Introductory Statistics: cram pack

Formula sheets, the key method steps for every topic and 12 practice problems, with full solutions at the back.

New problemsBack to the course

In the print window, choose “Save as PDF” to keep a copy on your device.

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: 28, 42, 38, 48, 51, 35, 15.

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

A quantity is normally distributed with mean 170 and standard deviation 15. What proportion P of values is above 147.5?

μ = 170σ = 15x = 147.5
P =
3.ProbabilityMedium
red: 8blue: 5

A bag holds 8 red and 5 blue balls. You draw two balls without putting the first one back. What is the probability P that both are red?

8 / 5
P =
4.ProbabilityHard
xp(x)10.3520.0530.340.3μ

X has the distribution shown, and Y = 10X + 20. Find the standard deviation σY of Y.

x: 1, 2, 3, 4p(x): 0.35, 0.05, 0.3, 0.3Y = 10X + 20
σY =
5.Normal distributionMedium
105 s114.6 sX + Y?

X is normal with mean 60 s and standard deviation 5 s; Y is independent and normal with mean 45 s and standard deviation 4 s. Find the probability P that X + Y is more than 114.6 s. Round z to two decimals.

μX = 60, σX = 5μY = 45, σY = 4X + Y > 114.6 s
P =
6.Normal distributionHard
μ = 178x = ?10%

The heights of adult men are normally distributed with mean 178 cm and standard deviation 7 cm. Find the value x below which 10% of them lie (the 10th percentile).

μ = 178σ = 7P = 10%
x = cm
7.Confidence intervalsMedium
n = 9, sd​ = 7.80LU = ?d̄ = −4.7E95%

The blood pressure of 9 patients is measured before and after a treatment. The differences d = before − after (mmHg) have mean −4.7 and standard deviation 7.8. Find the upper limit of the 95% confidence interval for the mean difference.

n = 9d̄ = −4.7sd = 7.895%t(8) = 2.306
upper limit = mmHg
8.Confidence intervalsHard
σ = 15, n = ?x̄ − Ex̄ + Ex̄E = 595%

The population standard deviation is 15. How large a sample n do you need so that a 95% confidence interval for the mean has a margin of error of at most 5?

σ = 15E = 595%
n =
9.Hypothesis tests & confidence intervalsMedium
t −1.9901.99

A sample of 81 students has a mean score of 50.53 and a sample standard deviation of 4. Find the test statistic t for H₀: μ = 50.

μ₀ = 50x̄ = 50.53s = 4n = 81
t =
10.Hypothesis tests & confidence intervalsHard
μ₀x̄c

A one-sided z test of H₀: μ = 200 against H₁: μ > 200 is done at α = 0.10 with a sample of 100; the standard deviation is known to be 5. In reality the mean is 201.05. Find the probability of a Type II error β in percent.

μ₀ = 200μ = 201.05σ = 5n = 100α = 0.10
β = %
11.Linear regression & correlationMedium
xy148367575797910211122xy51050100

The table gives 6 paired observations of x = hours studied and y = exam score. Find the coefficient of determination r² of the least-squares line.

n = 6
r² =
12.Linear regression & correlationHard
xy245357452559671774xy246506070x = 5.5

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 = 5.5.

n = 6x = 5.5
ŷ =

3Solutions

1.Descriptive statistics

Answer: s = 12.32

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 = 257/7 = 36.71

Squared deviations

Σ(x − x̄)² = (−8.714)² + 5.286² + 1.286² + 11.29² + 14.29² + (−1.714)² + (−21.71)² = 911.4

Standard deviation

s = √(911.4/6) = 12.32

Answer

s = 12.32
2.Descriptive statistics

Answer: P = 0.9332

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 = −1.5

Standard normal table

Φ(−1.5) = 0.06681
P = 1 − 0.06681 = 0.9332

Answer

P = 0.9332
3.Probability

Answer: P = 0.359

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 = 8/13 · 7/12 = 56/156 = 0.359

Answer

P = 0.359
4.Probability

Answer: σY = 12.44

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 = 2.55, Var(X) = 1.548, σX = 1.244

Linear change

σY = |10|·1.244 = 12.44

Answer

σY = 12.44

The mean does move: E(Y) = 10·2.55 + 20 = 45.5.

5.Normal distribution

Answer: P = 0.06681

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 sum

μ = 105 s, σ = √(5² + 4²) = 6.403 s

Standardise

z = (114.6 − 105)/6.403 = 1.5

Upper tail

P = 1 − Φ(1.5) = 1 − 0.9332 = 0.06681

Answer

P = 0.06681
6.Normal distribution

Answer: x = 169 cm

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) = 0.1 → z = −1.282

Back to x

x = 178 + −1.282·7 = 169.03 cm

Answer

x = 169.03 cm
7.Confidence intervals

Answer: upper limit = 1.296 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 = 2.306·7.8/√9 = 5.996

Interval

−4.7 ± 5.996 → [−10.7, 1.296]

Answer

upper limit = 1.296 mmHg
8.Confidence intervals

Answer: n = 35

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.

Solve for n

n = (1.96·15/5)² = 34.57

Round up

A sample size is a whole number, and rounding down would make the margin too large.

n = 35

Answer

n = 35
9.Hypothesis tests & confidence intervals

Answer: t = 1.193

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 = 4/√81 = 4/9 = 0.4444

Test statistic

t = (50.53 − 50)/0.4444 = 1.193

Answer

t = 1.193
t0.025; df = 80 = 1.99

|t| < 1.99, so H₀ is not rejected at the 5% level (two-sided).

10.Hypothesis tests & confidence intervals

Answer: β = 20.67 %

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 = 5/√100 = 0.5
x̄c = 200 + 1.282·0.5 = 200.6

z under the true mean

z = (200.6 − 201.05)/0.5 = −0.818

Type II error

power = P(Z > −0.818) = 79.33%
β = 100% − 79.33% = 20.67%

Answer

β = 20.67%
11.Linear regression & correlation

Answer: r² = 0.9816

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 = 36/6 = 6
ȳ = 511/6 = 85.17

Sums of squares and products

Sxx = Σ(x − x̄)² = 70
Sxy = Σ(x − x̄)(y − ȳ) = 497
Syy = Σ(y − ȳ)² = 3595

Coefficient of determination

r² = 497²/(70·3595) = 0.9816

Answer

r² = 0.9816

About 98.16% of the variation in y is explained by the line.

12.Linear regression & correlation

Answer: ŷ = 65.21

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 = 27/6 = 4.5
ȳ = 358/6 = 59.67

Sums of squares and products

Sxx = Σ(x − x̄)² = 17.5
Sxy = Σ(x − x̄)(y − ȳ) = 97

Regression line

b = 97/17.5 = 5.543
a = 59.67 − 5.543·4.5 = 34.72
ŷ = 34.72 + 5.543·x

Prediction

ŷ = 34.72 + 5.543·5.5 = 65.21

Answer

ŷ = 65.21

ExamProblems · examproblems.com/courses/statistics-1