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
20253035404550

Find the sample standard deviation s of the data: 50, 49, 18, 20, 37.

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

Exam scores are normally distributed with mean 500 and standard deviation 50. Find the z-score z of a score of 425.

μ = 500σ = 50x = 425
z =
3.ProbabilityMedium
red: 8blue: 6

A bag holds 8 red and 6 blue balls. You draw two balls without putting the first one back. What is the probability P that both are red?

8 / 6
P =
4.ProbabilityHard
n = 11, p = 0.3: P(X = 1) = ?k01234567891011

A test has 11 multiple-choice questions. A student who guesses gets each one right with probability 0.3, independently. What is the probability P of exactly 1 correct answers?

n = 11p = 0.3k = 1
P =
5.Normal distributionMedium
−3−2−10123zshaded area to the right of z = 0.0268

Drag z so that the shaded area under the standard normal curve, to the right of z, is 0.0268. Find z to two decimals.

A = 0.0268
z =
6.Normal distributionHard
10 g0X − Y?

The mass X of a product from line 1 and Y from line 2 are independent and normal: X has mean 500 g and standard deviation 8 g, Y has mean 490 g and standard deviation 6 g. Find the probability P that X is greater than Y. Round z to two decimals.

μX = 500 g, σX = 8 gμY = 490 g, σY = 6 g
P =
7.Confidence intervalsMedium
n = 8, s = 3.5LU = ?x̄ = 61.5E95%

A sample of 8 items has mean 61.5 and standard deviation s = 3.5. The population is roughly normal. Find the upper limit of the 95% confidence interval for the mean.

n = 8x̄ = 61.5s = 3.595%t(7) = 2.365
upper limit =
8.Confidence intervalsHard
n = 21σ = ?Us = 2.695%

The diameters of 21 machined shafts have a sample standard deviation of 2.6 µm. Assuming the diameters are normally distributed, find the lower limit σ of the 95% confidence interval for the population standard deviation. With df = 20, 2.5% of the χ² distribution lies below 9.591 and 2.5% lies above 34.17.

n = 21s = 2.6 µmχ² = 9.591, 34.17
σ = µm
9.Hypothesis tests & confidence intervalsMedium
t −1.99801.998

A sample of 64 students has a mean score of 47.4 and a sample standard deviation of 8. Find the test statistic t for H₀: μ = 50.

μ₀ = 50x̄ = 47.4s = 8n = 64
t =
10.Hypothesis tests & confidence intervalsHard
−1.961.96z = ?

In a trial, 100 of 250 patients on treatment A recover, and 70 of 250 patients on treatment B. Using the pooled proportion, find the test statistic z for H₀: p₁ = p₂.

x₁ = 100n₁ = 250x₂ = 70n₂ = 250
z =
11.Linear regression & correlationMedium
xy158258365476580xy24607080

The table gives 5 paired observations of x = hours studied and y = exam score. Find the coefficient of determination r² of the least-squares line.

n = 5
r² =
12.Linear regression & correlationHard
xy1457166318802089229424113xy1520256080100ŷ = 94

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. For which value x does the line predict ŷ = 94?

n = 6ŷ = 94
x =

3Solutions

1.Descriptive statistics

Answer: s = 15.32

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 = 174/5 = 34.8

Squared deviations

Σ(x − x̄)² = 15.2² + 14.2² + (−16.8)² + (−14.8)² + 2.2² = 938.8

Standard deviation

s = √(938.8/4) = 15.32

Answer

s = 15.32
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 = (425 − 500)/50 = −1.5

Answer

z = −1.5
3.Probability

Answer: P = 0.3077

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.

First and second draw

P = 8/14 · 7/13 = 56/182 = 0.3077

Answer

P = 0.3077
4.Probability

Answer: P = 0.09322

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.

Binomial formula

P(X = 1) = 11·0.31·0.710 = 11·0.3·0.02825 = 0.09322

Answer

P = 0.09322
5.Normal distribution

Answer: z = 1.93

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 − 0.0268 = 0.9732

Read the z-table backwards

Find 0.9732 inside the table and read off z.

z = 1.93

Answer

z = 1.93
6.Normal distribution

Answer: P = 0.8413

Sums and differences of normal variables: if X and Y are independent and normal, X + Y and X − Y are normal too. Means add or subtract: E(X ± Y) = μX ± μY. Variances always add, also for a difference: Var(X ± Y) = σX² + σY², so σ = √(σX² + σY²). Standard deviations never add directly. The total of n independent items with the same distribution has mean n·μ and standard deviation σ·√n. Then standardise as usual: z = (x − mean)/σ and read Φ(z).

Distribution of the difference

μ = 500 − 490 = 10 g, σ = 10 g

Standardise

z = (0 − 10)/10 = −1

Probability

P = 1 − Φ(−1) = 0.8413

Answer

P = 0.8413
7.Confidence intervals

Answer: upper limit = 64.43

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 = 7, t = 2.365

Margin of error

E = 2.365·3.5/√8 = 2.927

Interval

61.5 ± 2.927 → [58.57, 64.43]

Answer

upper limit = 64.43
8.Confidence intervals

Answer: σ = 1.989 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 = 20·2.6² = 135.2

Limits for σ²

135.2/34.17 = 3.957
135.2/9.591 = 14.1

Limit for σ

σ = √3.957 = 1.989 µm

Answer

σ = 1.989 µm

95% interval for σ: [1.989, 3.755] µm.

9.Hypothesis tests & confidence intervals

Answer: t = −2.6

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 = 8/√64 = 8/8 = 1

Test statistic

t = (47.4 − 50)/1 = −2.6

Answer

t = −2.6
t0.025; df = 63 = 1.998

|t| > 1.998, so H₀ is rejected at the 5% level (two-sided).

10.Hypothesis tests & confidence intervals

Answer: z = 2.832

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 = 100/250 = 0.4
p̂₂ = 70/250 = 0.28
p̂ = 170/500 = 0.34

Standard error under H₀

SE = √(0.34·0.66·(1/250 + 1/250)) = 0.04237

Test statistic

z = (0.4 − 0.28)/0.04237 = 2.832

Answer

z = 2.832

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

11.Linear regression & correlation

Answer: r² = 0.9258

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 = 337/5 = 67.4

Sums of squares and products

Sxx = Σ(x − x̄)² = 10
Sxy = Σ(x − x̄)(y − ȳ) = 62
Syy = Σ(y − ȳ)² = 415.2

Coefficient of determination

r² = 62²/(10·415.2) = 0.9258

Answer

r² = 0.9258

About 92.58% of the variation in y is explained by the line.

12.Linear regression & correlation

Answer: x = 21.08

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 = 114/6 = 19
ȳ = 496/6 = 82.67

Sums of squares and products

Sxx = Σ(x − x̄)² = 70
Sxy = Σ(x − x̄)(y − ȳ) = 382

Regression line

b = 382/70 = 5.457
a = 82.67 − 5.457·19 = −21.02
ŷ = −21.02 + 5.457·x

Solve for x

x = (94 − (−21.02))/5.457 = 21.08

Answer

x = 21.08

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