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: 45, 49, 18, 35, 23.

n = 5
s =
2.Descriptive statisticsHard
z −2−1012

Exam scores are normally distributed with mean 100 and standard deviation 5. Find the z-score z of a score of 96.

μ = 100σ = 5x = 96
z =
3.ProbabilityMedium
red: 3blue: 6

A bag holds 3 red and 6 blue balls. You draw two balls without putting the first one back. What is the probability P that both are red?

3 / 6
P =
4.ProbabilityHard
n = 8, p = 0.3: P(X ≤ 1) = ?k012345678

8 components are tested, and each one fails with probability 0.3, independently. What is the probability P that at most one fails?

n = 8p = 0.3
P =
5.Normal distributionMedium
−3−2−10123zshaded area between −z and z = 0.9265

Drag z so that the shaded area between −z and z under the standard normal curve is 0.9265. Find z to two decimals.

A = 0.9265
z =
6.Normal distributionHard
380420460500540580620hbottom 10%

The battery lifetimes are normally distributed with mean 500 h and standard deviation 40 h. Drag the cut-off so the shaded bottom 10% is to the left of it. Find the cut-off x.

μ = 500 hσ = 40 h10%
x = h
7.Confidence intervalsMedium
μ₁ − μ₂0LU = ?x̄₁ − x̄₂E99%

The battery life of two phone models is compared. Model 1: 100 phones, mean 149 h, known standard deviation 15 h. Model 2: 100 phones, mean 152 h, known standard deviation 19 h. Find the upper limit of the 99% confidence interval for μ₁ − μ₂.

n₁ = 100x̄₁ = 149 hσ₁ = 15 hn₂ = 100x̄₂ = 152 hσ₂ = 19 h99%
upper limit = h
8.Confidence intervalsHard
n = 30σ = ?Us = 10.795%

The diameters of 30 machined shafts have a sample standard deviation of 10.7 µm. Assuming the diameters are normally distributed, find the lower limit σ of the 95% confidence interval for the population standard deviation. With df = 29, 2.5% of the χ² distribution lies below 16.047 and 2.5% lies above 45.722.

n = 30s = 10.7 µmχ² = 16.047, 45.722
σ = µm
9.Hypothesis tests & confidence intervalsMedium
−2.2282.228t = ?df = 10

Two teaching methods are compared. Group 1: 6 students, mean score 118, standard deviation 3.5. Group 2: 6 students, mean score 111.7, standard deviation 4. 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 = 6x̄₁ = 118s₁ = 3.5n₂ = 6x̄₂ = 111.7s₂ = 4
t =
10.Hypothesis tests & confidence intervalsHard
H₀H₁μ₀x̄cμ₀ + 5α = 0.01βσ = 20, power 90%, n = ?

A one-sided z test at α = 0.01 must detect a shift of the mean of 5 units with a power of 90%. The standard deviation is 20. Find the minimum sample size n.

α = 0.01power = 90%δ = 5σ = 20zα = 2.326zβ = 1.282
n =
11.Linear regression & correlationMedium
xy24486148141017xy51051015

The table gives 5 paired observations of x = load in kN and y = extension in mm. Find the coefficient of determination r² of the least-squares line.

n = 5
r² =
12.Linear regression & correlationHard
xy123218317412514xy241520x = 3.5

The table gives 5 paired observations of x = car age in years and y = price in k€. Use the least-squares line ŷ = a + b·x. A new observation (3.5, 20) is made. Find its residual e = y − ŷ.

n = 5(3.5, 20)
e =

3Solutions

1.Descriptive statistics

Answer: s = 13.45

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 = 170/5 = 34

Squared deviations

Σ(x − x̄)² = 11² + 15² + (−16)² + 1² + (−11)² = 724

Standard deviation

s = √(724/4) = 13.45

Answer

s = 13.45
2.Descriptive statistics

Answer: z = −0.8

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 = (96 − 100)/5 = −0.8

Answer

z = −0.8
3.Probability

Answer: P = 0.08333

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.

First and second draw

P = 3/9 · 2/8 = 6/72 = 0.08333

Answer

P = 0.08333
4.Probability

Answer: P = 0.2553

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.

Zero and one failure

P(X = 0) = 0.78 = 0.05765
P(X = 1) = 8·0.3·0.77 = 0.1977

Add them

P = 0.05765 + 0.1977 = 0.2553

Answer

P = 0.2553
5.Normal distribution

Answer: z = 1.79

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 + 0.9265)/2 = 0.9633

Read the z-table backwards

Find 0.9633 inside the table and read off z.

z = 1.79

Answer

z = 1.79
6.Normal distribution

Answer: x = 449 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) = 0.1

z from the table

z ≈ −1.28

Back to the original units

x = 500 − 1.28·40 = 449 h

Answer

x = 449 h
7.Confidence intervals

Answer: upper limit = 3.236 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 = 149 − 152 = −3

Standard error

SE = √(15²/100 + 19²/100) = 2.421

Margin of error

E = 2.576·2.421 = 6.236

Interval

−3 ± 6.236 → [−9.236, 3.236]

Answer

upper limit = 3.236 h

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

8.Confidence intervals

Answer: σ = 8.522 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 = 29·10.7² = 3320

Limits for σ²

3320/45.722 = 72.62
3320/16.047 = 206.9

Limit for σ

σ = √72.62 = 8.522 µm

Answer

σ = 8.522 µm

95% interval for σ: [8.522, 14.38] µm.

9.Hypothesis tests & confidence intervals

Answer: t = 2.903

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 = (5·3.5² + 5·4²)/10 = 14.13
sp = 3.758

Standard error

SE = 3.758·√(1/6 + 1/6) = 2.17

Test statistic

t = (118 − 111.7)/2.17 = 2.903

Answer

t = 2.903

With df = 10, the two-sided 5% critical value is 2.228; |t| is larger, so H₀ is rejected.

10.Hypothesis tests & confidence intervals

Answer: n = 209

Sample size for a test: to detect a true shift δ with power 1 − β in a one-sided z test at level α, the standard error must satisfy δ = (z_α + z_β)·σ/√n, so n = ((z_α + z_β)·σ/δ)², rounded up.

Solve for n

n = ((2.326 + 1.282)·20/5)² = 208.3

Round up

n = 209

Answer

n = 209
11.Linear regression & correlation

Answer: r² = 0.9209

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) = 64
Syy = Σ(y − ȳ)² = 111.2

Coefficient of determination

r² = 64²/(40·111.2) = 0.9209

Answer

r² = 0.9209

About 92.09% of the variation in y is explained by the line.

12.Linear regression & correlation

Answer: e = 4.4

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 = 15/5 = 3
ȳ = 84/5 = 16.8

Sums of squares and products

Sxx = Σ(x − x̄)² = 10
Sxy = Σ(x − x̄)(y − ȳ) = −24

Regression line

b = −24/10 = −2.4
a = 16.8 − (−2.4)·3 = 24
ŷ = 24 − 2.4·x

Predicted value

ŷ = 24 + (−2.4)·3.5 = 15.6

Residual

e = 20 − 15.6 = 4.4

Answer

e = 4.4

The point lies above the line.

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