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
303540455055

Find the sample standard deviation s of the data: 29, 42, 59, 56, 37.

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

A quantity is normally distributed with mean 60 and standard deviation 8. What proportion P of values is above 44?

μ = 60σ = 8x = 44
P =
3.ProbabilityMedium
xp(x)00.220.340.3560.15μ = 2.9

For the distribution shown, find the variance Var(X) of X.

x: 0, 2, 4, 6p(x): 0.2, 0.3, 0.35, 0.15
Var(X) =
4.ProbabilityHard
xp(x)00.310.3520.35μ

X has the distribution shown, and Y = 3X + 10. Find the standard deviation σY of Y.

x: 0, 1, 2p(x): 0.3, 0.35, 0.35Y = 3X + 10
σY =
5.Normal distributionMedium
−3−2−10123zshaded area between −z and z = 0.9545

Drag z so that the shaded area between −z and z under the standard normal curve is 0.9545. Find z to two decimals.

A = 0.9545
z =
6.Normal distributionHard
0.03 mm0D − d?

Hole diameters D are normal with mean 25.03 mm and standard deviation 0.02 mm. Shaft diameters d are independent and normal with mean 25 mm and standard deviation 0.02 mm. A randomly chosen shaft fits when D − d > 0. Find the probability P that it fits. Round z to two decimals.

hole: μ = 25.03 mm, σ = 0.02 mmshaft: μ = 25 mm, σ = 0.02 mm
P =
7.Confidence intervalsMedium
μ₁ − μ₂0LU = ?x̄₁ − x̄₂E90%

The battery life of two phone models is compared. Model 1: 50 phones, mean 86.5 h, known standard deviation 10 h. Model 2: 40 phones, mean 76.5 h, known standard deviation 12 h. Find the upper limit of the 90% confidence interval for μ₁ − μ₂.

n₁ = 50x̄₁ = 86.5 hσ₁ = 10 hn₂ = 40x̄₂ = 76.5 hσ₂ = 12 h90%
upper limit = h
8.Confidence intervalsHard
σ = 35, n = ?x̄ − Ex̄ + Ex̄E = 190%

The population standard deviation is 35. How large a sample n do you need so that a 90% confidence interval for the mean has a margin of error of at most 1?

σ = 35E = 190%
n =
9.Hypothesis tests & confidence intervalsMedium
−1.8p = ?

The mean of a population is claimed to be 80; the population standard deviation is known to be 15. A random sample of 36 observations has a mean of 75.5. Test H₀: μ = 80 against H₁: μ < 80. Find the p-value p.

μ₀ = 80σ = 15n = 36x̄ = 75.5H₁: μ < μ₀
p =
10.Hypothesis tests & confidence intervalsHard
z −1.64501.645

A sample of 25 measurements has mean 46.8. The population standard deviation is 8. Find the upper limit U of the 90% confidence interval for the mean.

x̄ = 46.8σ = 8n = 2590%
U =
11.Linear regression & correlationMedium
xy3654705816887968106xy46880100

The table gives 6 paired observations of x = hours studied and y = exam score. Find Pearson's correlation coefficient r.

n = 6
r =
12.Linear regression & correlationHard
xy346451556651761xy4645505560ŷ = 52

The table gives 5 paired observations of x = advertising spend in k€ and y = sales in k€. Use the least-squares line ŷ = a + b·x. For which value x does the line predict ŷ = 52?

n = 5ŷ = 52
x =

3Solutions

1.Descriptive statistics

Answer: s = 12.7

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 = 223/5 = 44.6

Squared deviations

Σ(x − x̄)² = (−15.6)² + (−2.6)² + 14.4² + 11.4² + (−7.6)² = 645.2

Standard deviation

s = √(645.2/4) = 12.7

Answer

s = 12.7
2.Descriptive statistics

Answer: P = 0.9772

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 = −2

Standard normal table

Φ(−2) = 0.02275
P = 1 − 0.02275 = 0.9772

Answer

P = 0.9772
3.Probability

Answer: Var(X) = 3.79

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 = 0·0.2 + 2·0.3 + 4·0.35 + 6·0.15 = 2.9

Mean of the squares

E(X²) = 0·0.2 + 4·0.3 + 16·0.35 + 36·0.15 = 12.2

Variance

Var(X) = 12.2 − 2.9² = 3.79

Answer

Var(X) = 3.79
4.Probability

Answer: σY = 2.414

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 = 1.05, Var(X) = 0.6475, σX = 0.8047

Linear change

σY = |3|·0.8047 = 2.414

Answer

σY = 2.414

The mean does move: E(Y) = 3·1.05 + 10 = 13.15.

5.Normal distribution

Answer: z = 2

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 + 0.9545)/2 = 0.9772

Read the z-table backwards

Find 0.9772 inside the table and read off z.

z = 2.00

Answer

z = 2.00
6.Normal distribution

Answer: P = 0.8554

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 = 25.03 − 25 = 0.03 mm, σ = 0.02828 mm

Standardise

z = (0 − 0.03)/0.02828 = −1.06

Probability

P = 1 − Φ(−1.06) = 0.8554

Answer

P = 0.8554
7.Confidence intervals

Answer: upper limit = 13.89 h

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.

Point estimate

x̄₁ − x̄₂ = 86.5 − 76.5 = 10

Standard error

SE = √(10²/50 + 12²/40) = 2.366

Margin of error

E = 1.645·2.366 = 3.893

Interval

10 ± 3.893 → [6.107, 13.89]

Answer

upper limit = 13.89 h

The interval does not contain 0, so the means differ significantly.

8.Confidence intervals

Answer: n = 3315

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 = (1.645·35/1)² = 3315

Round up

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

n = 3315

Answer

n = 3315
9.Hypothesis tests & confidence intervals

Answer: p = 0.03593

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.

Test statistic

SE = σ/√n = 15/√36 = 2.5
z = (75.5 − 80)/2.5 = −1.8

Tail area

Φ(1.8) = 0.9641
1 − Φ(1.8) = 0.0359

p-value

p = 0.03593

Answer

p = 0.03593

p ≤ 0.05, so H₀ is rejected at the 5% level.

10.Hypothesis tests & confidence intervals

Answer: U = 49.43

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).

Margin of error

E = 1.645·8/√25 = 2.632

Interval limit

U = 46.8 + 2.632 = 49.43

Answer

U = 49.43
11.Linear regression & correlation

Answer: r = 0.9967

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 = 33/6 = 5.5
ȳ = 506/6 = 84.33

Sums of squares and products

Sxx = Σ(x − x̄)² = 17.5
Sxy = Σ(x − x̄)(y − ȳ) = 145
Syy = Σ(y − ȳ)² = 1209

Correlation coefficient

r = 145/√(17.5·1209) = 0.9967

Answer

r = 0.9967
12.Linear regression & correlation

Answer: x = 4.667

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 = 265/5 = 53

Sums of squares and products

Sxx = Σ(x − x̄)² = 10
Sxy = Σ(x − x̄)(y − ȳ) = 30

Regression line

b = 30/10 = 3
a = 53 − 3·5 = 38
ŷ = 38 + 3·x

Solve for x

x = (52 − 38)/3 = 4.667

Answer

x = 4.667

ExamProblems · examproblems.com/courses/statistics-1