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
3540455055

Find the sample standard deviation s of the data: 59, 42, 33, 58, 33, 40, 51.

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

Exam scores are normally distributed with mean 70 and standard deviation 8. Find the z-score z of a score of 78.

μ = 70σ = 8x = 78
z =
3.ProbabilityMedium
xp(x)100.3200.3300.25400.15μ = 22.5

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

x: 10, 20, 30, 40p(x): 0.3, 0.3, 0.25, 0.15
Var(X) =
4.ProbabilityHard
xp(x)00.3510.2520.230.2μ

X has the distribution shown, and Y = 0.5X − 3. Find the standard deviation σY of Y.

x: 0, 1, 2, 3p(x): 0.35, 0.25, 0.2, 0.2Y = 0.5X − 3
σY =
5.Normal distributionMedium
−3−2−10123zshaded area to the right of z = 0.1170

Drag z so that the shaded area under the standard normal curve, to the right of z, is 0.1170. Find z to two decimals.

A = 0.1170
z =
6.Normal distributionHard
380420460500540580620hbottom 40%

The battery lifetimes are normally distributed with mean 500 h and standard deviation 40 h. Drag the cut-off so the shaded bottom 40% is to the left of it. Find the cut-off x.

μ = 500 hσ = 40 h40%
x = h
7.Confidence intervalsMedium
n = 15, s = 1LU = ?x̄ = 71.5E95%

A sample of 15 items has mean 71.5 and standard deviation s = 1. The population is roughly normal. Find the upper limit of the 95% confidence interval for the mean.

n = 15x̄ = 71.5s = 195%t(14) = 2.145
upper limit =
8.Confidence intervalsHard
n = 25σ = ?Us = 11.595%

The diameters of 25 machined shafts have a sample standard deviation of 11.5 µm. Assuming the diameters are normally distributed, find the lower limit σ of the 95% confidence interval for the population standard deviation. With df = 24, 2.5% of the χ² distribution lies below 12.401 and 2.5% lies above 39.364.

n = 25s = 11.5 µmχ² = 12.401, 39.364
σ = µm
9.Hypothesis tests & confidence intervalsMedium
t −1.9901.99

A sample of 81 students has a mean score of 94.83 and a sample standard deviation of 15. Find the test statistic t for H₀: μ = 100.

μ₀ = 100x̄ = 94.83s = 15n = 81
t =
10.Hypothesis tests & confidence intervalsHard
z −1.9601.96

A sample of 25 measurements has mean 104.8. The population standard deviation is 20. Find the lower limit L of the 95% confidence interval for the mean.

x̄ = 104.8σ = 20n = 2595%
L =
11.Linear regression & correlationMedium
xy36357079299911114xy5106080100120

The table gives 5 paired observations of x = hours studied and y = exam score. Their sums are given. Find Pearson's correlation coefficient r from these sums.

n = 5Σx = 35Σy = 438Σx² = 285Σy² = 40130Σxy = 3328
r =
12.Linear regression & correlationHard
xy274116118171017xy5101015x = 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. Predict ŷ for x = 7.

n = 5x = 7
ŷ =

3Solutions

1.Descriptive statistics

Answer: s = 10.98

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 = 316/7 = 45.14

Squared deviations

Σ(x − x̄)² = 13.86² + (−3.143)² + (−12.14)² + 12.86² + (−12.14)² + (−5.143)² + 5.857² = 722.9

Standard deviation

s = √(722.9/6) = 10.98

Answer

s = 10.98
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 = (78 − 70)/8 = 1

Answer

z = 1
3.Probability

Answer: Var(X) = 108.8

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 = 10·0.3 + 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

Answer

Var(X) = 108.8
4.Probability

Answer: σY = 0.5673

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 = 1.25, Var(X) = 1.288, σX = 1.135

Linear change

σY = |0.5|·1.135 = 0.5673

Answer

σY = 0.5673

The mean does move: E(Y) = 0.5·1.25 − 3 = −2.375.

5.Normal distribution

Answer: z = 1.19

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 − 0.1170 = 0.8830

Read the z-table backwards

Find 0.8830 inside the table and read off z.

z = 1.19

Answer

z = 1.19
6.Normal distribution

Answer: x = 490 h

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) = 0.4

z from the table

z ≈ −0.25

Back to the original units

x = 500 − 0.25·40 = 490 h

Answer

x = 490 h
7.Confidence intervals

Answer: upper limit = 72.05

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 = 2.145·1/√15 = 0.5538

Interval

71.5 ± 0.5538 → [70.95, 72.05]

Answer

upper limit = 72.05
8.Confidence intervals

Answer: σ = 8.98 µm

Interval for a variance: (n − 1)s²/σ² has a χ² distribution with n − 1 degrees of freedom. The 95% interval for σ² runs from (n − 1)s² divided by the upper χ² value (2.5% above it) to (n − 1)s² divided by the lower χ² value (2.5% below it); the interval for σ is the square root of both limits. It is not symmetric around s.

Sum of squares

(n − 1)s² = 24·11.5² = 3174

Limits for σ²

3174/39.364 = 80.63
3174/12.401 = 255.9

Limit for σ

σ = √80.63 = 8.98 µm

Answer

σ = 8.98 µm

95% interval for σ: [8.98, 16] µm.

9.Hypothesis tests & confidence intervals

Answer: t = −3.102

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 = 15/√81 = 15/9 = 1.667

Test statistic

t = (94.83 − 100)/1.667 = −3.102

Answer

t = −3.102
t0.025; df = 80 = 1.99

|t| > 1.99, so H₀ is rejected at the 5% level (two-sided).

10.Hypothesis tests & confidence intervals

Answer: L = 96.96

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 = 1.96·20/√25 = 7.84

Interval limit

L = 104.8 − 7.84 = 96.96

Answer

L = 96.96
11.Linear regression & correlation

Answer: r = 0.9871

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 = 3328 − 35·438/5 = 262
Sxx = 285 − 35²/5 = 40
Syy = 40130 − 438²/5 = 1761

Correlation coefficient

r = 262/√(40·1761) = 0.9871

Answer

r = 0.9871
12.Linear regression & correlation

Answer: ŷ = 13.9

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 = 63/5 = 12.6

Sums of squares and products

Sxx = Σ(x − x̄)² = 40
Sxy = Σ(x − x̄)(y − ȳ) = 52

Regression line

b = 52/40 = 1.3
a = 12.6 − 1.3·6 = 4.8
ŷ = 4.8 + 1.3·x

Prediction

ŷ = 4.8 + 1.3·7 = 13.9

Answer

ŷ = 13.9

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