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: 23, 11, 15, 53, 40.

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

Exam scores are normally distributed with mean 100 and standard deviation 10. Find the z-score z of a score of 110.

μ = 100σ = 10x = 110
z =
3.ProbabilityMedium
xp(x)100.1200.35300.35400.2μ = 26.5

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

x: 10, 20, 30, 40p(x): 0.1, 0.35, 0.35, 0.2
σ =
4.ProbabilityHard
xp(x)00.3510.3520.3μ

X has the distribution shown, and Y = 5X + 5. Find the standard deviation σY of Y.

x: 0, 1, 2p(x): 0.35, 0.35, 0.3Y = 5X + 5
σY =
5.Normal distributionMedium
−3−2−10123zshaded area to the right of z = 0.2843

Drag z so that the shaded area under the standard normal curve, to the right of z, is 0.2843. Find z to two decimals.

A = 0.2843
z =
6.Normal distributionHard
235240245250255260265gbottom 15%

The bag weights are normally distributed with mean 250 g and standard deviation 5 g. Drag the cut-off so the shaded bottom 15% is to the left of it. Find the cut-off x.

μ = 250 gσ = 5 g15%
x = g
7.Confidence intervalsMedium
n = 10, s = 2LUx̄ = 37E = ?99%

A sample of 10 items has mean 37 and standard deviation s = 2. The population is roughly normal. Find the margin of error E of a 99% confidence interval for the mean.

n = 10x̄ = 37s = 299%t(9) = 3.25
E =
8.Confidence intervalsHard
μ₁ − μ₂, t(10) = 1.8120LUx̄₁ − x̄₂E = ?90%

Two fertilisers are tested. Plot group 1: 7 plots, mean yield 63 kg, standard deviation 5 kg. Plot group 2: 5 plots, mean 69.5 kg, standard deviation 3.5 kg. Assume equal variances (pooled t, df = n₁ + n₂ − 2). Find the margin of error E of the 90% confidence interval for μ₁ − μ₂.

n₁ = 7x̄₁ = 63s₁ = 5n₂ = 5x̄₂ = 69.5s₂ = 3.590%t(10) = 1.812
E = kg
9.Hypothesis tests & confidence intervalsMedium
−2.0642.064t = ?df = 24

Two teaching methods are compared. Group 1: 16 students, mean score 81.5, standard deviation 4. Group 2: 10 students, mean score 74.6, standard deviation 10.5. 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 = 16x̄₁ = 81.5s₁ = 4n₂ = 10x̄₂ = 74.6s₂ = 10.5
t =
10.Hypothesis tests & confidence intervalsHard
−1.961.96z = ?

In a trial, 35 of 100 patients on treatment A recover, and 129 of 300 patients on treatment B. Using the pooled proportion, find the test statistic z for H₀: p₁ = p₂.

x₁ = 35n₁ = 100x₂ = 129n₂ = 300
z =
11.Linear regression & correlationMedium
xy156260367475577xy246070

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
xy264761481610231226xy5101020x = 11

The table gives 6 paired observations of x = load in kN and y = extension in mm. Use the least-squares line ŷ = a + b·x. A new observation (11, 28) is made. Find its residual e = y − ŷ.

n = 6(11, 28)
e =

3Solutions

1.Descriptive statistics

Answer: s = 17.69

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 = 142/5 = 28.4

Squared deviations

Σ(x − x̄)² = (−5.4)² + (−17.4)² + (−13.4)² + 24.6² + 11.6² = 1251

Standard deviation

s = √(1251/4) = 17.69

Answer

s = 17.69
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 = (110 − 100)/10 = 1

Answer

z = 1
3.Probability

Answer: σ = 9.097

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 = 10·0.1 + 20·0.35 + 30·0.35 + 40·0.2 = 26.5

Mean of the squares

E(X²) = 100·0.1 + 400·0.35 + 900·0.35 + 1600·0.2 = 785

Variance

Var(X) = 785 − 26.5² = 82.75
σ = √82.75 = 9.097

Answer

σ = 9.097
4.Probability

Answer: σY = 4.023

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 = 0.95, Var(X) = 0.6475, σX = 0.8047

Linear change

σY = |5|·0.8047 = 4.023

Answer

σY = 4.023

The mean does move: E(Y) = 5·0.95 + 5 = 9.75.

5.Normal distribution

Answer: z = 0.57

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 − 0.2843 = 0.7157

Read the z-table backwards

Find 0.7157 inside the table and read off z.

z = 0.57

Answer

z = 0.57
6.Normal distribution

Answer: x = 244.8 g

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) = 0.15

z from the table

z ≈ −1.04

Back to the original units

x = 250 − 1.04·5 = 244.8 g

Answer

x = 244.8 g
7.Confidence intervals

Answer: E = 2.055

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.

Critical value

df = n − 1 = 9, t = 3.25

Margin of error

E = 3.25·2/√10 = 2.055

Answer

E = 2.055
8.Confidence intervals

Answer: E = 4.733 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 = (6·5² + 4·3.5²)/10 = 19.9
sp = 4.461

Margin of error

E = 1.812·4.461·√(1/7 + 1/5) = 4.733

Answer

E = 4.733 kg
9.Hypothesis tests & confidence intervals

Answer: t = 2.389

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 = (15·4² + 9·10.5²)/24 = 51.34
sp = 7.165

Standard error

SE = 7.165·√(1/16 + 1/10) = 2.888

Test statistic

t = (81.5 − 74.6)/2.888 = 2.389

Answer

t = 2.389

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: z = −1.409

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.

Sample proportions

p̂₁ = 35/100 = 0.35
p̂₂ = 129/300 = 0.43
p̂ = 164/400 = 0.41

Standard error under H₀

SE = √(0.41·0.59·(1/100 + 1/300)) = 0.05679

Test statistic

z = (0.35 − 0.43)/0.05679 = −1.409

Answer

z = −1.409

|z| < 1.96, so H₀ is not rejected at the 5% level (two-sided).

11.Linear regression & correlation

Answer: r = 0.9863

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 = 335/5 = 67

Sums of squares and products

Sxx = Σ(x − x̄)² = 10
Sxy = Σ(x − x̄)(y − ȳ) = 57
Syy = Σ(y − ȳ)² = 334

Correlation coefficient

r = 57/√(10·334) = 0.9863

Answer

r = 0.9863
12.Linear regression & correlation

Answer: e = 4.095

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 = 42/6 = 7
ȳ = 92/6 = 15.33

Sums of squares and products

Sxx = Σ(x − x̄)² = 70
Sxy = Σ(x − x̄)(y − ȳ) = 150

Regression line

b = 150/70 = 2.143
a = 15.33 − 2.143·7 = 0.3333
ŷ = 0.3333 + 2.143·x

Predicted value

ŷ = 0.3333 + 2.143·11 = 23.9

Residual

e = 28 − 23.9 = 4.095

Answer

e = 4.095

The point lies above the line.

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