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
2530354045

Find the sample standard deviation s of the data: 21, 30, 34, 46, 38, 28.

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

Exam scores are normally distributed with mean 60 and standard deviation 8. Find the z-score z of a score of 68.

μ = 60σ = 8x = 68
z =
3.ProbabilityMedium
xp(x)100.25200.35300.1400.3μ = 24.5

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

x: 10, 20, 30, 40p(x): 0.25, 0.35, 0.1, 0.3
Var(X) =
4.ProbabilityHard
n = 12, p = 0.3: P(X = 1) = ?k0123456789101112

A test has 12 multiple-choice questions. A student who guesses gets each one right with probability 0.3, independently. What is the probability P of exactly 1 correct answers?

n = 12p = 0.3k = 1
P =
5.Normal distributionMedium
−3−2−10123zshaded area between −z and z = 0.5527

Drag z so that the shaded area between −z and z under the standard normal curve is 0.5527. Find z to two decimals.

A = 0.5527
z =
6.Normal distributionHard
235240245250255260265gtop 90%

The bag weights are normally distributed with mean 250 g and standard deviation 5 g. Drag the cut-off so the shaded top 90% is to the right of it. Find the cut-off x.

μ = 250 gσ = 5 g90%
x = g
7.Confidence intervalsMedium
x̄E₁ = 2.699%, n = 36x̄E₂ = ?90%, n = 36

A 99% confidence interval for a mean, based on 36 observations and a known σ, has a margin of error of 2.6. What is the margin of error E₂ of a 90% interval from the same data?

99%n₁ = 36E₁ = 2.690%
E₂ =
8.Confidence intervalsHard
σ = 26, n = ?x̄ − Ex̄ + Ex̄E = 2.590%

The population standard deviation is 26. How large a sample n do you need so that a 90% confidence interval for the mean has a margin of error of at most 2.5?

σ = 26E = 2.590%
n =
9.Hypothesis tests & confidence intervalsMedium
−2.0642.064t = ?df = 24

Two teaching methods are compared. Group 1: 17 students, mean score 107.5, standard deviation 3.5. Group 2: 9 students, mean score 104.2, standard deviation 10.5. 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 = 17x̄₁ = 107.5s₁ = 3.5n₂ = 9x̄₂ = 104.2s₂ = 10.5
t =
10.Hypothesis tests & confidence intervalsHard
z −1.9601.96

A sample of 81 measurements has mean 198.9. The population standard deviation is 5. Find the lower limit L of the 95% confidence interval for the mean.

x̄ = 198.9σ = 5n = 8195%
L =
11.Linear regression & correlationMedium
xy15741895211092412927142xy152025100150

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
xy25486138151016xy51051015ŷ = 7

The table gives 5 paired observations of x = load in kN and y = extension in mm. Use the least-squares line ŷ = a + b·x. For which value x does the line predict ŷ = 7?

n = 5ŷ = 7
x =

3Solutions

1.Descriptive statistics

Answer: s = 8.635

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 = 197/6 = 32.83

Squared deviations

Σ(x − x̄)² = (−11.83)² + (−2.833)² + 1.167² + 13.17² + 5.167² + (−4.833)² = 372.8

Standard deviation

s = √(372.8/5) = 8.635

Answer

s = 8.635
2.Descriptive statistics

Answer: z = 1

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 = (68 − 60)/8 = 1

Answer

z = 1
3.Probability

Answer: Var(X) = 134.8

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 = 10·0.25 + 20·0.35 + 30·0.1 + 40·0.3 = 24.5

Mean of the squares

E(X²) = 100·0.25 + 400·0.35 + 900·0.1 + 1600·0.3 = 735

Variance

Var(X) = 735 − 24.5² = 134.8

Answer

Var(X) = 134.8
4.Probability

Answer: P = 0.07118

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 = 1) = 12·0.31·0.711 = 12·0.3·0.01977 = 0.07118

Answer

P = 0.07118
5.Normal distribution

Answer: z = 0.76

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 of z

Φ(z) = (1 + 0.5527)/2 = 0.7764

Read the z-table backwards

Find 0.7764 inside the table and read off z.

z = 0.76

Answer

z = 0.76
6.Normal distribution

Answer: x = 243.6 g

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) = 0.1

z from the table

z ≈ −1.28

Back to the original units

x = 250 − 1.28·5 = 243.6 g

Answer

x = 243.6 g
7.Confidence intervals

Answer: E₂ = 1.66

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.

Ratio of margins

E₂/E₁ = (1.645/2.576)·√(36/36) = 0.6386

New margin

E₂ = 2.6·0.6386 = 1.66

Answer

E₂ = 1.66
8.Confidence intervals

Answer: n = 293

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 = (1.645·26/2.5)² = 292.7

Round up

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

n = 293

Answer

n = 293
9.Hypothesis tests & confidence intervals

Answer: t = 1.194

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 = (16·3.5² + 8·10.5²)/24 = 44.92
sp = 6.702

Standard error

SE = 6.702·√(1/17 + 1/9) = 2.763

Test statistic

t = (107.5 − 104.2)/2.763 = 1.194

Answer

t = 1.194

With df = 24, the two-sided 5% critical value is 2.064; |t| is smaller, so H₀ is not rejected.

10.Hypothesis tests & confidence intervals

Answer: L = 197.8

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 = 1.96·5/√81 = 1.089

Interval limit

L = 198.9 − 1.089 = 197.8

Answer

L = 197.8
11.Linear regression & correlation

Answer: r = 0.9971

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 = 549/5 = 109.8

Sums of squares and products

Sxx = Σ(x − x̄)² = 90
Sxy = Σ(x − x̄)(y − ȳ) = 510
Syy = Σ(y − ȳ)² = 2907

Correlation coefficient

r = 510/√(90·2907) = 0.9971

Answer

r = 0.9971
12.Linear regression & correlation

Answer: x = 2.966

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 = 30/5 = 6
ȳ = 57/5 = 11.4

Sums of squares and products

Sxx = Σ(x − x̄)² = 40
Sxy = Σ(x − x̄)(y − ȳ) = 58

Regression line

b = 58/40 = 1.45
a = 11.4 − 1.45·6 = 2.7
ŷ = 2.7 + 1.45·x

Solve for x

x = (7 − 2.7)/1.45 = 2.966

Answer

x = 2.966

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