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
1015202530354045

Find the sample standard deviation s of the data: 35, 24, 45, 37, 39, 15, 10.

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

A quantity is normally distributed with mean 50 and standard deviation 8. What proportion P of values is above 42?

μ = 50σ = 8x = 42
P =
3.ProbabilityMedium
xp(x)00.220.340.2560.25μ = 3.1

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

x: 0, 2, 4, 6p(x): 0.2, 0.3, 0.25, 0.25
Var(X) =
4.ProbabilityHard
€p(x)−50.78100.22450.02

A game costs €5 to play. You win €15 with probability 0.2 and €250 with probability 0.02; otherwise you win nothing. The stake is not returned. Find the expected net gain E(G) per game.

stake = €5€15: p = 0.2€250: p = 0.02
E(G) = €
5.Normal distributionMedium
105 s114.6 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 114.6 s. Round z to two decimals.

μX = 60, σX = 5μY = 45, σY = 4X + Y > 114.6 s
P =
6.Normal distributionHard
μ = 500x = ?95%

The battery lifetimes are normally distributed with mean 500 h and standard deviation 40 h. Find the value x below which 95% of them lie (the 95th percentile).

μ = 500σ = 40P = 95%
x = h
7.Confidence intervalsMedium
μ₁ − μ₂0LU = ?x̄₁ − x̄₂E99%

The battery life of two phone models is compared. Model 1: 40 phones, mean 123.5 h, known standard deviation 17 h. Model 2: 50 phones, mean 112.5 h, known standard deviation 9 h. Find the upper limit of the 99% confidence interval for μ₁ − μ₂.

n₁ = 40x̄₁ = 123.5 hσ₁ = 17 hn₂ = 50x̄₂ = 112.5 hσ₂ = 9 h99%
upper limit = h
8.Confidence intervalsHard
σ = 34, n = ?x̄ − Ex̄ + Ex̄E = 595%

The population standard deviation is 34. How large a sample n do you need so that a 95% confidence interval for the mean has a margin of error of at most 5?

σ = 34E = 595%
n =
9.Hypothesis tests & confidence intervalsMedium
t −2.13102.131

A sample of 16 students has a mean score of 64 and a sample standard deviation of 20. Find the test statistic t for H₀: μ = 75.

μ₀ = 75x̄ = 64s = 20n = 16
t =
10.Hypothesis tests & confidence intervalsHard
μ₀x̄c

A one-sided z test of H₀: μ = 50 against H₁: μ > 50 is done at α = 0.05 with a sample of 25; the standard deviation is known to be 2.5. In reality the mean is 51.1. Find the power 1 − β of the test in percent.

μ₀ = 50μ = 51.1σ = 2.5n = 25α = 0.05
1 − β = %
11.Linear regression & correlationMedium
xy139251352452559660xy246405060

The table gives 6 paired observations of x = hours studied and y = exam score. Find Pearson's correlation coefficient r.

n = 6
r =
12.Linear regression & correlationHard
xy2104126228241033xy510102030x = 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, 8) is made. Find its residual e = y − ŷ.

n = 5(3, 8)
e =

3Solutions

1.Descriptive statistics

Answer: s = 13.15

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 = 205/7 = 29.29

Squared deviations

Σ(x − x̄)² = 5.714² + (−5.286)² + 15.71² + 7.714² + 9.714² + (−14.29)² + (−19.29)² = 1037

Standard deviation

s = √(1037/6) = 13.15

Answer

s = 13.15
2.Descriptive statistics

Answer: P = 0.8413

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) = 0.1587
P = 1 − 0.1587 = 0.8413

Answer

P = 0.8413
3.Probability

Answer: Var(X) = 4.59

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 = 0·0.2 + 2·0.3 + 4·0.25 + 6·0.25 = 3.1

Mean of the squares

E(X²) = 0·0.2 + 4·0.3 + 16·0.25 + 36·0.25 = 14.2

Variance

Var(X) = 14.2 − 3.1² = 4.59

Answer

Var(X) = 4.59
4.Probability

Answer: E(G) = 3 €

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.

Net outcomes

G: 10 (p = 0.2), 245 (p = 0.02), −5 (p = 0.78)

Expected value

E(G) = 10·0.2 + 245·0.02 − 5·0.78 = 3 €

Answer

E(G) = 3 €

A positive value means you win money on average.

5.Normal distribution

Answer: P = 0.06681

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 = (114.6 − 105)/6.403 = 1.5

Upper tail

P = 1 − Φ(1.5) = 1 − 0.9332 = 0.06681

Answer

P = 0.06681
6.Normal distribution

Answer: x = 565.8 h

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) = 0.95 → z = 1.645

Back to x

x = 500 + 1.645·40 = 565.8 h

Answer

x = 565.8 h
7.Confidence intervals

Answer: upper limit = 18.66 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 = 123.5 − 112.5 = 11

Standard error

SE = √(17²/40 + 9²/50) = 2.974

Margin of error

E = 2.576·2.974 = 7.661

Interval

11 ± 7.661 → [3.339, 18.66]

Answer

upper limit = 18.66 h

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

8.Confidence intervals

Answer: n = 178

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.

Solve for n

n = (1.96·34/5)² = 177.6

Round up

A sample size is a whole number, and rounding down would make the margin too large.

n = 178

Answer

n = 178
9.Hypothesis tests & confidence intervals

Answer: t = −2.2

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 = 20/√16 = 20/4 = 5

Test statistic

t = (64 − 75)/5 = −2.2

Answer

t = −2.2
t0.025; df = 15 = 2.131

|t| > 2.131, so H₀ is rejected at the 5% level (two-sided).

10.Hypothesis tests & confidence intervals

Answer: 1 − β = 71.06 %

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 = 2.5/√25 = 0.5
x̄c = 50 + 1.645·0.5 = 50.82

z under the true mean

z = (50.82 − 51.1)/0.5 = −0.555

Power

power = P(Z > −0.555) = 71.06%
1 − β = 71.06%

Answer

1 − β = 71.06%
11.Linear regression & correlation

Answer: r = 0.9168

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 = 21/6 = 3.5
ȳ = 313/6 = 52.17

Sums of squares and products

Sxx = Σ(x − x̄)² = 17.5
Sxy = Σ(x − x̄)(y − ȳ) = 64.5
Syy = Σ(y − ȳ)² = 282.8

Correlation coefficient

r = 64.5/√(17.5·282.8) = 0.9168

Answer

r = 0.9168
12.Linear regression & correlation

Answer: e = −3.5

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 = 101/5 = 20.2

Sums of squares and products

Sxx = Σ(x − x̄)² = 40
Sxy = Σ(x − x̄)(y − ȳ) = 116

Regression line

b = 116/40 = 2.9
a = 20.2 − 2.9·6 = 2.8
ŷ = 2.8 + 2.9·x

Predicted value

ŷ = 2.8 + 2.9·3 = 11.5

Residual

e = 8 − 11.5 = −3.5

Answer

e = −3.5

The point lies below the line.

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