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
40424446485052545658

Find the sample standard deviation s of the data: 40, 49, 44, 59, 59.

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

A quantity is normally distributed with mean 60 and standard deviation 15. What proportion P of values is above 75?

μ = 60σ = 15x = 75
P =
3.ProbabilityMedium
xp(x)10.320.230.340.1550.05μ = 2.45

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

x: 1, 2, 3, 4, 5p(x): 0.3, 0.2, 0.3, 0.15, 0.05
Var(X) =
4.ProbabilityHard
n = 8, p = 0.7: P(X = 5) = ?k012345678

A test has 8 multiple-choice questions. A student who guesses gets each one right with probability 0.7, independently. What is the probability P of exactly 5 correct answers?

n = 8p = 0.7k = 5
P =
5.Normal distributionMedium
−3−2−10123zshaded area between −z and z = 0.7199

Drag z so that the shaded area between −z and z under the standard normal curve is 0.7199. Find z to two decimals.

A = 0.7199
z =
6.Normal distributionHard
μ = 100x = ?20%

The IQ scores are normally distributed with mean 100 points and standard deviation 15 points. Find the value x below which 20% of them lie (the 20th percentile).

μ = 100σ = 15P = 20%
x = points
7.Confidence intervalsMedium
μ₁ − μ₂0L = ?Ux̄₁ − x̄₂E95%

The battery life of two phone models is compared. Model 1: 100 phones, mean 46 h, known standard deviation 8 h. Model 2: 100 phones, mean 32 h, known standard deviation 4 h. Find the lower limit of the 95% confidence interval for μ₁ − μ₂.

n₁ = 100x̄₁ = 46 hσ₁ = 8 hn₂ = 100x̄₂ = 32 hσ₂ = 4 h95%
lower limit = h
8.Confidence intervalsHard
n = 11σ = ?Us = 6.895%

The diameters of 11 machined shafts have a sample standard deviation of 6.8 µm. Assuming the diameters are normally distributed, find the lower limit σ of the 95% confidence interval for the population standard deviation. With df = 10, 2.5% of the χ² distribution lies below 3.247 and 2.5% lies above 20.483.

n = 11s = 6.8 µmχ² = 3.247, 20.483
σ = µm
9.Hypothesis tests & confidence intervalsMedium
1.18p = ?

The mean of a population is claimed to be 500; the population standard deviation is known to be 20. A random sample of 64 observations has a mean of 502.95. Test H₀: μ = 500 against H₁: μ > 500. Find the p-value p.

μ₀ = 500σ = 20n = 64x̄ = 502.95H₁: μ > μ₀
p =
10.Hypothesis tests & confidence intervalsHard
μ₀x̄c

A one-sided z test of H₀: μ = 50 against H₁: μ > 50 is done at α = 0.01 with a sample of 25; the standard deviation is known to be 7.5. In reality the mean is 54.5. Find the power 1 − β of the test in percent.

μ₀ = 50μ = 54.5σ = 7.5n = 25α = 0.01
1 − β = %
11.Linear regression & correlationMedium
xy18672171247627863010033112xy20253080100

The table gives 6 paired observations of x = temperature in °C and y = ice creams sold. Their sums are given. Find Pearson's correlation coefficient r from these sums.

n = 6Σx = 153Σy = 512Σx² = 4059Σy² = 45246Σxy = 13539
r =
12.Linear regression & correlationHard
xy274116148141020xy510101520x = 3

The table gives 5 paired observations of x = load in kN and y = extension in mm. Use the least-squares line ŷ = a + b·x. A new observation (3, 7) is made. Find its residual e = y − ŷ.

n = 5(3, 7)
e =

3Solutions

1.Descriptive statistics

Answer: s = 8.643

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 = 251/5 = 50.2

Squared deviations

Σ(x − x̄)² = (−10.2)² + (−1.2)² + (−6.2)² + 8.8² + 8.8² = 298.8

Standard deviation

s = √(298.8/4) = 8.643

Answer

s = 8.643
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: Var(X) = 1.448

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 + 3·0.3 + 4·0.15 + 5·0.05 = 2.45

Mean of the squares

E(X²) = 1·0.3 + 4·0.2 + 9·0.3 + 16·0.15 + 25·0.05 = 7.45

Variance

Var(X) = 7.45 − 2.45² = 1.448

Answer

Var(X) = 1.448
4.Probability

Answer: P = 0.2541

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 = 5) = 56·0.75·0.33 = 56·0.1681·0.027 = 0.2541

Answer

P = 0.2541
5.Normal distribution

Answer: z = 1.08

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 + 0.7199)/2 = 0.8599

Read the z-table backwards

Find 0.8599 inside the table and read off z.

z = 1.08

Answer

z = 1.08
6.Normal distribution

Answer: x = 87.38 points

Normal distribution: convert a value x to a z-score, z = (x − μ)/σ, which counts standard deviations from the mean. A z-table (or Φ(z)) gives the area to the left of z, which is P(X < x). The area to the right is 1 − Φ(z), and the area between two values is the difference of their two areas.

z for the percentile

Φ(z) = 0.2 → z = −0.8416

Back to x

x = 100 + −0.8416·15 = 87.38 points

Answer

x = 87.38 points
7.Confidence intervals

Answer: lower limit = 12.25 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 = 46 − 32 = 14

Standard error

SE = √(8²/100 + 4²/100) = 0.8944

Margin of error

E = 1.96·0.8944 = 1.753

Interval

14 ± 1.753 → [12.25, 15.75]

Answer

lower limit = 12.25 h

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

8.Confidence intervals

Answer: σ = 4.751 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 = 10·6.8² = 462.4

Limits for σ²

462.4/20.483 = 22.57
462.4/3.247 = 142.4

Limit for σ

σ = √22.57 = 4.751 µm

Answer

σ = 4.751 µm

95% interval for σ: [4.751, 11.93] µm.

9.Hypothesis tests & confidence intervals

Answer: p = 0.119

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 = 20/√64 = 2.5
z = (502.95 − 500)/2.5 = 1.18

Tail area

Φ(1.18) = 0.8810
1 − Φ(1.18) = 0.1190

p-value

p = 0.119

Answer

p = 0.119

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

10.Hypothesis tests & confidence intervals

Answer: 1 − β = 74.98 %

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 = 7.5/√25 = 1.5
x̄c = 50 + 2.326·1.5 = 53.49

z under the true mean

z = (53.49 − 54.5)/1.5 = −0.674

Power

power = P(Z > −0.674) = 74.98%
1 − β = 74.98%

Answer

1 − β = 74.98%
11.Linear regression & correlation

Answer: r = 0.9759

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 = 13539 − 153·512/6 = 483
Sxx = 4059 − 153²/6 = 157.5
Syy = 45246 − 512²/6 = 1555

Correlation coefficient

r = 483/√(157.5·1555) = 0.9759

Answer

r = 0.9759
12.Linear regression & correlation

Answer: e = −1.85

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 = 66/5 = 13.2

Sums of squares and products

Sxx = Σ(x − x̄)² = 40
Sxy = Σ(x − x̄)(y − ȳ) = 58

Regression line

b = 58/40 = 1.45
a = 13.2 − 1.45·6 = 4.5
ŷ = 4.5 + 1.45·x

Predicted value

ŷ = 4.5 + 1.45·3 = 8.85

Residual

e = 7 − 8.85 = −1.85

Answer

e = −1.85

The point lies below the line.

ExamProblems · examproblems.com/courses/statistics-1