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
3035404550

Find the sample standard deviation s of the data: 46, 33, 36, 26, 54.

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

A quantity is normally distributed with mean 100 and standard deviation 15. What proportion P of values is above 77.5?

μ = 100σ = 15x = 77.5
P =
3.ProbabilityMedium
xp(x)00.3510.0520.2530.35μ = 1.6

For the distribution shown, find the standard deviation σ of X.

x: 0, 1, 2, 3p(x): 0.35, 0.05, 0.25, 0.35
σ =
4.ProbabilityHard
n = 9, p = 0.2: P(X = 3) = ?k0123456789

A test has 9 multiple-choice questions. A student who guesses gets each one right with probability 0.2, independently. What is the probability P of exactly 3 correct answers?

n = 9p = 0.2k = 3
P =
5.Normal distributionMedium
−3−2−10123zshaded area to the right of z = 0.8849

Drag z so that the shaded area under the standard normal curve, to the right of z, is 0.8849. Find z to two decimals.

A = 0.8849
z =
6.Normal distributionHard
0.04 mm0D − d?

Hole diameters D are normal with mean 25.04 mm and standard deviation 0.02 mm. Shaft diameters d are independent and normal with mean 25 mm and standard deviation 0.01 mm. A randomly chosen shaft fits when D − d > 0. Find the probability P that it fits. Round z to two decimals.

hole: μ = 25.04 mm, σ = 0.02 mmshaft: μ = 25 mm, σ = 0.01 mm
P =
7.Confidence intervalsMedium
n = 20, sd​ = 4.10L = ?Ud̄ = 9.2E90%

The blood pressure of 20 patients is measured before and after a treatment. The differences d = before − after (mmHg) have mean 9.2 and standard deviation 4.1. Find the lower limit of the 90% confidence interval for the mean difference.

n = 20d̄ = 9.2sd = 4.190%t(19) = 1.729
lower limit = mmHg
8.Confidence intervalsHard
n = 25Lσ = ?s = 10.595%

The diameters of 25 machined shafts have a sample standard deviation of 10.5 µm. Assuming the diameters are normally distributed, find the upper limit σ of the 95% confidence interval for the population standard deviation. With df = 24, 2.5% of the χ² distribution lies below 12.401 and 2.5% lies above 39.364.

n = 25s = 10.5 µmχ² = 12.401, 39.364
σ = µm
9.Hypothesis tests & confidence intervalsMedium
t −2.06402.064

A sample of 25 students has a mean score of 80.04 and a sample standard deviation of 12. Find the test statistic t for H₀: μ = 75.

μ₀ = 75x̄ = 80.04s = 12n = 25
t =
10.Hypothesis tests & confidence intervalsHard
z −2.57602.576

A sample of 16 measurements has mean 120.3. The population standard deviation is 4. Find the lower limit L of the 99% confidence interval for the mean.

x̄ = 120.3σ = 4n = 1699%
L =
11.Linear regression & correlationMedium
xy258359470578682792xy2466080

The table gives 6 paired observations of x = hours studied and y = exam score. Their sums are given. Find Pearson's correlation coefficient r from these sums.

n = 6Σx = 27Σy = 439Σx² = 139Σy² = 33017Σxy = 2099
r =
12.Linear regression & correlationHard
xy148259360467571xy24506070x = 4.5

The table gives 5 paired observations of x = hours studied and y = exam score. Use the least-squares line ŷ = a + b·x. A new observation (4.5, 64) is made. Find its residual e = y − ŷ.

n = 5(4.5, 64)
e =

3Solutions

1.Descriptive statistics

Answer: s = 11.05

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 = 195/5 = 39

Squared deviations

Σ(x − x̄)² = 7² + (−6)² + (−3)² + (−13)² + 15² = 488

Standard deviation

s = √(488/4) = 11.05

Answer

s = 11.05
2.Descriptive statistics

Answer: P = 0.9332

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 = −1.5

Standard normal table

Φ(−1.5) = 0.06681
P = 1 − 0.06681 = 0.9332

Answer

P = 0.9332
3.Probability

Answer: σ = 1.281

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 = 0·0.35 + 1·0.05 + 2·0.25 + 3·0.35 = 1.6

Mean of the squares

E(X²) = 0·0.35 + 1·0.05 + 4·0.25 + 9·0.35 = 4.2

Variance

Var(X) = 4.2 − 1.6² = 1.64
σ = √1.64 = 1.281

Answer

σ = 1.281
4.Probability

Answer: P = 0.1762

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) = 84·0.23·0.86 = 84·0.008·0.2621 = 0.1762

Answer

P = 0.1762
5.Normal distribution

Answer: z = −1.2

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 − 0.8849 = 0.1151

Read the z-table backwards

Find 0.1151 inside the table and read off z.

z = −1.20

Answer

z = −1.20
6.Normal distribution

Answer: P = 0.9633

Sums and differences of normal variables: if X and Y are independent and normal, X + Y and X − Y are normal too. Means add or subtract: E(X ± Y) = μX ± μY. Variances always add, also for a difference: Var(X ± Y) = σX² + σY², so σ = √(σX² + σY²). Standard deviations never add directly. The total of n independent items with the same distribution has mean n·μ and standard deviation σ·√n. Then standardise as usual: z = (x − mean)/σ and read Φ(z).

Distribution of the difference

μ = 25.04 − 25 = 0.04 mm, σ = 0.02236 mm

Standardise

z = (0 − 0.04)/0.02236 = −1.79

Probability

P = 1 − Φ(−1.79) = 0.9633

Answer

P = 0.9633
7.Confidence intervals

Answer: lower limit = 7.615 mmHg

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.

Critical value

df = 19, t = 1.729

Margin of error

E = 1.729·4.1/√20 = 1.585

Interval

9.2 ± 1.585 → [7.615, 10.79]

Answer

lower limit = 7.615 mmHg
8.Confidence intervals

Answer: σ = 14.61 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 = 24·10.5² = 2646

Limits for σ²

2646/39.364 = 67.22
2646/12.401 = 213.4

Limit for σ

σ = √213.4 = 14.61 µm

Answer

σ = 14.61 µm

95% interval for σ: [8.199, 14.61] µm.

9.Hypothesis tests & confidence intervals

Answer: t = 2.1

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 = 12/√25 = 12/5 = 2.4

Test statistic

t = (80.04 − 75)/2.4 = 2.1

Answer

t = 2.1
t0.025; df = 24 = 2.064

|t| > 2.064, so H₀ is rejected at the 5% level (two-sided).

10.Hypothesis tests & confidence intervals

Answer: L = 117.7

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 = 2.576·4/√16 = 2.576

Interval limit

L = 120.3 − 2.576 = 117.7

Answer

L = 117.7
11.Linear regression & correlation

Answer: r = 0.9858

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 = 2099 − 27·439/6 = 123.5
Sxx = 139 − 27²/6 = 17.5
Syy = 33017 − 439²/6 = 896.8

Correlation coefficient

r = 123.5/√(17.5·896.8) = 0.9858

Answer

r = 0.9858
12.Linear regression & correlation

Answer: e = −5.1

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 = 305/5 = 61

Sums of squares and products

Sxx = Σ(x − x̄)² = 10
Sxy = Σ(x − x̄)(y − ȳ) = 54

Regression line

b = 54/10 = 5.4
a = 61 − 5.4·3 = 44.8
ŷ = 44.8 + 5.4·x

Predicted value

ŷ = 44.8 + 5.4·4.5 = 69.1

Residual

e = 64 − 69.1 = −5.1

Answer

e = −5.1

The point lies below the line.

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