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
152025303540455055

Find the sample standard deviation s of the data: 50, 51, 13, 56, 15, 12.

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

A quantity is normally distributed with mean 60 and standard deviation 12. What proportion P of values is above 64.8?

μ = 60σ = 12x = 64.8
P =
3.ProbabilityMedium
n = 15k = 3N = ?

In how many ways N can a committee of 3 people be chosen from a group of 15? The order does not matter.

n = 15k = 3
N =
4.ProbabilityHard
xp(x)00.1510.220.2530.2540.15μ

X has the distribution shown, and Y = 0.5X − 3. Find the standard deviation σY of Y.

x: 0, 1, 2, 3, 4p(x): 0.15, 0.2, 0.25, 0.25, 0.15Y = 0.5X − 3
σY =
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 = 178x = ?90%

The heights of adult men are normally distributed with mean 178 cm and standard deviation 7 cm. Find the value x below which 90% of them lie (the 90th percentile).

μ = 178σ = 7P = 90%
x = cm
7.Confidence intervalsMedium
n = 7, sd​ = 6.80LUd̄ = 1.1E = ?99%

The blood pressure of 7 patients is measured before and after a treatment. The differences d = before − after (mmHg) have mean 1.1 and standard deviation 6.8. Find the margin of error E of the 99% confidence interval for the mean difference.

n = 7d̄ = 1.1sd = 6.899%t(6) = 3.707
E = mmHg
8.Confidence intervalsHard
n = 20Lσ = ?s = 9.595%

The diameters of 20 machined shafts have a sample standard deviation of 9.5 µm. Assuming the diameters are normally distributed, find the upper limit σ of the 95% confidence interval for the population standard deviation. With df = 19, 2.5% of the χ² distribution lies below 8.907 and 2.5% lies above 32.852.

n = 20s = 9.5 µmχ² = 8.907, 32.852
σ = µm
9.Hypothesis tests & confidence intervalsMedium
t −2.0302.03

A sample of 36 students has a mean score of 505.3 and a sample standard deviation of 15. Find the test statistic t for H₀: μ = 500.

μ₀ = 500x̄ = 505.3s = 15n = 36
t =
10.Hypothesis tests & confidence intervalsHard
H₀H₁μ₀x̄cμ₀ + 2.5α = 0.05βσ = 18, power 95%, n = ?

A one-sided z test at α = 0.05 must detect a shift of the mean of 2.5 units with a power of 95%. The standard deviation is 18. Find the minimum sample size n.

α = 0.05power = 95%δ = 2.5σ = 18zα = 1.645zβ = 1.645
n =
11.Linear regression & correlationMedium
xy24445365686810781275xy5106080

The table gives 6 paired observations of x = hours studied and y = exam score. Find the coefficient of determination r² of the least-squares line.

n = 6
r² =
12.Linear regression & correlationHard
xy2514706748851010512118xy51050100ŷ = 110

The table gives 6 paired observations of x = hours studied and y = exam score. Use the least-squares line ŷ = a + b·x. For which value x does the line predict ŷ = 110?

n = 6ŷ = 110
x =

3Solutions

1.Descriptive statistics

Answer: s = 21.48

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 = 197/6 = 32.83

Squared deviations

Σ(x − x̄)² = 17.17² + 18.17² + (−19.83)² + 23.17² + (−17.83)² + (−20.83)² = 2307

Standard deviation

s = √(2307/5) = 21.48

Answer

s = 21.48
2.Descriptive statistics

Answer: P = 0.3446

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 = 0.4

Standard normal table

Φ(0.4) = 0.6554
P = 1 − 0.6554 = 0.3446

Answer

P = 0.3446
3.Probability

Answer: N = 455

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(15, 3) = (15·14·13)/(3·2·1) = 455

Answer

N = 455
4.Probability

Answer: σY = 0.6418

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 = 2.05, Var(X) = 1.647, σX = 1.284

Linear change

σY = |0.5|·1.284 = 0.6418

Answer

σY = 0.6418

The mean does move: E(Y) = 0.5·2.05 − 3 = −1.975.

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 = 187 cm

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 = 178 + 1.282·7 = 186.97 cm

Answer

x = 186.97 cm
7.Confidence intervals

Answer: E = 9.528 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 = 6, t = 3.707

Margin of error

E = 3.707·6.8/√7 = 9.528

Answer

E = 9.528 mmHg
8.Confidence intervals

Answer: σ = 13.88 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 = 19·9.5² = 1715

Limits for σ²

1715/32.852 = 52.2
1715/8.907 = 192.5

Limit for σ

σ = √192.5 = 13.88 µm

Answer

σ = 13.88 µm

95% interval for σ: [7.225, 13.88] µm.

9.Hypothesis tests & confidence intervals

Answer: t = 2.12

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 = (505.3 − 500)/2.5 = 2.12

Answer

t = 2.12
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: n = 562

Sample size for a test: to detect a true shift δ with power 1 − β in a one-sided z test at level α, the standard error must satisfy δ = (z_α + z_β)·σ/√n, so n = ((z_α + z_β)·σ/δ)², rounded up.

Solve for n

n = ((1.645 + 1.645)·18/2.5)² = 561.1

Round up

n = 562

Answer

n = 562
11.Linear regression & correlation

Answer: r² = 0.9282

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 = 42/6 = 7
ȳ = 374/6 = 62.33

Sums of squares and products

Sxx = Σ(x − x̄)² = 70
Sxy = Σ(x − x̄)(y − ȳ) = 242
Syy = Σ(y − ȳ)² = 901.3

Coefficient of determination

r² = 242²/(70·901.3) = 0.9282

Answer

r² = 0.9282

About 92.82% of the variation in y is explained by the line.

12.Linear regression & correlation

Answer: x = 11.06

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 = 42/6 = 7
ȳ = 503/6 = 83.83

Sums of squares and products

Sxx = Σ(x − x̄)² = 70
Sxy = Σ(x − x̄)(y − ȳ) = 451

Regression line

b = 451/70 = 6.443
a = 83.83 − 6.443·7 = 38.73
ŷ = 38.73 + 6.443·x

Solve for x

x = (110 − 38.73)/6.443 = 11.06

Answer

x = 11.06

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