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
36384042444648505254

Find the sample standard deviation s of the data: 55, 55, 38, 35, 47.

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

A quantity is normally distributed with mean 70 and standard deviation 10. What proportion P of values is below 65?

μ = 70σ = 10x = 65
P =
3.ProbabilityMedium
xp(x)100.3200.3300.25400.15μ = 22.5

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

x: 10, 20, 30, 40p(x): 0.3, 0.3, 0.25, 0.15
σ =
4.ProbabilityHard
n = 11, p = 0.4: P(X = 5) = ?k01234567891011

A test has 11 multiple-choice questions. A student who guesses gets each one right with probability 0.4, independently. What is the probability P of exactly 5 correct answers?

n = 11p = 0.4k = 5
P =
5.Normal distributionMedium
125 kg124.4 kgT = X₁ + … + X₅?

The masses of bags of cement are independent and normal with mean 25 kg and standard deviation 0.5 kg. Find the probability P that the total mass of 5 of them exceeds 124.4 kg. Round z to two decimals.

μ = 25 kgσ = 0.5 kgn = 5> 124.4 kg
P =
6.Normal distributionHard
35455565758595pointstop 40%

The exam scores are normally distributed with mean 65 points and standard deviation 10 points. Drag the cut-off so the shaded top 40% is to the right of it. Find the cut-off x.

μ = 65 pointsσ = 10 points40%
x = points
7.Confidence intervalsMedium
n = 15, s = 8.5L = ?Ux̄ = 89.5E95%

A sample of 15 items has mean 89.5 and standard deviation s = 8.5. The population is roughly normal. Find the lower limit of the 95% confidence interval for the mean.

n = 15x̄ = 89.5s = 8.595%t(14) = 2.145
lower limit =
8.Confidence intervalsHard
μ₁ − μ₂, t(20) = 2.0860L = ?Ux̄₁ − x̄₂E95%

Two fertilisers are tested. Plot group 1: 16 plots, mean yield 77.5 kg, standard deviation 1.5 kg. Plot group 2: 6 plots, mean 79 kg, standard deviation 1 kg. Assume equal variances (pooled t, df = n₁ + n₂ − 2). Find the lower limit of the 95% confidence interval for μ₁ − μ₂.

n₁ = 16x̄₁ = 77.5s₁ = 1.5n₂ = 6x̄₂ = 79s₂ = 195%t(20) = 2.086
lower limit = kg
9.Hypothesis tests & confidence intervalsMedium
−2.2282.228t = ?df = 10

Two teaching methods are compared. Group 1: 7 students, mean score 27.5, standard deviation 2.5. Group 2: 5 students, mean score 33.3, standard deviation 4.5. 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 = 7x̄₁ = 27.5s₁ = 2.5n₂ = 5x̄₂ = 33.3s₂ = 4.5
t =
10.Hypothesis tests & confidence intervalsHard
z −2.57602.576

A sample of 49 measurements has mean 70.9. The population standard deviation is 12. Find the upper limit U of the 99% confidence interval for the mean.

x̄ = 70.9σ = 12n = 4999%
U =
11.Linear regression & correlationMedium
xy1874208722952410526110xy202580100

The table gives 5 paired observations of x = temperature in °C and y = ice creams sold. Find the coefficient of determination r² of the least-squares line.

n = 5
r² =
12.Linear regression & correlationHard
xy143363572786995xy5406080100x = 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, 83) is made. Find its residual e = y − ŷ.

n = 5(6, 83)
e =

3Solutions

1.Descriptive statistics

Answer: s = 9.327

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 = 230/5 = 46

Squared deviations

Σ(x − x̄)² = 9² + 9² + (−8)² + (−11)² + 1² = 348

Standard deviation

s = √(348/4) = 9.327

Answer

s = 9.327
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: σ = 10.43

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 + 30·0.25 + 40·0.15 = 22.5

Mean of the squares

E(X²) = 100·0.3 + 400·0.3 + 900·0.25 + 1600·0.15 = 615

Variance

Var(X) = 615 − 22.5² = 108.8
σ = √108.8 = 10.43

Answer

σ = 10.43
4.Probability

Answer: P = 0.2207

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.

Binomial formula

P(X = 5) = 462·0.45·0.66 = 462·0.01024·0.04666 = 0.2207

Answer

P = 0.2207
5.Normal distribution

Answer: P = 0.7054

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 total

μT = 125 kg, σT = 0.5·√5 = 1.118 kg

Standardise

z = (124.4 − 125)/1.118 = −0.54

Probability

P = 1 − Φ(−0.54) = 0.7054

Answer

P = 0.7054
6.Normal distribution

Answer: x = 67.5 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) = 0.6

z from the table

z ≈ 0.25

Back to the original units

x = 65 + 0.25·10 = 67.5 points

Answer

x = 67.5 points
7.Confidence intervals

Answer: lower limit = 84.79

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 = 14, t = 2.145

Margin of error

E = 2.145·8.5/√15 = 4.708

Interval

89.5 ± 4.708 → [84.79, 94.21]

Answer

lower limit = 84.79
8.Confidence intervals

Answer: lower limit = −2.89 kg

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.

Pooled standard deviation

sp² = (15·1.5² + 5·1²)/20 = 1.938
sp = 1.392

Margin of error

E = 2.086·1.392·√(1/16 + 1/6) = 1.39

Interval

−1.5 ± 1.39 → [−2.89, −0.11]

Answer

lower limit = −2.89 kg
9.Hypothesis tests & confidence intervals

Answer: t = −2.877

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 = (6·2.5² + 4·4.5²)/10 = 11.85
sp = 3.442

Standard error

SE = 3.442·√(1/7 + 1/5) = 2.016

Test statistic

t = (27.5 − 33.3)/2.016 = −2.877

Answer

t = −2.877

With df = 10, the two-sided 5% critical value is 2.228; |t| is larger, so H₀ is rejected.

10.Hypothesis tests & confidence intervals

Answer: U = 75.32

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 = 70.9 + 4.416 = 75.32

Answer

U = 75.32
11.Linear regression & correlation

Answer: r² = 0.9797

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 = 471/5 = 94.2

Sums of squares and products

Sxx = Σ(x − x̄)² = 40
Sxy = Σ(x − x̄)(y − ȳ) = 180
Syy = Σ(y − ȳ)² = 826.8

Coefficient of determination

r² = 180²/(40·826.8) = 0.9797

Answer

r² = 0.9797

About 97.97% of the variation in y is explained by the line.

12.Linear regression & correlation

Answer: e = 4.85

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 = 359/5 = 71.8

Sums of squares and products

Sxx = Σ(x − x̄)² = 40
Sxy = Σ(x − x̄)(y − ȳ) = 254

Regression line

b = 254/40 = 6.35
a = 71.8 − 6.35·5 = 40.05
ŷ = 40.05 + 6.35·x

Predicted value

ŷ = 40.05 + 6.35·6 = 78.15

Residual

e = 83 − 78.15 = 4.85

Answer

e = 4.85

The point lies above the line.

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