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
152025303540455055

Find the sample standard deviation s of the data: 12, 44, 26, 54, 48, 57, 40.

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

Exam scores are normally distributed with mean 70 and standard deviation 10. Find the z-score z of a score of 85.

μ = 70σ = 10x = 85
z =
3.ProbabilityMedium
n = 7k = 3N = ?

In how many ways N can a committee of 3 people be chosen from a group of 7? The order does not matter.

n = 7k = 3
N =
4.ProbabilityHard
xp(x)10.3520.330.340.05μ

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

x: 1, 2, 3, 4p(x): 0.35, 0.3, 0.3, 0.05Y = 10X − 3
σY =
5.Normal distributionMedium
−3−2−10123zshaded area to the right of z = 0.2776

Drag z so that the shaded area under the standard normal curve, to the right of z, is 0.2776. Find z to two decimals.

A = 0.2776
z =
6.Normal distributionHard
μ = 250x = ?25%

The bag weights are normally distributed with mean 250 g and standard deviation 5 g. Find the value x below which 25% of them lie (the 25th percentile).

μ = 250σ = 5P = 25%
x = g
7.Confidence intervalsMedium
n = 10, sd​ = 4.60L = ?Ud̄ = 9E90%

The blood pressure of 10 patients is measured before and after a treatment. The differences d = before − after (mmHg) have mean 9 and standard deviation 4.6. Find the lower limit of the 90% confidence interval for the mean difference.

n = 10d̄ = 9sd = 4.690%t(9) = 1.833
lower limit = mmHg
8.Confidence intervalsHard
μ₁ − μ₂, t(19) = 2.0930L = ?Ux̄₁ − x̄₂E95%

Two fertilisers are tested. Plot group 1: 13 plots, mean yield 77 kg, standard deviation 5.5 kg. Plot group 2: 8 plots, mean 68 kg, standard deviation 4.5 kg. Assume equal variances (pooled t, df = n₁ + n₂ − 2). Find the lower limit of the 95% confidence interval for μ₁ − μ₂.

n₁ = 13x̄₁ = 77s₁ = 5.5n₂ = 8x̄₂ = 68s₂ = 4.595%t(19) = 2.093
lower limit = kg
9.Hypothesis tests & confidence intervalsMedium
−2.2282.228t = ?df = 10

Two teaching methods are compared. Group 1: 5 students, mean score 51, standard deviation 11. Group 2: 7 students, mean score 38.7, standard deviation 8.5. 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 = 5x̄₁ = 51s₁ = 11n₂ = 7x̄₂ = 38.7s₂ = 8.5
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 25; the standard deviation is known to be 2.5. In reality the mean is 100.9. Find the power 1 − β of the test in percent.

μ₀ = 100μ = 100.9σ = 2.5n = 25α = 0.10
1 − β = %
11.Linear regression & correlationMedium
xy2104216258351044xy5102040

The table gives 5 paired observations of x = load in kN and y = extension in mm. Find the coefficient of determination r² of the least-squares line.

n = 5
r² =
12.Linear regression & correlationHard
xy144417562061238026772988xy152025306080x = 15.5

The table gives 6 paired observations of x = temperature in °C and y = ice creams sold. Use the least-squares line ŷ = a + b·x. Predict ŷ for x = 15.5.

n = 6x = 15.5
ŷ =

3Solutions

1.Descriptive statistics

Answer: s = 16.05

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 = 281/7 = 40.14

Squared deviations

Σ(x − x̄)² = (−28.14)² + 3.857² + (−14.14)² + 13.86² + 7.857² + 16.86² + (−0.1429)² = 1545

Standard deviation

s = √(1545/6) = 16.05

Answer

s = 16.05
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 = (85 − 70)/10 = 1.5

Answer

z = 1.5
3.Probability

Answer: N = 35

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.

Combinations

C(7, 3) = (7·6·5)/(3·2·1) = 35

Answer

N = 35
4.Probability

Answer: σY = 9.206

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 = 2.05, Var(X) = 0.8475, σX = 0.9206

Linear change

σY = |10|·0.9206 = 9.206

Answer

σY = 9.206

The mean does move: E(Y) = 10·2.05 − 3 = 17.5.

5.Normal distribution

Answer: z = 0.59

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 − 0.2776 = 0.7224

Read the z-table backwards

Find 0.7224 inside the table and read off z.

z = 0.59

Answer

z = 0.59
6.Normal distribution

Answer: x = 246.6 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) = 0.25 → z = −0.6745

Back to x

x = 250 + −0.6745·5 = 246.63 g

Answer

x = 246.63 g
7.Confidence intervals

Answer: lower limit = 6.334 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 = 9, t = 1.833

Margin of error

E = 1.833·4.6/√10 = 2.666

Interval

9 ± 2.666 → [6.334, 11.67]

Answer

lower limit = 6.334 mmHg
8.Confidence intervals

Answer: lower limit = 4.152 kg

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.

Pooled standard deviation

sp² = (12·5.5² + 7·4.5²)/19 = 26.57
sp = 5.154

Margin of error

E = 2.093·5.154·√(1/13 + 1/8) = 4.848

Interval

9 ± 4.848 → [4.152, 13.85]

Answer

lower limit = 4.152 kg
9.Hypothesis tests & confidence intervals

Answer: t = 2.193

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 = (4·11² + 6·8.5²)/10 = 91.75
sp = 9.579

Standard error

SE = 9.579·√(1/5 + 1/7) = 5.609

Test statistic

t = (51 − 38.7)/5.609 = 2.193

Answer

t = 2.193

With df = 10, the two-sided 5% critical value is 2.228; |t| is smaller, so H₀ is not rejected.

10.Hypothesis tests & confidence intervals

Answer: 1 − β = 69.78 %

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 = 2.5/√25 = 0.5
x̄c = 100 + 1.282·0.5 = 100.6

z under the true mean

z = (100.6 − 100.9)/0.5 = −0.518

Power

power = P(Z > −0.518) = 69.78%
1 − β = 69.78%

Answer

1 − β = 69.78%
11.Linear regression & correlation

Answer: r² = 0.9859

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 = 135/5 = 27

Sums of squares and products

Sxx = Σ(x − x̄)² = 40
Sxy = Σ(x − x̄)(y − ȳ) = 164
Syy = Σ(y − ȳ)² = 682

Coefficient of determination

r² = 164²/(40·682) = 0.9859

Answer

r² = 0.9859

About 98.59% of the variation in y is explained by the line.

12.Linear regression & correlation

Answer: ŷ = 50.41

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 = 129/6 = 21.5
ȳ = 406/6 = 67.67

Sums of squares and products

Sxx = Σ(x − x̄)² = 157.5
Sxy = Σ(x − x̄)(y − ȳ) = 453

Regression line

b = 453/157.5 = 2.876
a = 67.67 − 2.876·21.5 = 5.829
ŷ = 5.829 + 2.876·x

Prediction

ŷ = 5.829 + 2.876·15.5 = 50.41

Answer

ŷ = 50.41

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