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
253035404550

Find the sample standard deviation s of the data: 53, 49, 37, 42, 24.

n = 5
s =
2.Descriptive statisticsHard
z −2−1012

Exam scores are normally distributed with mean 170 and standard deviation 10. Find the z-score z of a score of 180.

μ = 170σ = 10x = 180
z =
3.ProbabilityMedium
xp(x)00.3510.320.35μ = 1

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

x: 0, 1, 2p(x): 0.35, 0.3, 0.35
Var(X) =
4.ProbabilityHard
n = 10, p = 0.2: P(X ≤ 1) = ?k012345678910

10 components are tested, and each one fails with probability 0.2, independently. What is the probability P that at most one fails?

n = 10p = 0.2
P =
5.Normal distributionMedium
−3−2−10123zshaded area to the right of z = 0.9406

Drag z so that the shaded area under the standard normal curve, to the right of z, is 0.9406. Find z to two decimals.

A = 0.9406
z =
6.Normal distributionHard
557085100115130145pointstop 20%

The IQ scores are normally distributed with mean 100 points and standard deviation 15 points. Drag the cut-off so the shaded top 20% is to the right of it. Find the cut-off x.

μ = 100 pointsσ = 15 points20%
x = points
7.Confidence intervalsMedium
n = 8, sd​ = 5.40LU = ?d̄ = 5.4E99%

The blood pressure of 8 patients is measured before and after a treatment. The differences d = before − after (mmHg) have mean 5.4 and standard deviation 5.4. Find the upper limit of the 99% confidence interval for the mean difference.

n = 8d̄ = 5.4sd = 5.499%t(7) = 3.499
upper limit = mmHg
8.Confidence intervalsHard
σ = 34, n = ?x̄ − Ex̄ + Ex̄E = 495%

The population standard deviation is 34. How large a sample n do you need so that a 95% confidence interval for the mean has a margin of error of at most 4?

σ = 34E = 495%
n =
9.Hypothesis tests & confidence intervalsMedium
−2.0642.064t = ?df = 24

Two teaching methods are compared. Group 1: 6 students, mean score 49, standard deviation 4. Group 2: 20 students, mean score 42.2, standard deviation 5.5. 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 = 6x̄₁ = 49s₁ = 4n₂ = 20x̄₂ = 42.2s₂ = 5.5
t =
10.Hypothesis tests & confidence intervalsHard
μ₀x̄c

A one-sided z test of H₀: μ = 100 against H₁: μ > 100 is done at α = 0.10 with a sample of 64; the standard deviation is known to be 12. In reality the mean is 103.3. Find the power 1 − β of the test in percent.

μ₀ = 100μ = 103.3σ = 12n = 64α = 0.10
1 − β = %
11.Linear regression & correlationMedium
xy17812011123121261392915132176xy202530100150

The table gives 6 paired observations of x = temperature in °C and y = ice creams sold. Find the coefficient of determination r² of the least-squares line.

n = 6
r² =
12.Linear regression & correlationHard
xy122223321415514613xy2461520ŷ = 16

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. For which value x does the line predict ŷ = 16?

n = 6ŷ = 16
x =

3Solutions

1.Descriptive statistics

Answer: s = 11.34

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 = 205/5 = 41

Squared deviations

Σ(x − x̄)² = 12² + 8² + (−4)² + 1² + (−17)² = 514

Standard deviation

s = √(514/4) = 11.34

Answer

s = 11.34
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 = (180 − 170)/10 = 1

Answer

z = 1
3.Probability

Answer: Var(X) = 0.7

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 + 1·0.3 + 2·0.35 = 1

Mean of the squares

E(X²) = 0·0.35 + 1·0.3 + 4·0.35 = 1.7

Variance

Var(X) = 1.7 − 1² = 0.7

Answer

Var(X) = 0.7
4.Probability

Answer: P = 0.3758

Probability: with equally likely outcomes, P(A) = favourable/total. For independent events multiply: P(A and B) = P(A)·P(B). The complement often helps: P(at least one) = 1 − P(none). Binomial: in n independent trials with success probability p, P(X = k) = C(n, k)·pᵏ·(1 − p)ⁿ⁻ᵏ.

Zero and one failure

P(X = 0) = 0.810 = 0.1074
P(X = 1) = 10·0.2·0.89 = 0.2684

Add them

P = 0.1074 + 0.2684 = 0.3758

Answer

P = 0.3758
5.Normal distribution

Answer: z = −1.56

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 − 0.9406 = 0.0594

Read the z-table backwards

Find 0.0594 inside the table and read off z.

z = −1.56

Answer

z = −1.56
6.Normal distribution

Answer: x = 112.6 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) = 0.8

z from the table

z ≈ 0.84

Back to the original units

x = 100 + 0.84·15 = 112.6 points

Answer

x = 112.6 points
7.Confidence intervals

Answer: upper limit = 12.08 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 = 7, t = 3.499

Margin of error

E = 3.499·5.4/√8 = 6.68

Interval

5.4 ± 6.68 → [−1.28, 12.08]

Answer

upper limit = 12.08 mmHg
8.Confidence intervals

Answer: n = 278

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.

Solve for n

n = (1.96·34/4)² = 277.6

Round up

A sample size is a whole number, and rounding down would make the margin too large.

n = 278

Answer

n = 278
9.Hypothesis tests & confidence intervals

Answer: t = 2.797

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 = (5·4² + 19·5.5²)/24 = 27.28
sp = 5.223

Standard error

SE = 5.223·√(1/6 + 1/20) = 2.431

Test statistic

t = (49 − 42.2)/2.431 = 2.797

Answer

t = 2.797

With df = 24, the two-sided 5% critical value is 2.064; |t| is larger, so H₀ is rejected.

10.Hypothesis tests & confidence intervals

Answer: 1 − β = 82.07 %

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 = 12/√64 = 1.5
x̄c = 100 + 1.282·1.5 = 101.9

z under the true mean

z = (101.9 − 103.3)/1.5 = −0.918

Power

power = P(Z > −0.918) = 82.07%
1 − β = 82.07%

Answer

1 − β = 82.07%
11.Linear regression & correlation

Answer: r² = 0.9794

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 = 147/6 = 24.5
ȳ = 779/6 = 129.8

Sums of squares and products

Sxx = Σ(x − x̄)² = 157.5
Sxy = Σ(x − x̄)(y − ȳ) = 919.5
Syy = Σ(y − ȳ)² = 5481

Coefficient of determination

r² = 919.5²/(157.5·5481) = 0.9794

Answer

r² = 0.9794

About 97.94% of the variation in y is explained by the line.

12.Linear regression & correlation

Answer: x = 4.397

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 = 108/6 = 18

Sums of squares and products

Sxx = Σ(x − x̄)² = 17.5
Sxy = Σ(x − x̄)(y − ȳ) = −39

Regression line

b = −39/17.5 = −2.229
a = 18 − (−2.229)·3.5 = 25.8
ŷ = 25.8 − 2.229·x

Solve for x

x = (16 − 25.8)/(−2.229) = 4.397

Answer

x = 4.397

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