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: 51, 15, 29, 23, 21.

n = 5
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)00.3510.320.130.25μ = 1.25

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

x: 0, 1, 2, 3p(x): 0.35, 0.3, 0.1, 0.25
Var(X) =
4.ProbabilityHard
n = 11, p = 0.2: P(X = 5) = ?k01234567891011

A test has 11 multiple-choice questions. A student who guesses gets each one right with probability 0.2, independently. What is the probability P of exactly 5 correct answers?

n = 11p = 0.2k = 5
P =
5.Normal distributionMedium
225 kg223.5 kgT = X₁ + … + X₉?

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 9 of them is below 223.5 kg. Round z to two decimals.

μ = 25 kgσ = 0.5 kgn = 9< 223.5 kg
P =
6.Normal distributionHard
μ = 178x = ?5%

The heights of adult men are normally distributed with mean 178 cm and standard deviation 7 cm. Find the value x below which 5% of them lie (the 5th percentile).

μ = 178σ = 7P = 5%
x = cm
7.Confidence intervalsMedium
n = 13, s = 1L = ?Ux̄ = 95E90%

A sample of 13 items has mean 95 and standard deviation s = 1. The population is roughly normal. Find the lower limit of the 90% confidence interval for the mean.

n = 13x̄ = 95s = 190%t(12) = 1.782
lower limit =
8.Confidence intervalsHard
σ = 24, n = ?x̄ − Ex̄ + Ex̄E = 1.599%

The population standard deviation is 24. How large a sample n do you need so that a 99% confidence interval for the mean has a margin of error of at most 1.5?

σ = 24E = 1.599%
n =
9.Hypothesis tests & confidence intervalsMedium
−2.3062.306t = ?df = 8

9 employees take a typing test before and after a training course. For each person the difference d = after − before (words per minute) is recorded. The mean difference is 0.95 and the standard deviation of the differences is 2.2. Find the paired test statistic t for H₀: μd = 0.

n = 9d̄ = 0.95sd = 2.2
t =
10.Hypothesis tests & confidence intervalsHard
z −2.57602.576

A sample of 100 measurements has mean 73.8. The population standard deviation is 4. Find the upper limit U of the 99% confidence interval for the mean.

x̄ = 73.8σ = 4n = 10099%
U =
11.Linear regression & correlationMedium
xy158318100211182414127163xy152025100150

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
xy247362463574679xy24650607080ŷ = 54

The table gives 5 paired observations of x = hours studied and y = exam score. Use the least-squares line ŷ = a + b·x. For which value x does the line predict ŷ = 54?

n = 5ŷ = 54
x =

3Solutions

1.Descriptive statistics

Answer: s = 13.9

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 = 139/5 = 27.8

Squared deviations

Σ(x − x̄)² = 23.2² + (−12.8)² + 1.2² + (−4.8)² + (−6.8)² = 772.8

Standard deviation

s = √(772.8/4) = 13.9

Answer

s = 13.9
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: Var(X) = 1.388

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 = 0·0.35 + 1·0.3 + 2·0.1 + 3·0.25 = 1.25

Mean of the squares

E(X²) = 0·0.35 + 1·0.3 + 4·0.1 + 9·0.25 = 2.95

Variance

Var(X) = 2.95 − 1.25² = 1.388

Answer

Var(X) = 1.388
4.Probability

Answer: P = 0.03876

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) = 462·0.25·0.86 = 462·0.00032·0.2621 = 0.03876

Answer

P = 0.03876
5.Normal distribution

Answer: P = 0.1587

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 = 225 kg, σT = 0.5·√9 = 1.5 kg

Standardise

z = (223.5 − 225)/1.5 = −1

Probability

P = Φ(−1) = 0.1587

Answer

P = 0.1587
6.Normal distribution

Answer: x = 166.5 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) = 0.05 → z = −1.645

Back to x

x = 178 + −1.645·7 = 166.49 cm

Answer

x = 166.49 cm
7.Confidence intervals

Answer: lower limit = 94.51

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 = 12, t = 1.782

Margin of error

E = 1.782·1/√13 = 0.4942

Interval

95 ± 0.4942 → [94.51, 95.49]

Answer

lower limit = 94.51
8.Confidence intervals

Answer: n = 1699

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 = (2.576·24/1.5)² = 1699

Round up

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

n = 1699

Answer

n = 1699
9.Hypothesis tests & confidence intervals

Answer: t = 1.295

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).

Standard error of the mean difference

SE = 2.2/√9 = 0.7333

Test statistic

t = 0.95/0.7333 = 1.295

Answer

t = 1.295

With df = 8, the two-sided 5% critical value is 2.306; |t| is smaller, so H₀ is not rejected.

10.Hypothesis tests & confidence intervals

Answer: U = 74.83

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 = 2.576·4/√100 = 1.03

Interval limit

U = 73.8 + 1.03 = 74.83

Answer

U = 74.83
11.Linear regression & correlation

Answer: r² = 0.9956

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 = 105/5 = 21
ȳ = 605/5 = 121

Sums of squares and products

Sxx = Σ(x − x̄)² = 90
Sxy = Σ(x − x̄)(y − ȳ) = 603
Syy = Σ(y − ȳ)² = 4058

Coefficient of determination

r² = 603²/(90·4058) = 0.9956

Answer

r² = 0.9956

About 99.56% of the variation in y is explained by the line.

12.Linear regression & correlation

Answer: x = 2.553

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 = 20/5 = 4
ȳ = 325/5 = 65

Sums of squares and products

Sxx = Σ(x − x̄)² = 10
Sxy = Σ(x − x̄)(y − ȳ) = 76

Regression line

b = 76/10 = 7.6
a = 65 − 7.6·4 = 34.6
ŷ = 34.6 + 7.6·x

Solve for x

x = (54 − 34.6)/7.6 = 2.553

Answer

x = 2.553

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