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
15202530354045

Find the sample standard deviation s of the data: 46, 13, 13, 32, 39, 34.

n = 6
s =
2.Descriptive statisticsHard
z −2−1012

Exam scores are normally distributed with mean 60 and standard deviation 15. Find the z-score z of a score of 90.

μ = 60σ = 15x = 90
z =
3.ProbabilityMedium
xp(x)10.320.3530.2540.1μ = 2.15

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

x: 1, 2, 3, 4p(x): 0.3, 0.35, 0.25, 0.1
σ =
4.ProbabilityHard
xp(x)100.35200.25300.35400.05μ

X has the distribution shown, and Y = 0.5X + 4. Find the standard deviation σY of Y.

x: 10, 20, 30, 40p(x): 0.35, 0.25, 0.35, 0.05Y = 0.5X + 4
σY =
5.Normal distributionMedium
−3−2−10123zshaded area between −z and z = 0.7243

Drag z so that the shaded area between −z and z under the standard normal curve is 0.7243. Find z to two decimals.

A = 0.7243
z =
6.Normal distributionHard
235240245250255260265gbottom 80%

The bag weights are normally distributed with mean 250 g and standard deviation 5 g. Drag the cut-off so the shaded bottom 80% is to the left of it. Find the cut-off x.

μ = 250 gσ = 5 g80%
x = g
7.Confidence intervalsMedium
n = 25, s = 7.5L = ?Ux̄ = 20.5E99%

A sample of 25 items has mean 20.5 and standard deviation s = 7.5. The population is roughly normal. Find the lower limit of the 99% confidence interval for the mean.

n = 25x̄ = 20.5s = 7.599%t(24) = 2.797
lower limit =
8.Confidence intervalsHard
μ₁ − μ₂, t(15) = 2.9470LU = ?x̄₁ − x̄₂E99%

Two fertilisers are tested. Plot group 1: 10 plots, mean yield 10 kg, standard deviation 6.5 kg. Plot group 2: 7 plots, mean 14.5 kg, standard deviation 2.5 kg. Assume equal variances (pooled t, df = n₁ + n₂ − 2). Find the upper limit of the 99% confidence interval for μ₁ − μ₂.

n₁ = 10x̄₁ = 10s₁ = 6.5n₂ = 7x̄₂ = 14.5s₂ = 2.599%t(15) = 2.947
upper limit = kg
9.Hypothesis tests & confidence intervalsMedium
t −2.0302.03

A sample of 36 students has a mean score of 94.5 and a sample standard deviation of 15. Find the test statistic t for H₀: μ = 100.

μ₀ = 100x̄ = 94.5s = 15n = 36
t =
10.Hypothesis tests & confidence intervalsHard
H₀H₁μ₀x̄cμ₀ + 4α = 0.01βσ = 21, power 95%, n = ?

A one-sided z test at α = 0.01 must detect a shift of the mean of 4 units with a power of 95%. The standard deviation is 21. Find the minimum sample size n.

α = 0.01power = 95%δ = 4σ = 21zα = 2.326zβ = 1.645
n =
11.Linear regression & correlationMedium
xy241347451547659xy246405060

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
xy36857478691041111413124xy51080100120x = 10

The table gives 6 paired observations of x = hours studied and y = exam score. Use the least-squares line ŷ = a + b·x. Predict ŷ for x = 10.

n = 6x = 10
ŷ =

3Solutions

1.Descriptive statistics

Answer: s = 13.66

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 = 177/6 = 29.5

Squared deviations

Σ(x − x̄)² = 16.5² + (−16.5)² + (−16.5)² + 2.5² + 9.5² + 4.5² = 933.5

Standard deviation

s = √(933.5/5) = 13.66

Answer

s = 13.66
2.Descriptive statistics

Answer: z = 2

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 = (90 − 60)/15 = 2

Answer

z = 2
3.Probability

Answer: σ = 0.9631

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 = 1·0.3 + 2·0.35 + 3·0.25 + 4·0.1 = 2.15

Mean of the squares

E(X²) = 1·0.3 + 4·0.35 + 9·0.25 + 16·0.1 = 5.55

Variance

Var(X) = 5.55 − 2.15² = 0.9275
σ = √0.9275 = 0.9631

Answer

σ = 0.9631
4.Probability

Answer: σY = 4.717

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 = 21, Var(X) = 89, σX = 9.434

Linear change

σY = |0.5|·9.434 = 4.717

Answer

σY = 4.717

The mean does move: E(Y) = 0.5·21 + 4 = 14.5.

5.Normal distribution

Answer: z = 1.09

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 + 0.7243)/2 = 0.8621

Read the z-table backwards

Find 0.8621 inside the table and read off z.

z = 1.09

Answer

z = 1.09
6.Normal distribution

Answer: x = 254.2 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) = 0.8

z from the table

z ≈ 0.84

Back to the original units

x = 250 + 0.84·5 = 254.2 g

Answer

x = 254.2 g
7.Confidence intervals

Answer: lower limit = 16.3

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 = 24, t = 2.797

Margin of error

E = 2.797·7.5/√25 = 4.196

Interval

20.5 ± 4.196 → [16.3, 24.7]

Answer

lower limit = 16.3
8.Confidence intervals

Answer: upper limit = 3.164 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 = (9·6.5² + 6·2.5²)/15 = 27.85
sp = 5.277

Margin of error

E = 2.947·5.277·√(1/10 + 1/7) = 7.664

Interval

−4.5 ± 7.664 → [−12.16, 3.164]

Answer

upper limit = 3.164 kg
9.Hypothesis tests & confidence intervals

Answer: t = −2.2

Testing a mean: the test statistic measures how many standard errors the sample mean lies from the value claimed in H₀: z = (x̄ − μ₀)/(σ/√n) when σ is known, t = (x̄ − μ₀)/(s/√n) when it is estimated from the sample. A confidence interval is x̄ ± (critical value)·(standard error).

Standard error

SE = 15/√36 = 15/6 = 2.5

Test statistic

t = (94.5 − 100)/2.5 = −2.2

Answer

t = −2.2
t0.025; df = 35 = 2.03

|t| > 2.03, so H₀ is rejected at the 5% level (two-sided).

10.Hypothesis tests & confidence intervals

Answer: n = 435

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 + 1.645)·21/4)² = 434.6

Round up

n = 435

Answer

n = 435
11.Linear regression & correlation

Answer: r = 0.8581

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 = 20/5 = 4
ȳ = 245/5 = 49

Sums of squares and products

Sxx = Σ(x − x̄)² = 10
Sxy = Σ(x − x̄)(y − ȳ) = 36
Syy = Σ(y − ȳ)² = 176

Correlation coefficient

r = 36/√(10·176) = 0.8581

Answer

r = 0.8581
12.Linear regression & correlation

Answer: ŷ = 106.9

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 = 48/6 = 8
ȳ = 570/6 = 95

Sums of squares and products

Sxx = Σ(x − x̄)² = 70
Sxy = Σ(x − x̄)(y − ȳ) = 418

Regression line

b = 418/70 = 5.971
a = 95 − 5.971·8 = 47.23
ŷ = 47.23 + 5.971·x

Prediction

ŷ = 47.23 + 5.971·10 = 106.9

Answer

ŷ = 106.9

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