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
101520253035404550

Find the sample standard deviation s of the data: 32, 41, 50, 19, 10, 20.

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

A quantity is normally distributed with mean 50 and standard deviation 12. What proportion P of values is above 65?

μ = 50σ = 12x = 65
P =
3.ProbabilityMedium
xp(x)10.2520.1530.240.150.3μ = 3.05

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

x: 1, 2, 3, 4, 5p(x): 0.25, 0.15, 0.2, 0.1, 0.3
Var(X) =
4.ProbabilityHard
n = 8, p = 0.3: P(X ≤ 1) = ?k012345678

8 components are tested, and each one fails with probability 0.3, independently. What is the probability P that at most one fails?

n = 8p = 0.3
P =
5.Normal distributionMedium
105 s108.2 sX + Y?

X is normal with mean 60 s and standard deviation 5 s; Y is independent and normal with mean 45 s and standard deviation 4 s. Find the probability P that X + Y is more than 108.2 s. Round z to two decimals.

μX = 60, σX = 5μY = 45, σY = 4X + Y > 108.2 s
P =
6.Normal distributionHard
10 g0X − Y?

The mass X of a product from line 1 and Y from line 2 are independent and normal: X has mean 500 g and standard deviation 8 g, Y has mean 490 g and standard deviation 6 g. Find the probability P that X is greater than Y. Round z to two decimals.

μX = 500 g, σX = 8 gμY = 490 g, σY = 6 g
P =
7.Confidence intervalsMedium
n = 16, sd​ = 2.80LU = ?d̄ = 2.7E99%

The blood pressure of 16 patients is measured before and after a treatment. The differences d = before − after (mmHg) have mean 2.7 and standard deviation 2.8. Find the upper limit of the 99% confidence interval for the mean difference.

n = 16d̄ = 2.7sd = 2.899%t(15) = 2.947
upper limit = mmHg
8.Confidence intervalsHard
n = 11Lσ = ?s = 8.295%

The diameters of 11 machined shafts have a sample standard deviation of 8.2 µm. Assuming the diameters are normally distributed, find the upper 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 = 8.2 µmχ² = 3.247, 20.483
σ = µm
9.Hypothesis tests & confidence intervalsMedium
t −2.0302.03

A sample of 36 students has a mean score of 92.25 and a sample standard deviation of 15. Find the test statistic t for H₀: μ = 100.

μ₀ = 100x̄ = 92.25s = 15n = 36
t =
10.Hypothesis tests & confidence intervalsHard
μ₀x̄c

A one-sided z test of H₀: μ = 200 against H₁: μ > 200 is done at α = 0.05 with a sample of 64; the standard deviation is known to be 4. In reality the mean is 201.3. Find the power 1 − β of the test in percent.

μ₀ = 200μ = 201.3σ = 4n = 64α = 0.05
1 − β = %
11.Linear regression & correlationMedium
xy15435957078899511110xy5106080100

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 = 36Σy = 476Σx² = 286Σy² = 40166Σxy = 3262
r =
12.Linear regression & correlationHard
xy34454675595111561365xy5105060x = 8

The table gives 6 paired observations of x = advertising spend in k€ and y = sales in k€. Use the least-squares line ŷ = a + b·x. A new observation (8, 48) is made. Find its residual e = y − ŷ.

n = 6(8, 48)
e =

3Solutions

1.Descriptive statistics

Answer: s = 15.07

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 = 172/6 = 28.67

Squared deviations

Σ(x − x̄)² = 3.333² + 12.33² + 21.33² + (−9.667)² + (−18.67)² + (−8.667)² = 1135

Standard deviation

s = √(1135/5) = 15.07

Answer

s = 15.07
2.Descriptive statistics

Answer: P = 0.1056

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 = 1.25

Standard normal table

Φ(1.25) = 0.8944
P = 1 − 0.8944 = 0.1056

Answer

P = 0.1056
3.Probability

Answer: Var(X) = 2.447

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 = 1·0.25 + 2·0.15 + 3·0.2 + 4·0.1 + 5·0.3 = 3.05

Mean of the squares

E(X²) = 1·0.25 + 4·0.15 + 9·0.2 + 16·0.1 + 25·0.3 = 11.75

Variance

Var(X) = 11.75 − 3.05² = 2.447

Answer

Var(X) = 2.447
4.Probability

Answer: P = 0.2553

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.

Zero and one failure

P(X = 0) = 0.78 = 0.05765
P(X = 1) = 8·0.3·0.77 = 0.1977

Add them

P = 0.05765 + 0.1977 = 0.2553

Answer

P = 0.2553
5.Normal distribution

Answer: P = 0.3085

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 sum

μ = 105 s, σ = √(5² + 4²) = 6.403 s

Standardise

z = (108.2 − 105)/6.403 = 0.5

Upper tail

P = 1 − Φ(0.5) = 1 − 0.6915 = 0.3085

Answer

P = 0.3085
6.Normal distribution

Answer: P = 0.8413

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 = 500 − 490 = 10 g, σ = 10 g

Standardise

z = (0 − 10)/10 = −1

Probability

P = 1 − Φ(−1) = 0.8413

Answer

P = 0.8413
7.Confidence intervals

Answer: upper limit = 4.763 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 = 15, t = 2.947

Margin of error

E = 2.947·2.8/√16 = 2.063

Interval

2.7 ± 2.063 → [0.6371, 4.763]

Answer

upper limit = 4.763 mmHg
8.Confidence intervals

Answer: σ = 14.39 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 = 10·8.2² = 672.4

Limits for σ²

672.4/20.483 = 32.83
672.4/3.247 = 207.1

Limit for σ

σ = √207.1 = 14.39 µm

Answer

σ = 14.39 µm

95% interval for σ: [5.73, 14.39] µm.

9.Hypothesis tests & confidence intervals

Answer: t = −3.1

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 = 15/√36 = 15/6 = 2.5

Test statistic

t = (92.25 − 100)/2.5 = −3.1

Answer

t = −3.1
t0.025; df = 35 = 2.03

|t| > 2.03, so H₀ is rejected at the 5% level (two-sided).

10.Hypothesis tests & confidence intervals

Answer: 1 − β = 83.02 %

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 = 4/√64 = 0.5
x̄c = 200 + 1.645·0.5 = 200.8

z under the true mean

z = (200.8 − 201.3)/0.5 = −0.955

Power

power = P(Z > −0.955) = 83.02%
1 − β = 83.02%

Answer

1 − β = 83.02%
11.Linear regression & correlation

Answer: r = 0.9899

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 = 3262 − 36·476/6 = 406
Sxx = 286 − 36²/6 = 70
Syy = 40166 − 476²/6 = 2403

Correlation coefficient

r = 406/√(70·2403) = 0.9899

Answer

r = 0.9899
12.Linear regression & correlation

Answer: e = −4.833

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 = 48/6 = 8
ȳ = 317/6 = 52.83

Sums of squares and products

Sxx = Σ(x − x̄)² = 70
Sxy = Σ(x − x̄)(y − ȳ) = 131

Regression line

b = 131/70 = 1.871
a = 52.83 − 1.871·8 = 37.86
ŷ = 37.86 + 1.871·x

Predicted value

ŷ = 37.86 + 1.871·8 = 52.83

Residual

e = 48 − 52.83 = −4.833

Answer

e = −4.833

The point lies below the line.

ExamProblems · examproblems.com/courses/statistics-1