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
1015202530354045

Find the sample standard deviation s of the data: 31, 10, 39, 19, 48, 32.

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

A quantity is normally distributed with mean 170 and standard deviation 15. What proportion P of values is below 162.5?

μ = 170σ = 15x = 162.5
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)10.1520.3530.340.2μ

X has the distribution shown, and Y = −2X + 20. Find the standard deviation σY of Y.

x: 1, 2, 3, 4p(x): 0.15, 0.35, 0.3, 0.2Y = −2X + 20
σY =
5.Normal distributionMedium
−3−2−10123zshaded area to the right of z = 0.9429

Drag z so that the shaded area under the standard normal curve, to the right of z, is 0.9429. Find z to two decimals.

A = 0.9429
z =
6.Normal distributionHard
μ = 32x = ?5%

The delivery times are normally distributed with mean 32 min and standard deviation 6 min. Find the value x below which 5% of them lie (the 5th percentile).

μ = 32σ = 6P = 5%
x = min
7.Confidence intervalsMedium
μ₁ − μ₂0LUx̄₁ − x̄₂E = ?99%

The battery life of two phone models is compared. Model 1: 50 phones, mean 131.5 h, known standard deviation 9 h. Model 2: 30 phones, mean 137 h, known standard deviation 18 h. Find the margin of error E of the 99% confidence interval for μ₁ − μ₂.

n₁ = 50x̄₁ = 131.5 hσ₁ = 9 hn₂ = 30x̄₂ = 137 hσ₂ = 18 h99%
E = h
8.Confidence intervalsHard
σ = 31, n = ?x̄ − Ex̄ + Ex̄E = 390%

The population standard deviation is 31. 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 3?

σ = 31E = 390%
n =
9.Hypothesis tests & confidence intervalsMedium
t −2.0302.03

A sample of 36 students has a mean score of 115.9 and a sample standard deviation of 8. Find the test statistic t for H₀: μ = 120.

μ₀ = 120x̄ = 115.9s = 8n = 36
t =
10.Hypothesis tests & confidence intervalsHard
σ = 15, n = ?x̄ − Ex̄ + Ex̄E = 2.599%

How large must a sample be so that the 99% confidence interval for a mean has a margin of error of at most 2.5? The standard deviation is 15. Find the smallest sample size n.

σ = 15E = 2.599%
n =
11.Linear regression & correlationMedium
xy346445550649757xy46455055

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

n = 5Σx = 25Σy = 247Σx² = 135Σy² = 12291Σxy = 1261
r =
12.Linear regression & correlationHard
xy142350562762977xy5406080x = 4

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 (4, 58) is made. Find its residual e = y − ŷ.

n = 5(4, 58)
e =

3Solutions

1.Descriptive statistics

Answer: s = 13.64

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 = 179/6 = 29.83

Squared deviations

Σ(x − x̄)² = 1.167² + (−19.83)² + 9.167² + (−10.83)² + 18.17² + 2.167² = 930.8

Standard deviation

s = √(930.8/5) = 13.64

Answer

s = 13.64
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: 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 = 1.947

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 = |−2|·0.9734 = 1.947

Answer

σY = 1.947

The mean does move: E(Y) = −2·2.55 + 20 = 14.9.

5.Normal distribution

Answer: z = −1.58

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 − 0.9429 = 0.0571

Read the z-table backwards

Find 0.0571 inside the table and read off z.

z = −1.58

Answer

z = −1.58
6.Normal distribution

Answer: x = 22.13 min

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 = 32 + −1.645·6 = 22.13 min

Answer

x = 22.13 min
7.Confidence intervals

Answer: E = 9.078 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 = 131.5 − 137 = −5.5

Standard error

SE = √(9²/50 + 18²/30) = 3.524

Margin of error

E = 2.576·3.524 = 9.078

Answer

E = 9.078 h

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

8.Confidence intervals

Answer: n = 289

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 = (1.645·31/3)² = 288.9

Round up

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

n = 289

Answer

n = 289
9.Hypothesis tests & confidence intervals

Answer: t = −3.075

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 = 8/√36 = 8/6 = 1.333

Test statistic

t = (115.9 − 120)/1.333 = −3.075

Answer

t = −3.075
t0.025; df = 35 = 2.03

|t| > 2.03, so H₀ is rejected at the 5% level (two-sided).

10.Hypothesis tests & confidence intervals

Answer: n = 239

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 = (2.576·15/2.5)² = 238.9

Round up

n = 239

Answer

n = 239
11.Linear regression & correlation

Answer: r = 0.8705

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 = 1261 − 25·247/5 = 26
Sxx = 135 − 25²/5 = 10
Syy = 12291 − 247²/5 = 89.2

Correlation coefficient

r = 26/√(10·89.2) = 0.8705

Answer

r = 0.8705
12.Linear regression & correlation

Answer: e = 3.5

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 = 293/5 = 58.6

Sums of squares and products

Sxx = Σ(x − x̄)² = 40
Sxy = Σ(x − x̄)(y − ȳ) = 164

Regression line

b = 164/40 = 4.1
a = 58.6 − 4.1·5 = 38.1
ŷ = 38.1 + 4.1·x

Predicted value

ŷ = 38.1 + 4.1·4 = 54.5

Residual

e = 58 − 54.5 = 3.5

Answer

e = 3.5

The point lies above the line.

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