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
20253035404550

Find the sample standard deviation s of the data: 50, 41, 37, 19, 21.

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

Exam scores are normally distributed with mean 60 and standard deviation 15. Find the z-score z of a score of 66.

μ = 60σ = 15x = 66
z =
3.ProbabilityMedium
xp(x)10.320.230.1540.35μ = 2.55

For the distribution shown, find the variance Var(X) of X.

x: 1, 2, 3, 4p(x): 0.3, 0.2, 0.15, 0.35
Var(X) =
4.ProbabilityHard
xp(x)00.1510.1520.1530.2540.3μ

X has the distribution shown, and Y = 3X + 10. Find the standard deviation σY of Y.

x: 0, 1, 2, 3, 4p(x): 0.15, 0.15, 0.15, 0.25, 0.3Y = 3X + 10
σY =
5.Normal distributionMedium
−3−2−10123zshaded area to the right of z = 0.2005

Drag z so that the shaded area under the standard normal curve, to the right of z, is 0.2005. Find z to two decimals.

A = 0.2005
z =
6.Normal distributionHard
0.03 mm0D − d?

Hole diameters D are normal with mean 25.03 mm and standard deviation 0.015 mm. Shaft diameters d are independent and normal with mean 25 mm and standard deviation 0.015 mm. A randomly chosen shaft fits when D − d > 0. Find the probability P that it fits. Round z to two decimals.

hole: μ = 25.03 mm, σ = 0.015 mmshaft: μ = 25 mm, σ = 0.015 mm
P =
7.Confidence intervalsMedium
n = 10, s = 15LUx̄ = 74.5E = ?99%

A sample of 10 items has mean 74.5 and standard deviation s = 15. The population is roughly normal. Find the margin of error E of a 99% confidence interval for the mean.

n = 10x̄ = 74.5s = 1599%t(9) = 3.25
E =
8.Confidence intervalsHard
n = 15σ = ?Us = 3.695%

The diameters of 15 machined shafts have a sample standard deviation of 3.6 µm. Assuming the diameters are normally distributed, find the lower limit σ of the 95% confidence interval for the population standard deviation. With df = 14, 2.5% of the χ² distribution lies below 5.629 and 2.5% lies above 26.119.

n = 15s = 3.6 µmχ² = 5.629, 26.119
σ = µm
9.Hypothesis tests & confidence intervalsMedium
−2.1312.131t = ?df = 15

Two teaching methods are compared. Group 1: 11 students, mean score 88.5, standard deviation 7.5. Group 2: 6 students, mean score 98.6, standard deviation 7. 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 = 11x̄₁ = 88.5s₁ = 7.5n₂ = 6x̄₂ = 98.6s₂ = 7
t =
10.Hypothesis tests & confidence intervalsHard
μ₀x̄c

A one-sided z test of H₀: μ = 100 against H₁: μ < 100 is done at α = 0.05 with a sample of 64; the standard deviation is known to be 16. In reality the mean is 94. Find the power 1 − β of the test in percent.

μ₀ = 100μ = 94σ = 16n = 64α = 0.05
1 − β = %
11.Linear regression & correlationMedium
xy14731789201042312526137xy15202580100120140

The table gives 5 paired observations of x = temperature in °C and y = ice creams sold. Find Pearson's correlation coefficient r.

n = 5
r =
12.Linear regression & correlationHard
xy152718382153245627653073xy152025304060x = 16.5

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. A new observation (16.5, 36) is made. Find its residual e = y − ŷ.

n = 6(16.5, 36)
e =

3Solutions

1.Descriptive statistics

Answer: s = 13.3

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 = 168/5 = 33.6

Squared deviations

Σ(x − x̄)² = 16.4² + 7.4² + 3.4² + (−14.6)² + (−12.6)² = 707.2

Standard deviation

s = √(707.2/4) = 13.3

Answer

s = 13.3
2.Descriptive statistics

Answer: z = 0.4

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 = (66 − 60)/15 = 0.4

Answer

z = 0.4
3.Probability

Answer: Var(X) = 1.548

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 = 1·0.3 + 2·0.2 + 3·0.15 + 4·0.35 = 2.55

Mean of the squares

E(X²) = 1·0.3 + 4·0.2 + 9·0.15 + 16·0.35 = 8.05

Variance

Var(X) = 8.05 − 2.55² = 1.548

Answer

Var(X) = 1.548
4.Probability

Answer: σY = 4.285

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.

Spread of X

μ = 2.4, Var(X) = 2.04, σX = 1.428

Linear change

σY = |3|·1.428 = 4.285

Answer

σY = 4.285

The mean does move: E(Y) = 3·2.4 + 10 = 17.2.

5.Normal distribution

Answer: z = 0.84

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 − 0.2005 = 0.7995

Read the z-table backwards

Find 0.7995 inside the table and read off z.

z = 0.84

Answer

z = 0.84
6.Normal distribution

Answer: P = 0.9207

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 = 25.03 − 25 = 0.03 mm, σ = 0.02121 mm

Standardise

z = (0 − 0.03)/0.02121 = −1.41

Probability

P = 1 − Φ(−1.41) = 0.9207

Answer

P = 0.9207
7.Confidence intervals

Answer: E = 15.42

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 = 9, t = 3.25

Margin of error

E = 3.25·15/√10 = 15.42

Answer

E = 15.42
8.Confidence intervals

Answer: σ = 2.636 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 = 14·3.6² = 181.4

Limits for σ²

181.4/26.119 = 6.947
181.4/5.629 = 32.23

Limit for σ

σ = √6.947 = 2.636 µm

Answer

σ = 2.636 µm

95% interval for σ: [2.636, 5.677] µm.

9.Hypothesis tests & confidence intervals

Answer: t = −2.712

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 = (10·7.5² + 5·7²)/15 = 53.83
sp = 7.337

Standard error

SE = 7.337·√(1/11 + 1/6) = 3.724

Test statistic

t = (88.5 − 98.6)/3.724 = −2.712

Answer

t = −2.712

With df = 15, the two-sided 5% critical value is 2.131; |t| is larger, so H₀ is rejected.

10.Hypothesis tests & confidence intervals

Answer: 1 − β = 91.23 %

Power of a test: power = 1 − β is the probability of rejecting H₀ when it is false. Find the cutoff x̄c from μ₀ and z_α, then compute the probability of landing beyond x̄c using the true mean: z = (x̄c − μ)/(σ/√n).

Critical sample mean

SE = 16/√64 = 2
x̄c = 100 − 1.645·2 = 96.71

z under the true mean

z = (96.71 − 94)/2 = 1.355

Power

power = P(Z < 1.355) = 91.23%
1 − β = 91.23%

Answer

1 − β = 91.23%
11.Linear regression & correlation

Answer: r = 0.9975

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 = 100/5 = 20
ȳ = 528/5 = 105.6

Sums of squares and products

Sxx = Σ(x − x̄)² = 90
Sxy = Σ(x − x̄)(y − ȳ) = 492
Syy = Σ(y − ȳ)² = 2703

Correlation coefficient

r = 492/√(90·2703) = 0.9975

Answer

r = 0.9975
12.Linear regression & correlation

Answer: e = 1.943

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 = 135/6 = 22.5
ȳ = 312/6 = 52

Sums of squares and products

Sxx = Σ(x − x̄)² = 157.5
Sxy = Σ(x − x̄)(y − ȳ) = 471

Regression line

b = 471/157.5 = 2.99
a = 52 − 2.99·22.5 = −15.29
ŷ = −15.29 + 2.99·x

Predicted value

ŷ = −15.29 + 2.99·16.5 = 34.06

Residual

e = 36 − 34.06 = 1.943

Answer

e = 1.943

The point lies above the line.

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