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: 44, 54, 23, 19, 22, 40.

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

A quantity is normally distributed with mean 70 and standard deviation 15. What proportion P of values is below 76?

μ = 70σ = 15x = 76
P =
3.ProbabilityMedium
xp(x)00.1510.1520.2530.340.15μ = 2.15

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

x: 0, 1, 2, 3, 4p(x): 0.15, 0.15, 0.25, 0.3, 0.15
σ =
4.ProbabilityHard
xp(x)00.3510.320.35μ

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

x: 0, 1, 2p(x): 0.35, 0.3, 0.35Y = 3X − 3
σY =
5.Normal distributionMedium
2344.3P = ?

The delivery times are normally distributed with mean 32 min and standard deviation 6 min. What fraction lies between 23 and 44.3 min? Give P as a decimal.

μ = 32σ = 6x1 = 23x2 = 44.3
P =
6.Normal distributionHard
μ = 178x = ?90%

The heights of adult men are normally distributed with mean 178 cm and standard deviation 7 cm. Find the value x below which 90% of them lie (the 90th percentile).

μ = 178σ = 7P = 90%
x = cm
7.Confidence intervalsMedium
n = 8, s = 9LUx̄ = 71.5E = ?90%

A sample of 8 items has mean 71.5 and standard deviation s = 9. The population is roughly normal. Find the margin of error E of a 90% confidence interval for the mean.

n = 8x̄ = 71.5s = 990%t(7) = 1.895
E =
8.Confidence intervalsHard
σ = 14, n = ?x̄ − Ex̄ + Ex̄E = 390%

The population standard deviation is 14. 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 3?

σ = 14E = 390%
n =
9.Hypothesis tests & confidence intervalsMedium
t −2.06402.064

A sample of 25 students has a mean score of 102.1 and a sample standard deviation of 15. Find the test statistic t for H₀: μ = 100.

μ₀ = 100x̄ = 102.1s = 15n = 25
t =
10.Hypothesis tests & confidence intervalsHard
μ₀x̄c

A one-sided z test of H₀: μ = 100 against H₁: μ < 100 is done at α = 0.01 with a sample of 36; the standard deviation is known to be 3. In reality the mean is 98.4. Find the power 1 − β of the test in percent.

μ₀ = 100μ = 98.4σ = 3n = 36α = 0.01
1 − β = %
11.Linear regression & correlationMedium
xy294136208301035xy510102030

The table gives 5 paired observations of x = load in kN and y = extension in mm. Find Pearson's correlation coefficient r.

n = 5
r =
12.Linear regression & correlationHard
xy3535597709891198xy5106080100ŷ = 68

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 = 68?

n = 5ŷ = 68
x =

3Solutions

1.Descriptive statistics

Answer: s = 14.32

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 = 202/6 = 33.67

Squared deviations

Σ(x − x̄)² = 10.33² + 20.33² + (−10.67)² + (−14.67)² + (−11.67)² + 6.333² = 1025

Standard deviation

s = √(1025/5) = 14.32

Answer

s = 14.32
2.Descriptive statistics

Answer: P = 0.6554

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 = 0.4

Standard normal table

Φ(0.4) = 0.6554
P = 0.6554

Answer

P = 0.6554
3.Probability

Answer: σ = 1.276

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 = 0·0.15 + 1·0.15 + 2·0.25 + 3·0.3 + 4·0.15 = 2.15

Mean of the squares

E(X²) = 0·0.15 + 1·0.15 + 4·0.25 + 9·0.3 + 16·0.15 = 6.25

Variance

Var(X) = 6.25 − 2.15² = 1.628
σ = √1.628 = 1.276

Answer

σ = 1.276
4.Probability

Answer: σY = 2.51

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, Var(X) = 0.7, σX = 0.8367

Linear change

σY = |3|·0.8367 = 2.51

Answer

σY = 2.51

The mean does move: E(Y) = 3·1 − 3 = 0.

5.Normal distribution

Answer: P = 0.913

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 = (23 − 32)/6 = −1.5
z₂ = (44.3 − 32)/6 = 2.05

Area between

P = Φ(2.05) − Φ(−1.5) = 0.9798 − 0.06681 = 0.913

Answer

P = 0.913
6.Normal distribution

Answer: x = 187 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) = 0.9 → z = 1.282

Back to x

x = 178 + 1.282·7 = 186.97 cm

Answer

x = 186.97 cm
7.Confidence intervals

Answer: E = 6.03

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 = 1.895

Margin of error

E = 1.895·9/√8 = 6.03

Answer

E = 6.03
8.Confidence intervals

Answer: n = 59

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 = (1.645·14/3)² = 58.93

Round up

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

n = 59

Answer

n = 59
9.Hypothesis tests & confidence intervals

Answer: t = 0.7

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/√25 = 15/5 = 3

Test statistic

t = (102.1 − 100)/3 = 0.7

Answer

t = 0.7
t0.025; df = 24 = 2.064

|t| < 2.064, so H₀ is not rejected at the 5% level (two-sided).

10.Hypothesis tests & confidence intervals

Answer: 1 − β = 80.89 %

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 = 3/√36 = 0.5
x̄c = 100 − 2.326·0.5 = 98.84

z under the true mean

z = (98.84 − 98.4)/0.5 = 0.874

Power

power = P(Z < 0.874) = 80.89%
1 − β = 80.89%

Answer

1 − β = 80.89%
11.Linear regression & correlation

Answer: r = 0.9906

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 = 107/5 = 21.4

Sums of squares and products

Sxx = Σ(x − x̄)² = 40
Sxy = Σ(x − x̄)(y − ȳ) = 138
Syy = Σ(y − ȳ)² = 485.2

Correlation coefficient

r = 138/√(40·485.2) = 0.9906

Answer

r = 0.9906
12.Linear regression & correlation

Answer: x = 6.033

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 = 35/5 = 7
ȳ = 369/5 = 73.8

Sums of squares and products

Sxx = Σ(x − x̄)² = 40
Sxy = Σ(x − x̄)(y − ȳ) = 240

Regression line

b = 240/40 = 6
a = 73.8 − 6·7 = 31.8
ŷ = 31.8 + 6·x

Solve for x

x = (68 − 31.8)/6 = 6.033

Answer

x = 6.033

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