Introductory Statistics: cram pack

Formula sheets, the key method steps for every topic and 12 practice problems, with full solutions at the back.

New problemsBack to the course

In the print window, choose “Save as PDF” to keep a copy on your device.

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
1520253035404550

Find the sample standard deviation s of the data: 12, 52, 26, 42, 20, 53, 16.

n = 7
s =
2.Descriptive statisticsHard
z −2−1012

Exam scores are normally distributed with mean 170 and standard deviation 20. Find the z-score z of a score of 146.

μ = 170σ = 20x = 146
z =
3.ProbabilityMedium
P(at least one six) = ??1?2?3P(six) = 1/6 on each roll

A fair die is rolled 3 times. What is the probability P of getting at least one six?

n = 3
P =
4.ProbabilityHard
n = 9, p = 0.25: P(X = 1) = ?k0123456789

A test has 9 multiple-choice questions. A student who guesses gets each one right with probability 0.25, independently. What is the probability P of exactly 1 correct answers?

n = 9p = 0.25k = 1
P =
5.Normal distributionMedium
240 g237.1 gT = X₁ + … + X₆?

The masses of bolts are independent and normal with mean 40 g and standard deviation 1.2 g. Find the probability P that the total mass of 6 of them is below 237.1 g. Round z to two decimals.

μ = 40 gσ = 1.2 gn = 6< 237.1 g
P =
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 0.01 mm. A randomly chosen shaft fits when D − d > 0. Find the probability P that it fits. Round z to two decimals.

hole: μ = 25.03 mm, σ = 0.015 mmshaft: μ = 25 mm, σ = 0.01 mm
P =
7.Confidence intervalsMedium
μ₁ − μ₂0LU = ?x̄₁ − x̄₂E90%

The battery life of two phone models is compared. Model 1: 30 phones, mean 140.5 h, known standard deviation 18 h. Model 2: 30 phones, mean 130 h, known standard deviation 5 h. Find the upper limit of the 90% confidence interval for μ₁ − μ₂.

n₁ = 30x̄₁ = 140.5 hσ₁ = 18 hn₂ = 30x̄₂ = 130 hσ₂ = 5 h90%
upper limit = h
8.Confidence intervalsHard
n = 16Lσ = ?s = 995%

The diameters of 16 machined shafts have a sample standard deviation of 9 µm. Assuming the diameters are normally distributed, find the upper limit σ of the 95% confidence interval for the population standard deviation. With df = 15, 2.5% of the χ² distribution lies below 6.262 and 2.5% lies above 27.488.

n = 16s = 9 µmχ² = 6.262, 27.488
σ = µm
9.Hypothesis tests & confidence intervalsMedium
−2.0932.093t = ?df = 19

Two teaching methods are compared. Group 1: 9 students, mean score 44, standard deviation 10. Group 2: 12 students, mean score 48.3, standard deviation 5. 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 = 9x̄₁ = 44s₁ = 10n₂ = 12x̄₂ = 48.3s₂ = 5
t =
10.Hypothesis tests & confidence intervalsHard
μ₀x̄c

A one-sided z test of H₀: μ = 500 against H₁: μ > 500 is done at α = 0.01 with a sample of 36; the standard deviation is known to be 9. In reality the mean is 503. Find the probability of a Type II error β in percent.

μ₀ = 500μ = 503σ = 9n = 36α = 0.01
β = %
11.Linear regression & correlationMedium
xy356459565660765xy466065

The table gives 5 paired observations of x = hours studied and y = exam score. Find Pearson's correlation coefficient r.

n = 5
r =
12.Linear regression & correlationHard
xy125225321420520xy24202224x = 1.5

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 (1.5, 20) is made. Find its residual e = y − ŷ.

n = 5(1.5, 20)
e =

3Solutions

1.Descriptive statistics

Answer: s = 17.2

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 = 221/7 = 31.57

Squared deviations

Σ(x − x̄)² = (−19.57)² + 20.43² + (−5.571)² + 10.43² + (−11.57)² + 21.43² + (−15.57)² = 1776

Standard deviation

s = √(1776/6) = 17.2

Answer

s = 17.2
2.Descriptive statistics

Answer: z = −1.2

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 = (146 − 170)/20 = −1.2

Answer

z = −1.2
3.Probability

Answer: P = 0.4213

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)3 = 0.5787
P = 1 − 0.5787 = 0.4213

Answer

P = 0.4213
4.Probability

Answer: P = 0.2253

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 = 1) = 9·0.251·0.758 = 9·0.25·0.1001 = 0.2253

Answer

P = 0.2253
5.Normal distribution

Answer: P = 0.1611

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 total

μT = 240 g, σT = 1.2·√6 = 2.939 g

Standardise

z = (237.1 − 240)/2.939 = −0.99

Probability

P = Φ(−0.99) = 0.1611

Answer

P = 0.1611
6.Normal distribution

Answer: P = 0.9515

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 = 0.01803 mm

Standardise

z = (0 − 0.03)/0.01803 = −1.66

Probability

P = 1 − Φ(−1.66) = 0.9515

Answer

P = 0.9515
7.Confidence intervals

Answer: upper limit = 16.11 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 = 140.5 − 130 = 10.5

Standard error

SE = √(18²/30 + 5²/30) = 3.411

Margin of error

E = 1.645·3.411 = 5.611

Interval

10.5 ± 5.611 → [4.889, 16.11]

Answer

upper limit = 16.11 h

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

8.Confidence intervals

Answer: σ = 13.93 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 = 15·9² = 1215

Limits for σ²

1215/27.488 = 44.2
1215/6.262 = 194

Limit for σ

σ = √194 = 13.93 µm

Answer

σ = 13.93 µm

95% interval for σ: [6.648, 13.93] µm.

9.Hypothesis tests & confidence intervals

Answer: t = −1.296

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 = (8·10² + 11·5²)/19 = 56.58
sp = 7.522

Standard error

SE = 7.522·√(1/9 + 1/12) = 3.317

Test statistic

t = (44 − 48.3)/3.317 = −1.296

Answer

t = −1.296

With df = 19, the two-sided 5% critical value is 2.093; |t| is smaller, so H₀ is not rejected.

10.Hypothesis tests & confidence intervals

Answer: β = 62.78 %

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 = 9/√36 = 1.5
x̄c = 500 + 2.326·1.5 = 503.5

z under the true mean

z = (503.5 − 503)/1.5 = 0.326

Type II error

power = P(Z > 0.326) = 37.22%
β = 100% − 37.22% = 62.78%

Answer

β = 62.78%
11.Linear regression & correlation

Answer: r = 0.7631

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 = 25/5 = 5
ȳ = 305/5 = 61

Sums of squares and products

Sxx = Σ(x − x̄)² = 10
Sxy = Σ(x − x̄)(y − ȳ) = 19
Syy = Σ(y − ȳ)² = 62

Correlation coefficient

r = 19/√(10·62) = 0.7631

Answer

r = 0.7631
12.Linear regression & correlation

Answer: e = −4.45

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 = 111/5 = 22.2

Sums of squares and products

Sxx = Σ(x − x̄)² = 10
Sxy = Σ(x − x̄)(y − ȳ) = −15

Regression line

b = −15/10 = −1.5
a = 22.2 − (−1.5)·3 = 26.7
ŷ = 26.7 − 1.5·x

Predicted value

ŷ = 26.7 + (−1.5)·1.5 = 24.45

Residual

e = 20 − 24.45 = −4.45

Answer

e = −4.45

The point lies below the line.

ExamProblems · examproblems.com/courses/statistics-1