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)·p^k(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

Convert the value to a z-score: z = (x − μ)/σ.
Look up Φ(z), the area to the left of z.
For the area to the right use 1 − Φ(z); for an interval subtract the two left areas.
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

Decide between z (population σ known) and t (only the sample s, with n − 1 degrees of freedom).
Find the critical value for the confidence level.
Margin of error: E = critical value × standard deviation/√n.
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

State H₀ and H₁ and decide whether the test is one- or two-sided.
Calculate the test statistic: (x̄ − μ₀)/(σ/√n), or with s and t.
Compare with the critical value, or find the p-value.
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: S_xx = Σ(x − x̄)² = Σx² − (Σx)²/n, S_xy = Σxy − ΣxΣy/n
Slope: b = S_xy/S_xx
Intercept: a = ȳ − b·x̄
Correlation: r = S_xy/√(S_xxS_yy)
Residual: e = y − ŷ

2Practice problems

1.Descriptive statisticsMedium

Find the sample standard deviation s of the data: 37, 38, 35, 59, 36, 31, 46.

n = 7

s =

2.Descriptive statisticsHard

Exam scores are normally distributed with mean 500 and standard deviation 60. Find the z-score z of a score of 428.

μ = 500σ = 60x = 428

z =

3.ProbabilityMedium

A fair die is rolled 5 times. What is the probability P of getting at least one six?

n = 5

P =

4.ProbabilityHard

10 components are tested, and each one fails with probability 0.25, independently. What is the probability P that at most one fails?

n = 10p = 0.25

P =

5.Normal distributionMedium

Drag z so that the shaded area under the standard normal curve, to the right of z, is 0.2843. Find z to two decimals.

A = 0.2843

z =

6.Normal distributionHard

The heights of adult men are normally distributed with mean 178 cm and standard deviation 7 cm. Find the value x below which 25% of them lie (the 25th percentile).

μ = 178σ = 7P = 25%

x = cm

7.Confidence intervalsMedium

A sample of 13 items has mean 47.5 and standard deviation s = 7.5. The population is roughly normal. Find the upper limit of the 95% confidence interval for the mean.

n = 13x̄ = 47.5s = 7.595%t(12) = 2.179

upper limit =

8.Confidence intervalsHard

Nothing is known in advance about the proportion of defective products. How many products n must be inspected so that a 95% confidence interval for the proportion has a margin of error of at most 0.04?

p = 0.5E = 0.0495%

n =

9.Hypothesis tests & confidence intervalsMedium

A sample of 49 students has a mean score of 54.86 and a sample standard deviation of 20. Find the test statistic t for H₀: μ = 50.

μ₀ = 50x̄ = 54.86s = 20n = 49

t =

10.Hypothesis tests & confidence intervalsHard

A sample of 49 measurements has mean 119. The population standard deviation is 10. Find the upper limit U of the 99% confidence interval for the mean.

x̄ = 119σ = 10n = 4999%

U =

11.Linear regression & correlationMedium

The table gives 6 paired observations of x = advertising spend in k€ and y = sales in k€. Find the coefficient of determination r² of the least-squares line.

n = 6

r² =

12.Linear regression & correlationHard

The table gives 6 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 = 63?

n = 6ŷ = 63

x =

3Solutions

1.Descriptive statistics

Answer: s = 9.411

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 = 282/7 = 40.29

Squared deviations

Σ(x − x̄)² = (−3.286)² + (−2.286)² + (−5.286)² + 18.71² + (−4.286)² + (−9.286)² + 5.714² = 531.4

Standard deviation

s = √(531.4/6) = 9.411

Answer

s = 9.411

2.Descriptive statistics

Answer: z = −1.2

Standardise

z = (428 − 500)/60 = −1.2

Answer

z = −1.2

3.Probability

Answer: P = 0.5981

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)⁵ = 0.4019

P = 1 − 0.4019 = 0.5981

Answer

P = 0.5981

4.Probability

Answer: P = 0.244

Zero and one failure

P(X = 0) = 0.75¹⁰ = 0.05631

P(X = 1) = 10·0.25·0.75⁹ = 0.1877

Add them

P = 0.05631 + 0.1877 = 0.244

Answer

P = 0.244

5.Normal distribution

Answer: z = 0.57

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 − 0.2843 = 0.7157

Read the z-table backwards

Find 0.7157 inside the table and read off z.

z = 0.57

Answer

z = 0.57

6.Normal distribution

Answer: x = 173.3 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) = 0.25 → z = −0.6745

Back to x

x = 178 + −0.6745·7 = 173.28 cm

Answer

x = 173.28 cm

7.Confidence intervals

Answer: upper limit = 52.03

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 = 2.179

Margin of error

E = 2.179·7.5/√13 = 4.533

Interval

47.5 ± 4.533 → [42.97, 52.03]

Answer

upper limit = 52.03

8.Confidence intervals

Answer: n = 601

Confidence interval for a proportion: p̂ ± z·√(p̂(1 − p̂)/n), where p̂ = x/n. To reach a margin E you need n ≥ z²·p(1 − p)/E², rounded up; without a prior estimate use p = 0.5, which gives the largest n.

Solve for n

n = 1.96²·0.5·0.5/0.04² = 600.2

Round up

n = 601

Answer

n = 601

9.Hypothesis tests & confidence intervals

Answer: t = 1.701

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 = 20/√49 = 20/7 = 2.857

Test statistic

t = (54.86 − 50)/2.857 = 1.701

Answer

t = 1.701

t_{0.025; df = 48} = 2.011

|t| < 2.011, so H₀ is not rejected at the 5% level (two-sided).

10.Hypothesis tests & confidence intervals

Answer: U = 122.7

Margin of error

E = 2.576·10/√49 = 3.68

Interval limit

U = 119 + 3.68 = 122.7

Answer

U = 122.7

11.Linear regression & correlation

Answer: r² = 0.77

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 = 33/6 = 5.5

ȳ = 265/6 = 44.17

Sums of squares and products

S_xx = Σ(x − x̄)² = 17.5

S_xy = Σ(x − x̄)(y − ȳ) = 51.5

S_yy = Σ(y − ȳ)² = 196.8

Coefficient of determination

r² = 51.5²/(17.5·196.8) = 0.77

Answer

r² = 0.77

About 77% of the variation in y is explained by the line.

12.Linear regression & correlation

Answer: x = 3.97

Means

x̄ = 48/6 = 8

ȳ = 549/6 = 91.5

Sums of squares and products

S_xx = Σ(x − x̄)² = 70

S_xy = Σ(x − x̄)(y − ȳ) = 495

Regression line

b = 495/70 = 7.071

a = 91.5 − 7.071·8 = 34.93

ŷ = 34.93 + 7.071·x

Solve for x

x = (63 − 34.93)/7.071 = 3.97

Answer

x = 3.97

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