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
2025303540455055

Find the sample standard deviation s of the data: 56, 33, 40, 44, 36, 20, 20.

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

Exam scores are normally distributed with mean 70 and standard deviation 15. Find the z-score z of a score of 88.75.

μ = 70σ = 15x = 88.75
z =
3.ProbabilityMedium
xp(x)100.3200.3300.1400.3μ = 24

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

x: 10, 20, 30, 40p(x): 0.3, 0.3, 0.1, 0.3
σ =
4.ProbabilityHard
xp(x)10.1520.3530.340.2μ

X has the distribution shown, and Y = 0.5X + 20. Find the standard deviation σY of Y.

x: 1, 2, 3, 4p(x): 0.15, 0.35, 0.3, 0.2Y = 0.5X + 20
σY =
5.Normal distributionMedium
20.338P = ?

The delivery times are normally distributed with mean 32 min and standard deviation 6 min. What fraction lies between 20.3 and 38 min? Give P as a decimal.

μ = 32σ = 6x1 = 20.3x2 = 38
P =
6.Normal distributionHard
35455565758595pointsbottom 90%

The exam scores are normally distributed with mean 65 points and standard deviation 10 points. Drag the cut-off so the shaded bottom 90% is to the left of it. Find the cut-off x.

μ = 65 pointsσ = 10 points90%
x = points
7.Confidence intervalsMedium
μ₁ − μ₂0LU = ?x̄₁ − x̄₂E90%

The battery life of two phone models is compared. Model 1: 30 phones, mean 111 h, known standard deviation 18 h. Model 2: 40 phones, mean 111.5 h, known standard deviation 5 h. Find the upper limit of the 90% confidence interval for μ₁ − μ₂.

n₁ = 30x̄₁ = 111 hσ₁ = 18 hn₂ = 40x̄₂ = 111.5 hσ₂ = 5 h90%
upper limit = h
8.Confidence intervalsHard
p = 0.5, n = ?p̂ − Ep̂ + Ep̂E = 0.02595%

Nothing is known in advance about the proportion of defective products. How many products n must be inspected so that a 95% confidence interval for the proportion has a margin of error of at most 0.025?

p = 0.5E = 0.02595%
n =
9.Hypothesis tests & confidence intervalsMedium
t −1.98401.984

A sample of 100 students has a mean score of 500.3 and a sample standard deviation of 4. Find the test statistic t for H₀: μ = 500.

μ₀ = 500x̄ = 500.3s = 4n = 100
t =
10.Hypothesis tests & confidence intervalsHard
σ = 10, n = ?x̄ − Ex̄ + Ex̄E = 1.599%

How large must a sample be so that the 99% confidence interval for a mean has a margin of error of at most 1.5? The standard deviation is 10. Find the smallest sample size n.

σ = 10E = 1.599%
n =
11.Linear regression & correlationMedium
xy1463166618802083229224106xy1520256080100

The table gives 6 paired observations of x = temperature in °C and y = ice creams sold. Their sums are given. Find Pearson's correlation coefficient r from these sums.

n = 6Σx = 114Σy = 490Σx² = 2236Σy² = 41314Σxy = 9606
r =
12.Linear regression & correlationHard
xy3535657669731185xy510607080x = 6

The table gives 5 paired observations of x = hours studied and y = exam score. Use the least-squares line ŷ = a + b·x. A new observation (6, 60) is made. Find its residual e = y − ŷ.

n = 5(6, 60)
e =

3Solutions

1.Descriptive statistics

Answer: s = 12.91

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 = 249/7 = 35.57

Squared deviations

Σ(x − x̄)² = 20.43² + (−2.571)² + 4.429² + 8.429² + 0.4286² + (−15.57)² + (−15.57)² = 999.7

Standard deviation

s = √(999.7/6) = 12.91

Answer

s = 12.91
2.Descriptive statistics

Answer: z = 1.25

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 = (88.75 − 70)/15 = 1.25

Answer

z = 1.25
3.Probability

Answer: σ = 12

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 = 10·0.3 + 20·0.3 + 30·0.1 + 40·0.3 = 24

Mean of the squares

E(X²) = 100·0.3 + 400·0.3 + 900·0.1 + 1600·0.3 = 720

Variance

Var(X) = 720 − 24² = 144
σ = √144 = 12

Answer

σ = 12
4.Probability

Answer: σY = 0.4867

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) = 0.9475, σX = 0.9734

Linear change

σY = |0.5|·0.9734 = 0.4867

Answer

σY = 0.4867

The mean does move: E(Y) = 0.5·2.55 + 20 = 21.27.

5.Normal distribution

Answer: P = 0.8158

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 = (20.3 − 32)/6 = −1.95
z₂ = (38 − 32)/6 = 1

Area between

P = Φ(1) − Φ(−1.95) = 0.8413 − 0.02559 = 0.8158

Answer

P = 0.8158
6.Normal distribution

Answer: x = 77.8 points

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) = 0.9

z from the table

z ≈ 1.28

Back to the original units

x = 65 + 1.28·10 = 77.8 points

Answer

x = 77.8 points
7.Confidence intervals

Answer: upper limit = 5.06 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 = 111 − 111.5 = −0.5

Standard error

SE = √(18²/30 + 5²/40) = 3.38

Margin of error

E = 1.645·3.38 = 5.56

Interval

−0.5 ± 5.56 → [−6.06, 5.06]

Answer

upper limit = 5.06 h

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

8.Confidence intervals

Answer: n = 1537

Confidence interval for a proportion: p̂ ± z·√(p̂(1 − p̂)/n), where p̂ = x/n. To reach a margin E you need n ≥ z²·p(1 − p)/E², rounded up; without a prior estimate use p = 0.5, which gives the largest n.

Solve for n

n = 1.96²·0.5·0.5/0.025² = 1537

Round up

n = 1537

Answer

n = 1537
9.Hypothesis tests & confidence intervals

Answer: t = 0.75

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/√100 = 4/10 = 0.4

Test statistic

t = (500.3 − 500)/0.4 = 0.75

Answer

t = 0.75
t0.025; df = 99 = 1.984

|t| < 1.984, so H₀ is not rejected at the 5% level (two-sided).

10.Hypothesis tests & confidence intervals

Answer: n = 295

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 = (2.576·10/1.5)² = 294.9

Round up

n = 295

Answer

n = 295
11.Linear regression & correlation

Answer: r = 0.9822

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.

Centred sums

Sxy = 9606 − 114·490/6 = 296
Sxx = 2236 − 114²/6 = 70
Syy = 41314 − 490²/6 = 1297

Correlation coefficient

r = 296/√(70·1297) = 0.9822

Answer

r = 0.9822
12.Linear regression & correlation

Answer: e = −4.8

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 = 342/5 = 68.4

Sums of squares and products

Sxx = Σ(x − x̄)² = 40
Sxy = Σ(x − x̄)(y − ȳ) = 144

Regression line

b = 144/40 = 3.6
a = 68.4 − 3.6·7 = 43.2
ŷ = 43.2 + 3.6·x

Predicted value

ŷ = 43.2 + 3.6·6 = 64.8

Residual

e = 60 − 64.8 = −4.8

Answer

e = −4.8

The point lies below the line.

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