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
30354045505560

Find the sample standard deviation s of the data: 30, 38, 37, 60, 31, 27.

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

Exam scores are normally distributed with mean 70 and standard deviation 12. Find the z-score z of a score of 88.

μ = 70σ = 12x = 88
z =
3.ProbabilityMedium
P(at least one six) = ??1?2?3?4?5?6?7P(six) = 1/6 on each roll

A fair die is rolled 7 times. What is the probability P of getting at least one six?

n = 7
P =
4.ProbabilityHard
n = 5, p = 0.5: P(X = 3) = ?k012345

A test has 5 multiple-choice questions. A student who guesses gets each one right with probability 0.5, independently. What is the probability P of exactly 3 correct answers?

n = 5p = 0.5k = 3
P =
5.Normal distributionMedium
167.5179.75P = ?

The heights of adult men are normally distributed with mean 178 cm and standard deviation 7 cm. What fraction lies between 167.5 and 179.75 cm? Give P as a decimal.

μ = 178σ = 7x1 = 167.5x2 = 179.75
P =
6.Normal distributionHard
0.04 mm0D − d?

Hole diameters D are normal with mean 25.04 mm and standard deviation 0.015 mm. Shaft diameters d are independent and normal with mean 25 mm and standard deviation 0.01 mm. A randomly chosen shaft fits when D − d > 0. Find the probability P that it fits. Round z to two decimals.

hole: μ = 25.04 mm, σ = 0.015 mmshaft: μ = 25 mm, σ = 0.01 mm
P =
7.Confidence intervalsMedium
n = 9, sd​ = 5.20L = ?Ud̄ = 6.2E95%

The blood pressure of 9 patients is measured before and after a treatment. The differences d = before − after (mmHg) have mean 6.2 and standard deviation 5.2. Find the lower limit of the 95% confidence interval for the mean difference.

n = 9d̄ = 6.2sd = 5.295%t(8) = 2.306
lower limit = mmHg
8.Confidence intervalsHard
n = 16σ = ?Us = 11.895%

The diameters of 16 machined shafts have a sample standard deviation of 11.8 µm. Assuming the diameters are normally distributed, find the lower limit σ of the 95% confidence interval for the population standard deviation. With df = 15, 2.5% of the χ² distribution lies below 6.262 and 2.5% lies above 27.488.

n = 16s = 11.8 µmχ² = 6.262, 27.488
σ = µm
9.Hypothesis tests & confidence intervalsMedium
t −2.13102.131

A sample of 16 students has a mean score of 494.8 and a sample standard deviation of 8. Find the test statistic t for H₀: μ = 500.

μ₀ = 500x̄ = 494.8s = 8n = 16
t =
10.Hypothesis tests & confidence intervalsHard
−1.961.96z = ?

In a trial, 24 of 80 patients on treatment A recover, and 28 of 80 patients on treatment B. Using the pooled proportion, find the test statistic z for H₀: p₁ = p₂.

x₁ = 24n₁ = 80x₂ = 28n₂ = 80
z =
11.Linear regression & correlationMedium
xy137351558758965xy5405060

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
xy25496148131018xy51051015ŷ = 7

The table gives 5 paired observations of x = load in kN and y = extension in mm. Use the least-squares line ŷ = a + b·x. For which value x does the line predict ŷ = 7?

n = 5ŷ = 7
x =

3Solutions

1.Descriptive statistics

Answer: s = 11.96

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 = 223/6 = 37.17

Squared deviations

Σ(x − x̄)² = (−7.167)² + 0.8333² + (−0.1667)² + 22.83² + (−6.167)² + (−10.17)² = 714.8

Standard deviation

s = √(714.8/5) = 11.96

Answer

s = 11.96
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 = (88 − 70)/12 = 1.5

Answer

z = 1.5
3.Probability

Answer: P = 0.7209

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.

Complement

P(no six) = (5/6)7 = 0.2791
P = 1 − 0.2791 = 0.7209

Answer

P = 0.7209
4.Probability

Answer: P = 0.3125

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 = 3) = 10·0.53·0.52 = 10·0.125·0.25 = 0.3125

Answer

P = 0.3125
5.Normal distribution

Answer: P = 0.5319

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-scores

z₁ = (167.5 − 178)/7 = −1.5
z₂ = (179.75 − 178)/7 = 0.25

Area between

P = Φ(0.25) − Φ(−1.5) = 0.5987 − 0.06681 = 0.5319

Answer

P = 0.5319
6.Normal distribution

Answer: P = 0.9868

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 = 25.04 − 25 = 0.04 mm, σ = 0.01803 mm

Standardise

z = (0 − 0.04)/0.01803 = −2.22

Probability

P = 1 − Φ(−2.22) = 0.9868

Answer

P = 0.9868
7.Confidence intervals

Answer: lower limit = 2.203 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 = 8, t = 2.306

Margin of error

E = 2.306·5.2/√9 = 3.997

Interval

6.2 ± 3.997 → [2.203, 10.2]

Answer

lower limit = 2.203 mmHg
8.Confidence intervals

Answer: σ = 8.717 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 = 15·11.8² = 2089

Limits for σ²

2089/27.488 = 75.98
2089/6.262 = 333.5

Limit for σ

σ = √75.98 = 8.717 µm

Answer

σ = 8.717 µm

95% interval for σ: [8.717, 18.26] 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/√16 = 8/4 = 2

Test statistic

t = (494.8 − 500)/2 = −2.6

Answer

t = −2.6
t0.025; df = 15 = 2.131

|t| > 2.131, so H₀ is rejected at the 5% level (two-sided).

10.Hypothesis tests & confidence intervals

Answer: z = −0.6752

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 = 24/80 = 0.3
p̂₂ = 28/80 = 0.35
p̂ = 52/160 = 0.325

Standard error under H₀

SE = √(0.325·0.675·(1/80 + 1/80)) = 0.07406

Test statistic

z = (0.3 − 0.35)/0.07406 = −0.6752

Answer

z = −0.6752

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

11.Linear regression & correlation

Answer: r = 0.9383

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 = 25/5 = 5
ȳ = 269/5 = 53.8

Sums of squares and products

Sxx = Σ(x − x̄)² = 40
Sxy = Σ(x − x̄)(y − ȳ) = 126
Syy = Σ(y − ȳ)² = 450.8

Correlation coefficient

r = 126/√(40·450.8) = 0.9383

Answer

r = 0.9383
12.Linear regression & correlation

Answer: x = 2.8

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 = 59/5 = 11.8

Sums of squares and products

Sxx = Σ(x − x̄)² = 40
Sxy = Σ(x − x̄)(y − ȳ) = 60

Regression line

b = 60/40 = 1.5
a = 11.8 − 1.5·6 = 2.8
ŷ = 2.8 + 1.5·x

Solve for x

x = (7 − 2.8)/1.5 = 2.8

Answer

x = 2.8

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