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
1520253035404550

Find the sample standard deviation s of the data: 54, 14, 48, 23, 50, 54.

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

A quantity is normally distributed with mean 70 and standard deviation 8. What proportion P of values is below 86?

μ = 70σ = 8x = 86
P =
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
xp(x)10.320.0530.0540.3550.25μ

X has the distribution shown, and Y = −2X + 5. Find the standard deviation σY of Y.

x: 1, 2, 3, 4, 5p(x): 0.3, 0.05, 0.05, 0.35, 0.25Y = −2X + 5
σY =
5.Normal distributionMedium
2643.1P = ?

The delivery times are normally distributed with mean 32 min and standard deviation 6 min. What fraction lies between 26 and 43.1 min? Give P as a decimal.

μ = 32σ = 6x1 = 26x2 = 43.1
P =
6.Normal distributionHard
380420460500540580620hbottom 5%

The battery lifetimes are normally distributed with mean 500 h and standard deviation 40 h. Drag the cut-off so the shaded bottom 5% is to the left of it. Find the cut-off x.

μ = 500 hσ = 40 h5%
x = h
7.Confidence intervalsMedium
μ₁ − μ₂0L = ?Ux̄₁ − x̄₂E99%

The battery life of two phone models is compared. Model 1: 60 phones, mean 130.5 h, known standard deviation 8 h. Model 2: 100 phones, mean 123 h, known standard deviation 20 h. Find the lower limit of the 99% confidence interval for μ₁ − μ₂.

n₁ = 60x̄₁ = 130.5 hσ₁ = 8 hn₂ = 100x̄₂ = 123 hσ₂ = 20 h99%
lower limit = h
8.Confidence intervalsHard
n = 20σ = ?Us = 1.995%

The diameters of 20 machined shafts have a sample standard deviation of 1.9 µm. Assuming the diameters are normally distributed, find the lower limit σ of the 95% confidence interval for the population standard deviation. With df = 19, 2.5% of the χ² distribution lies below 8.907 and 2.5% lies above 32.852.

n = 20s = 1.9 µmχ² = 8.907, 32.852
σ = µm
9.Hypothesis tests & confidence intervalsMedium
−2.0932.093t = ?df = 19

Two teaching methods are compared. Group 1: 14 students, mean score 56.5, standard deviation 6. Group 2: 7 students, mean score 61.5, standard deviation 2.5. Assume equal population variances and use the pooled two-sample t test. Find the test statistic t for H₀: μ₁ = μ₂, taking the difference as x̄₁ − x̄₂ (group 1 minus group 2).

n₁ = 14x̄₁ = 56.5s₁ = 6n₂ = 7x̄₂ = 61.5s₂ = 2.5
t =
10.Hypothesis tests & confidence intervalsHard
σ = 20, n = ?x̄ − Ex̄ + Ex̄E = 0.590%

How large must a sample be so that the 90% confidence interval for a mean has a margin of error of at most 0.5? The standard deviation is 20. Find the smallest sample size n.

σ = 20E = 0.590%
n =
11.Linear regression & correlationMedium
xy136338547756954xy54050

The table gives 5 paired observations of x = advertising spend in k€ and y = sales in k€. Find Pearson's correlation coefficient r.

n = 5
r =
12.Linear regression & correlationHard
xy129225322419516xy2415202530ŷ = 24

The table gives 5 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 = 24?

n = 5ŷ = 24
x =

3Solutions

1.Descriptive statistics

Answer: s = 17.43

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 = 243/6 = 40.5

Squared deviations

Σ(x − x̄)² = 13.5² + (−26.5)² + 7.5² + (−17.5)² + 9.5² + 13.5² = 1520

Standard deviation

s = √(1520/5) = 17.43

Answer

s = 17.43
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) = 0.9772
P = 0.9772

Answer

P = 0.9772
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: σY = 3.2

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 = 3.2, Var(X) = 2.56, σX = 1.6

Linear change

σY = |−2|·1.6 = 3.2

Answer

σY = 3.2

The mean does move: E(Y) = −2·3.2 + 5 = −1.4.

5.Normal distribution

Answer: P = 0.8092

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 = (26 − 32)/6 = −1
z₂ = (43.1 − 32)/6 = 1.85

Area between

P = Φ(1.85) − Φ(−1) = 0.9678 − 0.1587 = 0.8092

Answer

P = 0.8092
6.Normal distribution

Answer: x = 434 h

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

Φ(z) = 0.05

z from the table

z ≈ −1.64

Back to the original units

x = 500 − 1.64·40 = 434 h

Answer

x = 434 h
7.Confidence intervals

Answer: lower limit = 1.702 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 = 130.5 − 123 = 7.5

Standard error

SE = √(8²/60 + 20²/100) = 2.251

Margin of error

E = 2.576·2.251 = 5.798

Interval

7.5 ± 5.798 → [1.702, 13.3]

Answer

lower limit = 1.702 h

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

8.Confidence intervals

Answer: σ = 1.445 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 = 19·1.9² = 68.59

Limits for σ²

68.59/32.852 = 2.088
68.59/8.907 = 7.701

Limit for σ

σ = √2.088 = 1.445 µm

Answer

σ = 1.445 µm

95% interval for σ: [1.445, 2.775] µm.

9.Hypothesis tests & confidence intervals

Answer: t = −2.094

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.

Pooled variance

sp² = (13·6² + 6·2.5²)/19 = 26.61
sp = 5.158

Standard error

SE = 5.158·√(1/14 + 1/7) = 2.388

Test statistic

t = (56.5 − 61.5)/2.388 = −2.094

Answer

t = −2.094

With df = 19, the two-sided 5% critical value is 2.093; |t| is larger, so H₀ is rejected.

10.Hypothesis tests & confidence intervals

Answer: n = 4330

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).

Solve for n

n = (1.645·20/0.5)² = 4330

Round up

n = 4330

Answer

n = 4330
11.Linear regression & correlation

Answer: r = 0.9417

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 = 231/5 = 46.2

Sums of squares and products

Sxx = Σ(x − x̄)² = 40
Sxy = Σ(x − x̄)(y − ȳ) = 108
Syy = Σ(y − ȳ)² = 328.8

Correlation coefficient

r = 108/√(40·328.8) = 0.9417

Answer

r = 0.9417
12.Linear regression & correlation

Answer: x = 2.438

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 = 15/5 = 3
ȳ = 111/5 = 22.2

Sums of squares and products

Sxx = Σ(x − x̄)² = 10
Sxy = Σ(x − x̄)(y − ȳ) = −32

Regression line

b = −32/10 = −3.2
a = 22.2 − (−3.2)·3 = 31.8
ŷ = 31.8 − 3.2·x

Solve for x

x = (24 − 31.8)/(−3.2) = 2.438

Answer

x = 2.438

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