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
202530354045

Find the sample standard deviation s of the data: 29, 46, 30, 36, 17, 35.

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

A quantity is normally distributed with mean 50 and standard deviation 8. What proportion P of values is below 43.6?

μ = 50σ = 8x = 43.6
P =
3.ProbabilityMedium
P(at least one six) = ??1?2?3?4?5?6?7?8P(six) = 1/6 on each roll

A fair die is rolled 8 times. What is the probability P of getting at least one six?

n = 8
P =
4.ProbabilityHard
n = 8, p = 0.4: P(X = 2) = ?k012345678

A test has 8 multiple-choice questions. A student who guesses gets each one right with probability 0.4, independently. What is the probability P of exactly 2 correct answers?

n = 8p = 0.4k = 2
P =
5.Normal distributionMedium
−3−2−10123zshaded area to the right of z = 0.0170

Drag z so that the shaded area under the standard normal curve, to the right of z, is 0.0170. Find z to two decimals.

A = 0.0170
z =
6.Normal distributionHard
μ = 250x = ?5%

The bag weights are normally distributed with mean 250 g and standard deviation 5 g. Find the value x below which 5% of them lie (the 5th percentile).

μ = 250σ = 5P = 5%
x = g
7.Confidence intervalsMedium
μ₁ − μ₂0L = ?Ux̄₁ − x̄₂E95%

The battery life of two phone models is compared. Model 1: 60 phones, mean 109.5 h, known standard deviation 20 h. Model 2: 60 phones, mean 122.5 h, known standard deviation 12 h. Find the lower limit of the 95% confidence interval for μ₁ − μ₂.

n₁ = 60x̄₁ = 109.5 hσ₁ = 20 hn₂ = 60x̄₂ = 122.5 hσ₂ = 12 h95%
lower limit = h
8.Confidence intervalsHard
n = 16σ = ?Us = 6.395%

The diameters of 16 machined shafts have a sample standard deviation of 6.3 µm. Assuming the diameters are normally distributed, find the lower limit σ of the 95% confidence interval for the population standard deviation. With df = 15, 2.5% of the χ² distribution lies below 6.262 and 2.5% lies above 27.488.

n = 16s = 6.3 µmχ² = 6.262, 27.488
σ = µm
9.Hypothesis tests & confidence intervalsMedium
−1.961.96z = ?

Two production lines make the same part. Line 1: a sample of 80 parts has mean length 54.5 mm, and the line's known standard deviation is 20 mm. Line 2: a sample of 100 parts has mean 47.7 mm, known standard deviation 3 mm. Find the test statistic z for H₀: μ₁ = μ₂, taking the difference as x̄₁ − x̄₂ (line 1 minus line 2).

n₁ = 80x̄₁ = 54.5 mmσ₁ = 20 mmn₂ = 100x̄₂ = 47.7 mmσ₂ = 3 mm
z =
10.Hypothesis tests & confidence intervalsHard
z −1.9601.96

A sample of 100 measurements has mean 196.6. The population standard deviation is 12. Find the lower limit L of the 95% confidence interval for the mean.

x̄ = 196.6σ = 12n = 10095%
L =
11.Linear regression & correlationMedium
xy2941862583110401244xy5102040

The table gives 6 paired observations of x = load in kN and y = extension in mm. Their sums are given. Find Pearson's correlation coefficient r from these sums.

n = 6Σx = 42Σy = 167Σx² = 364Σy² = 5527Σxy = 1416
r =
12.Linear regression & correlationHard
xy2941662382810381245xy5102040x = 7

The table gives 6 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 = 6x = 7
ŷ =

3Solutions

1.Descriptive statistics

Answer: s = 9.579

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 = 193/6 = 32.17

Squared deviations

Σ(x − x̄)² = (−3.167)² + 13.83² + (−2.167)² + 3.833² + (−15.17)² + 2.833² = 458.8

Standard deviation

s = √(458.8/5) = 9.579

Answer

s = 9.579
2.Descriptive statistics

Answer: P = 0.2119

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 = −0.8

Standard normal table

Φ(−0.8) = 0.2119
P = 0.2119

Answer

P = 0.2119
3.Probability

Answer: P = 0.7674

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)8 = 0.2326
P = 1 − 0.2326 = 0.7674

Answer

P = 0.7674
4.Probability

Answer: P = 0.209

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 = 2) = 28·0.42·0.66 = 28·0.16·0.04666 = 0.209

Answer

P = 0.209
5.Normal distribution

Answer: z = 2.12

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 − 0.0170 = 0.9830

Read the z-table backwards

Find 0.9830 inside the table and read off z.

z = 2.12

Answer

z = 2.12
6.Normal distribution

Answer: x = 241.8 g

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 = 250 + −1.645·5 = 241.78 g

Answer

x = 241.78 g
7.Confidence intervals

Answer: lower limit = −18.9 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 = 109.5 − 122.5 = −13

Standard error

SE = √(20²/60 + 12²/60) = 3.011

Margin of error

E = 1.96·3.011 = 5.902

Interval

−13 ± 5.902 → [−18.9, −7.098]

Answer

lower limit = −18.9 h

The interval does not contain 0, so the means differ significantly.

8.Confidence intervals

Answer: σ = 4.654 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 = 15·6.3² = 595.4

Limits for σ²

595.4/27.488 = 21.66
595.4/6.262 = 95.07

Limit for σ

σ = √21.66 = 4.654 µm

Answer

σ = 4.654 µm

95% interval for σ: [4.654, 9.751] µm.

9.Hypothesis tests & confidence intervals

Answer: z = 3.014

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.

Standard error of the difference

SE = √(20²/80 + 3²/100) = √5.09 = 2.256

Test statistic

z = (54.5 − 47.7)/2.256 = 3.014

Answer

z = 3.014

|z| > 1.96, so H₀ is rejected at the 5% level (two-sided).

10.Hypothesis tests & confidence intervals

Answer: L = 194.2

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 = 1.96·12/√100 = 2.352

Interval limit

L = 196.6 − 2.352 = 194.2

Answer

L = 194.2
11.Linear regression & correlation

Answer: r = 0.9959

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 = 1416 − 42·167/6 = 247
Sxx = 364 − 42²/6 = 70
Syy = 5527 − 167²/6 = 878.8

Correlation coefficient

r = 247/√(70·878.8) = 0.9959

Answer

r = 0.9959
12.Linear regression & correlation

Answer: ŷ = 26.5

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 = 42/6 = 7
ȳ = 159/6 = 26.5

Sums of squares and products

Sxx = Σ(x − x̄)² = 70
Sxy = Σ(x − x̄)(y − ȳ) = 251

Regression line

b = 251/70 = 3.586
a = 26.5 − 3.586·7 = 1.4
ŷ = 1.4 + 3.586·x

Prediction

ŷ = 1.4 + 3.586·7 = 26.5

Answer

ŷ = 26.5

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