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
15202530354045

Find the sample standard deviation s of the data: 39, 39, 48, 22, 14, 20.

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

A quantity is normally distributed with mean 60 and standard deviation 8. What proportion P of values is above 68?

μ = 60σ = 8x = 68
P =
3.ProbabilityMedium
red: 7blue: 7

A bag holds 7 red and 7 blue balls. You draw two balls without putting the first one back. What is the probability P that both are red?

7 / 7
P =
4.ProbabilityHard
n = 8, p = 0.4: P(X = 3) = ?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 3 correct answers?

n = 8p = 0.4k = 3
P =
5.Normal distributionMedium
−3−2−10123zshaded area to the right of z = 0.1112

Drag z so that the shaded area under the standard normal curve, to the right of z, is 0.1112. Find z to two decimals.

A = 0.1112
z =
6.Normal distributionHard
235240245250255260265gtop 95%

The bag weights are normally distributed with mean 250 g and standard deviation 5 g. Drag the cut-off so the shaded top 95% is to the right of it. Find the cut-off x.

μ = 250 gσ = 5 g95%
x = g
7.Confidence intervalsMedium
μ₁ − μ₂0LUx̄₁ − x̄₂E = ?95%

The battery life of two phone models is compared. Model 1: 80 phones, mean 112 h, known standard deviation 11 h. Model 2: 80 phones, mean 122 h, known standard deviation 8 h. Find the margin of error E of the 95% confidence interval for μ₁ − μ₂.

n₁ = 80x̄₁ = 112 hσ₁ = 11 hn₂ = 80x̄₂ = 122 hσ₂ = 8 h95%
E = h
8.Confidence intervalsHard
n = 20Lσ = ?s = 9.595%

The diameters of 20 machined shafts have a sample standard deviation of 9.5 µm. Assuming the diameters are normally distributed, find the upper limit σ of the 95% confidence interval for the population standard deviation. With df = 19, 2.5% of the χ² distribution lies below 8.907 and 2.5% lies above 32.852.

n = 20s = 9.5 µmχ² = 8.907, 32.852
σ = µm
9.Hypothesis tests & confidence intervalsMedium
1.44p = ?

The mean of a population is claimed to be 120; the population standard deviation is known to be 32. A random sample of 64 observations has a mean of 125.76. Test H₀: μ = 120 against H₁: μ > 120. Find the p-value p.

μ₀ = 120σ = 32n = 64x̄ = 125.76H₁: μ > μ₀
p =
10.Hypothesis tests & confidence intervalsHard
σ = 15, n = ?x̄ − Ex̄ + Ex̄E = 1.595%

How large must a sample be so that the 95% confidence interval for a mean has a margin of error of at most 1.5? The standard deviation is 15. Find the smallest sample size n.

σ = 15E = 1.595%
n =
11.Linear regression & correlationMedium
xy2104166288321041xy51010203040

The table gives 5 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 = 5Σx = 30Σy = 127Σx² = 220Σy² = 3845Σxy = 918
r =
12.Linear regression & correlationHard
xy14811710020117231262614429174xy15202530100150x = 21.5

The table gives 6 paired observations of x = temperature in °C and y = ice creams sold. Use the least-squares line ŷ = a + b·x. Predict ŷ for x = 21.5.

n = 6x = 21.5
ŷ =

3Solutions

1.Descriptive statistics

Answer: s = 13.46

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 = 182/6 = 30.33

Squared deviations

Σ(x − x̄)² = 8.667² + 8.667² + 17.67² + (−8.333)² + (−16.33)² + (−10.33)² = 905.3

Standard deviation

s = √(905.3/5) = 13.46

Answer

s = 13.46
2.Descriptive statistics

Answer: P = 0.1587

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 = 1

Standard normal table

Φ(1) = 0.8413
P = 1 − 0.8413 = 0.1587

Answer

P = 0.1587
3.Probability

Answer: P = 0.2308

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 = 7/14 · 6/13 = 42/182 = 0.2308

Answer

P = 0.2308
4.Probability

Answer: P = 0.2787

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 = 3) = 56·0.43·0.65 = 56·0.064·0.07776 = 0.2787

Answer

P = 0.2787
5.Normal distribution

Answer: z = 1.22

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 − 0.1112 = 0.8888

Read the z-table backwards

Find 0.8888 inside the table and read off z.

z = 1.22

Answer

z = 1.22
6.Normal distribution

Answer: x = 241.8 g

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) = 0.05

z from the table

z ≈ −1.64

Back to the original units

x = 250 − 1.64·5 = 241.8 g

Answer

x = 241.8 g
7.Confidence intervals

Answer: E = 2.981 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 = 112 − 122 = −10

Standard error

SE = √(11²/80 + 8²/80) = 1.521

Margin of error

E = 1.96·1.521 = 2.981

Answer

E = 2.981 h

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

8.Confidence intervals

Answer: σ = 13.88 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 = 19·9.5² = 1715

Limits for σ²

1715/32.852 = 52.2
1715/8.907 = 192.5

Limit for σ

σ = √192.5 = 13.88 µm

Answer

σ = 13.88 µm

95% interval for σ: [7.225, 13.88] µm.

9.Hypothesis tests & confidence intervals

Answer: p = 0.07493

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.

Test statistic

SE = σ/√n = 32/√64 = 4
z = (125.76 − 120)/4 = 1.44

Tail area

Φ(1.44) = 0.9251
1 − Φ(1.44) = 0.0749

p-value

p = 0.07493

Answer

p = 0.07493

p > 0.05, so H₀ is not rejected at the 5% level.

10.Hypothesis tests & confidence intervals

Answer: n = 385

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 = (1.96·15/1.5)² = 384.2

Round up

n = 385

Answer

n = 385
11.Linear regression & correlation

Answer: r = 0.9912

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 = 918 − 30·127/5 = 156
Sxx = 220 − 30²/5 = 40
Syy = 3845 − 127²/5 = 619.2

Correlation coefficient

r = 156/√(40·619.2) = 0.9912

Answer

r = 0.9912
12.Linear regression & correlation

Answer: ŷ = 123.7

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 = 129/6 = 21.5
ȳ = 742/6 = 123.7

Sums of squares and products

Sxx = Σ(x − x̄)² = 157.5
Sxy = Σ(x − x̄)(y − ȳ) = 909

Regression line

b = 909/157.5 = 5.771
a = 123.7 − 5.771·21.5 = −0.419
ŷ = −0.419 + 5.771·x

Prediction

ŷ = −0.419 + 5.771·21.5 = 123.7

Answer

ŷ = 123.7

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