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
152025303540455055

Find the sample standard deviation s of the data: 59, 24, 14, 27, 26, 46, 25.

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

A quantity is normally distributed with mean 100 and standard deviation 8. What proportion P of values is below 96?

μ = 100σ = 8x = 96
P =
3.ProbabilityMedium
xp(x)100.15200.25300.35400.25μ = 27

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

x: 10, 20, 30, 40p(x): 0.15, 0.25, 0.35, 0.25
Var(X) =
4.ProbabilityHard
n = 11, p = 0.4: P(X ≤ 1) = ?k01234567891011

11 components are tested, and each one fails with probability 0.4, independently. What is the probability P that at most one fails?

n = 11p = 0.4
P =
5.Normal distributionMedium
100108.25P = ?

The IQ scores are normally distributed with mean 100 points and standard deviation 15 points. What fraction lies between 100 and 108.25 points? Give P as a decimal.

μ = 100σ = 15x1 = 100x2 = 108.25
P =
6.Normal distributionHard
μ = 65x = ?25%

The exam scores are normally distributed with mean 65 points and standard deviation 10 points. Find the value x below which 25% of them lie (the 25th percentile).

μ = 65σ = 10P = 25%
x = points
7.Confidence intervalsMedium
μ₁ − μ₂0LU = ?x̄₁ − x̄₂E99%

The battery life of two phone models is compared. Model 1: 40 phones, mean 134 h, known standard deviation 10 h. Model 2: 80 phones, mean 138 h, known standard deviation 9 h. Find the upper limit of the 99% confidence interval for μ₁ − μ₂.

n₁ = 40x̄₁ = 134 hσ₁ = 10 hn₂ = 80x̄₂ = 138 hσ₂ = 9 h99%
upper limit = h
8.Confidence intervalsHard
n = 11σ = ?Us = 11.395%

The diameters of 11 machined shafts have a sample standard deviation of 11.3 µm. Assuming the diameters are normally distributed, find the lower limit σ of the 95% confidence interval for the population standard deviation. With df = 10, 2.5% of the χ² distribution lies below 3.247 and 2.5% lies above 20.483.

n = 11s = 11.3 µmχ² = 3.247, 20.483
σ = µm
9.Hypothesis tests & confidence intervalsMedium
t −2.13102.131

A sample of 16 students has a mean score of 114.5 and a sample standard deviation of 20. Find the test statistic t for H₀: μ = 100.

μ₀ = 100x̄ = 114.5s = 20n = 16
t =
10.Hypothesis tests & confidence intervalsHard
z −2.57602.576

A sample of 49 measurements has mean 104.8. The population standard deviation is 12. Find the upper limit U of the 99% confidence interval for the mean.

x̄ = 104.8σ = 12n = 4999%
U =
11.Linear regression & correlationMedium
xy166319722276258428102xy20256080100

The table gives 5 paired observations of x = temperature in °C and y = ice creams sold. Find Pearson's correlation coefficient r.

n = 5
r =
12.Linear regression & correlationHard
xy124225319419514612xy246152025ŷ = 16

The table gives 6 paired observations of x = car age in years and y = price in k€. Use the least-squares line ŷ = a + b·x. For which value x does the line predict ŷ = 16?

n = 6ŷ = 16
x =

3Solutions

1.Descriptive statistics

Answer: s = 15.39

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 = 221/7 = 31.57

Squared deviations

Σ(x − x̄)² = 27.43² + (−7.571)² + (−17.57)² + (−4.571)² + (−5.571)² + 14.43² + (−6.571)² = 1422

Standard deviation

s = √(1422/6) = 15.39

Answer

s = 15.39
2.Descriptive statistics

Answer: P = 0.3085

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 = −0.5

Standard normal table

Φ(−0.5) = 0.3085
P = 0.3085

Answer

P = 0.3085
3.Probability

Answer: Var(X) = 101

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 = 10·0.15 + 20·0.25 + 30·0.35 + 40·0.25 = 27

Mean of the squares

E(X²) = 100·0.15 + 400·0.25 + 900·0.35 + 1600·0.25 = 830

Variance

Var(X) = 830 − 27² = 101

Answer

Var(X) = 101
4.Probability

Answer: P = 0.03023

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.

Zero and one failure

P(X = 0) = 0.611 = 0.003628
P(X = 1) = 11·0.4·0.610 = 0.02661

Add them

P = 0.003628 + 0.02661 = 0.03023

Answer

P = 0.03023
5.Normal distribution

Answer: P = 0.2088

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 = (100 − 100)/15 = 0
z₂ = (108.25 − 100)/15 = 0.55

Area between

P = Φ(0.55) − Φ(0) = 0.7088 − 0.5 = 0.2088

Answer

P = 0.2088
6.Normal distribution

Answer: x = 58.26 points

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) = 0.25 → z = −0.6745

Back to x

x = 65 + −0.6745·10 = 58.26 points

Answer

x = 58.26 points
7.Confidence intervals

Answer: upper limit = 0.8279 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 = 134 − 138 = −4

Standard error

SE = √(10²/40 + 9²/80) = 1.874

Margin of error

E = 2.576·1.874 = 4.828

Interval

−4 ± 4.828 → [−8.828, 0.8279]

Answer

upper limit = 0.8279 h

The interval contains 0, so the data do not show a significant difference.

8.Confidence intervals

Answer: σ = 7.896 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 = 10·11.3² = 1277

Limits for σ²

1277/20.483 = 62.34
1277/3.247 = 393.3

Limit for σ

σ = √62.34 = 7.896 µm

Answer

σ = 7.896 µm

95% interval for σ: [7.896, 19.83] µm.

9.Hypothesis tests & confidence intervals

Answer: t = 2.9

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 = 20/√16 = 20/4 = 5

Test statistic

t = (114.5 − 100)/5 = 2.9

Answer

t = 2.9
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: U = 109.2

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 = 2.576·12/√49 = 4.416

Interval limit

U = 104.8 + 4.416 = 109.2

Answer

U = 109.2
11.Linear regression & correlation

Answer: r = 0.9665

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 = 110/5 = 22
ȳ = 397/5 = 79.4

Sums of squares and products

Sxx = Σ(x − x̄)² = 90
Sxy = Σ(x − x̄)(y − ȳ) = 270
Syy = Σ(y − ȳ)² = 867.2

Correlation coefficient

r = 270/√(90·867.2) = 0.9665

Answer

r = 0.9665
12.Linear regression & correlation

Answer: x = 4.566

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 = 21/6 = 3.5
ȳ = 113/6 = 18.83

Sums of squares and products

Sxx = Σ(x − x̄)² = 17.5
Sxy = Σ(x − x̄)(y − ȳ) = −46.5

Regression line

b = −46.5/17.5 = −2.657
a = 18.83 − (−2.657)·3.5 = 28.13
ŷ = 28.13 − 2.657·x

Solve for x

x = (16 − 28.13)/(−2.657) = 4.566

Answer

x = 4.566

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