafts have a sample standard deviation of 9.9 µm. Assuming the diameters are normally distributed, find the upper limit σ of the 95% confidence interval for the population standard deviation. With df = 29, 2.5% of the χ² distribution lies below 16.047 and 2.5% lies above 45.722.

n = 30s = 9.9 µmχ² = 16.047, 45.722
σ = µm
9.Hypothesis tests & confidence intervalsMedium
−1.961.96z = ?

Two production lines make the same part. Line 1: a sample of 50 parts has mean length 122 mm, and the line's known standard deviation is 12 mm. Line 2: a sample of 30 parts has mean 113.2 mm, known standard deviation 16 mm. Find the test statistic z for H₀: μ₁ = μ₂, taking the difference as x̄₁ − x̄₂ (line 1 minus line 2).

n₁ = 50x̄₁ = 122 mmσ₁ = 12 mmn₂ = 30x̄₂ = 113.2 mmσ₂ = 16 mm
z =
10.Hypothesis tests & confidence intervalsHard
μ₀x̄c

A one-sided z test of H₀: μ = 200 against H₁: μ < 200 is done at α = 0.10 with a sample of 25; the standard deviation is known to be 7.5. In reality the mean is 195.65. Find the probability of a Type II error β in percent.

μ₀ = 200μ = 195.65σ = 7.5n = 25α = 0.10
β = %
11.Linear regression & correlationMedium
xy146358562769969xy5506070

The table gives 5 paired observations of x = hours studied and y = exam score. Find Pearson's correlation coefficient r.

n = 5
r =
12.Linear regression & correlationHard
xy1459167518862010222108xy15206080100ŷ = 92

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. For which value x does the line predict ŷ = 92?

n = 5ŷ = 92
x =

3Solutions

1.Descriptive statistics

Answer: s = 12.32

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 = 219/7 = 31.29

Squared deviations

Σ(x − x̄)² = 22.71² + (−6.286)² + 2.714² + (−18.29)² + 1.714² + 0.7143² + (−3.286)² = 911.4

Standard deviation

s = √(911.4/6) = 12.32

Answer

s = 12.32
2.Descriptive statistics

Answer: z = −1

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 = (85 − 100)/15 = −1

Answer

z = −1
3.Probability

Answer: Var(X) = 1.15

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 = 1·0.2 + 2·0.35 + 3·0.2 + 4·0.25 = 2.5

Mean of the squares

E(X²) = 1·0.2 + 4·0.35 + 9·0.2 + 16·0.25 = 7.4

Variance

Var(X) = 7.4 − 2.5² = 1.15

Answer

Var(X) = 1.15
4.Probability

Answer: E(G) = −0.3 €

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: 12 (p = 0.1), 57 (p = 0.02), −3 (p = 0.88)

Expected value

E(G) = 12·0.1 + 57·0.02 − 3·0.88 = −0.3 €

Answer

E(G) = −0.3 €

A negative value means you lose money on average.

5.Normal distribution

Answer: z = 0.96

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 − 0.1685 = 0.8315

Read the z-table backwards

Find 0.8315 inside the table and read off z.

z = 0.96

Answer

z = 0.96
6.Normal distribution

Answer: x = 56.6 points

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) = 0.2

z from the table

z ≈ −0.84

Back to the original units

x = 65 − 0.84·10 = 56.6 points

Answer

x = 56.6 points
7.Confidence intervals

Answer: upper limit = −3.409 mmHg

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.

Critical value

df = 19, t = 2.861

Margin of error

E = 2.861·2.8/√20 = 1.791

Interval

−5.2 ± 1.791 → [−6.991, −3.409]

Answer

upper limit = −3.409 mmHg
8.Confidence intervals

Answer: σ = 13.31 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 = 29·9.9² = 2842

Limits for σ²

2842/45.722 = 62.16
2842/16.047 = 177.1

Limit for σ

σ = √177.1 = 13.31 µm

Answer

σ = 13.31 µm

95% interval for σ: [7.884, 13.31] µm.

9.Hypothesis tests & confidence intervals

Answer: z = 2.605

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.

Standard error of the difference

SE = √(12²/50 + 16²/30) = √11.41 = 3.378

Test statistic

z = (122 − 113.2)/3.378 = 2.605

Answer

z = 2.605

|z| > 1.96, so H₀ is rejected at the 5% level (two-sided).

10.Hypothesis tests & confidence intervals

Answer: β = 5.283 %

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 = 200 − 1.282·1.5 = 198.1

z under the true mean

z = (198.1 − 195.65)/1.5 = 1.618

Type II error

power = P(Z < 1.618) = 94.72%
β = 100% − 94.72% = 5.283%

Answer

β = 5.283%
11.Linear regression & correlation

Answer: r = 0.9463

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 = 25/5 = 5
ȳ = 304/5 = 60.8

Sums of squares and products

Sxx = Σ(x − x̄)² = 40
Sxy = Σ(x − x̄)(y − ȳ) = 114
Syy = Σ(y − ȳ)² = 362.8

Correlation coefficient

r = 114/√(40·362.8) = 0.9463

Answer

r = 0.9463
12.Linear regression & correlation

Answer: x = 18.96

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 = 90/5 = 18
ȳ = 430/5 = 86

Sums of squares and products

Sxx = Σ(x − x̄)² = 40
Sxy = Σ(x − x̄)(y − ȳ) = 250

Regression line

b = 250/40 = 6.25
a = 86 − 6.25·18 = −26.5
ŷ = −26.5 + 6.25·x

Solve for x

x = (92 − (−26.5))/6.25 = 18.96

Answer

x = 18.96

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