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
202530354045505560

Find the sample standard deviation s of the data: 60, 42, 20, 41, 45, 29, 31.

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

Exam scores are normally distributed with mean 500 and standard deviation 50. Find the z-score z of a score of 600.

μ = 500σ = 50x = 600
z =
3.ProbabilityMedium
n = 14k = 2N = ?

In how many ways N can a committee of 2 people be chosen from a group of 14? The order does not matter.

n = 14k = 2
N =
4.ProbabilityHard
xp(x)00.3510.3520.3μ

X has the distribution shown, and Y = 0.5X + 1. Find the standard deviation σY of Y.

x: 0, 1, 2p(x): 0.35, 0.35, 0.3Y = 0.5X + 1
σY =
5.Normal distributionMedium
−3−2−10123zshaded area to the right of z = 0.8289

Drag z so that the shaded area under the standard normal curve, to the right of z, is 0.8289. Find z to two decimals.

A = 0.8289
z =
6.Normal distributionHard
μ = 65x = ?90%

The exam scores are normally distributed with mean 65 points and standard deviation 10 points. Find the value x below which 90% of them lie (the 90th percentile).

μ = 65σ = 10P = 90%
x = points
7.Confidence intervalsMedium
n = 20, s = 12.5L = ?Ux̄ = 37.5E99%

A sample of 20 items has mean 37.5 and standard deviation s = 12.5. The population is roughly normal. Find the lower limit of the 99% confidence interval for the mean.

n = 20x̄ = 37.5s = 12.599%t(19) = 2.861
lower limit =
8.Confidence intervalsHard
n = 15σ = ?Us = 8.995%

The diameters of 15 machined shafts have a sample standard deviation of 8.9 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 = 8.9 µmχ² = 5.629, 26.119
σ = µm
9.Hypothesis tests & confidence intervalsMedium
−2.1312.131t = ?df = 15

Two teaching methods are compared. Group 1: 12 students, mean score 67, standard deviation 11.5. Group 2: 5 students, mean score 63.1, standard deviation 7.5. 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 = 12x̄₁ = 67s₁ = 11.5n₂ = 5x̄₂ = 63.1s₂ = 7.5
t =
10.Hypothesis tests & confidence intervalsHard
μ₀x̄c

A one-sided z test of H₀: μ = 80 against H₁: μ > 80 is done at α = 0.10 with a sample of 64; the standard deviation is known to be 4. In reality the mean is 81.1. Find the probability of a Type II error β in percent.

μ₀ = 80μ = 81.1σ = 4n = 64α = 0.10
β = %
11.Linear regression & correlationMedium
xy123225321417518xy242025

The table gives 5 paired observations of x = car age in years and y = price in k€. Their sums are given. Find Pearson's correlation coefficient r from these sums.

n = 5Σx = 15Σy = 104Σx² = 55Σy² = 2208Σxy = 294
r =
12.Linear regression & correlationHard
xy184921492456276730793385xy2025306080x = 19.5

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 (19.5, 50) is made. Find its residual e = y − ŷ.

n = 6(19.5, 50)
e =

3Solutions

1.Descriptive statistics

Answer: s = 12.98

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 = 268/7 = 38.29

Squared deviations

Σ(x − x̄)² = 21.71² + 3.714² + (−18.29)² + 2.714² + 6.714² + (−9.286)² + (−7.286)² = 1011

Standard deviation

s = √(1011/6) = 12.98

Answer

s = 12.98
2.Descriptive statistics

Answer: z = 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 = (600 − 500)/50 = 2

Answer

z = 2
3.Probability

Answer: N = 91

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.

Combinations

C(14, 2) = (14·13)/(2·1) = 91

Answer

N = 91
4.Probability

Answer: σY = 0.4023

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 = 0.95, Var(X) = 0.6475, σX = 0.8047

Linear change

σY = |0.5|·0.8047 = 0.4023

Answer

σY = 0.4023

The mean does move: E(Y) = 0.5·0.95 + 1 = 1.475.

5.Normal distribution

Answer: z = −0.95

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 − 0.8289 = 0.1711

Read the z-table backwards

Find 0.1711 inside the table and read off z.

z = −0.95

Answer

z = −0.95
6.Normal distribution

Answer: x = 77.82 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) = 0.9 → z = 1.282

Back to x

x = 65 + 1.282·10 = 77.82 points

Answer

x = 77.82 points
7.Confidence intervals

Answer: lower limit = 29.5

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 = 19, t = 2.861

Margin of error

E = 2.861·12.5/√20 = 7.997

Interval

37.5 ± 7.997 → [29.5, 45.5]

Answer

lower limit = 29.5
8.Confidence intervals

Answer: σ = 6.516 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 = 14·8.9² = 1109

Limits for σ²

1109/26.119 = 42.46
1109/5.629 = 197

Limit for σ

σ = √42.46 = 6.516 µm

Answer

σ = 6.516 µm

95% interval for σ: [6.516, 14.04] µm.

9.Hypothesis tests & confidence intervals

Answer: t = 0.6924

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 = (11·11.5² + 4·7.5²)/15 = 112
sp = 10.58

Standard error

SE = 10.58·√(1/12 + 1/5) = 5.633

Test statistic

t = (67 − 63.1)/5.633 = 0.6924

Answer

t = 0.6924

With df = 15, the two-sided 5% critical value is 2.131; |t| is smaller, so H₀ is not rejected.

10.Hypothesis tests & confidence intervals

Answer: β = 17.93 %

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 = 80 + 1.282·0.5 = 80.64

z under the true mean

z = (80.64 − 81.1)/0.5 = −0.918

Type II error

power = P(Z > −0.918) = 82.07%
β = 100% − 82.07% = 17.93%

Answer

β = 17.93%
11.Linear regression & correlation

Answer: r = −0.8504

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 = 294 − 15·104/5 = −18
Sxx = 55 − 15²/5 = 10
Syy = 2208 − 104²/5 = 44.8

Correlation coefficient

r = −18/√(10·44.8) = −0.8504

Answer

r = −0.8504
12.Linear regression & correlation

Answer: e = 1.89

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 = 153/6 = 25.5
ȳ = 385/6 = 64.17

Sums of squares and products

Sxx = Σ(x − x̄)² = 157.5
Sxy = Σ(x − x̄)(y − ȳ) = 421.5

Regression line

b = 421.5/157.5 = 2.676
a = 64.17 − 2.676·25.5 = −4.076
ŷ = −4.076 + 2.676·x

Predicted value

ŷ = −4.076 + 2.676·19.5 = 48.11

Residual

e = 50 − 48.11 = 1.89

Answer

e = 1.89

The point lies above the line.

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