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
1520253035404550

Find the sample standard deviation s of the data: 38, 14, 44, 52, 49, 26, 21.

n = 7
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 below 32?

μ = 50σ = 12x = 32
P =
3.ProbabilityMedium
xp(x)00.3510.220.230.25μ = 1.35

For the distribution shown, find the standard deviation σ of X.

x: 0, 1, 2, 3p(x): 0.35, 0.2, 0.2, 0.25
σ =
4.ProbabilityHard
n = 7, p = 0.2: P(X ≤ 1) = ?k01234567

7 components are tested, and each one fails with probability 0.2, independently. What is the probability P that at most one fails?

n = 7p = 0.2
P =
5.Normal distributionMedium
750 g745.1 gX + Y?

X is normal with mean 250 g and standard deviation 4 g; Y is independent and normal with mean 500 g and standard deviation 9 g. Find the probability P that X + Y is more than 745.1 g. Round z to two decimals.

μX = 250, σX = 4μY = 500, σY = 9X + Y > 745.1 g
P =
6.Normal distributionHard
380420460500540580620hbottom 70%

The battery lifetimes are normally distributed with mean 500 h and standard deviation 40 h. Drag the cut-off so the shaded bottom 70% is to the left of it. Find the cut-off x.

μ = 500 hσ = 40 h70%
x = h
7.Confidence intervalsMedium
n = 8, s = 2L = ?Ux̄ = 112E95%

A sample of 8 items has mean 112 and standard deviation s = 2. The population is roughly normal. Find the lower limit of the 95% confidence interval for the mean.

n = 8x̄ = 112s = 295%t(7) = 2.365
lower limit =
8.Confidence intervalsHard
n = 11Lσ = ?s = 10.595%

The diameters of 11 machined shafts have a sample standard deviation of 10.5 µm. Assuming the diameters are normally distributed, find the upper limit σ of the 95% confidence interval for the population standard deviation. With df = 10, 2.5% of the χ² distribution lies below 3.247 and 2.5% lies above 20.483.

n = 11s = 10.5 µmχ² = 3.247, 20.483
σ = µm
9.Hypothesis tests & confidence intervalsMedium
−1.081.08p = ?

The mean of a population is claimed to be 50; the population standard deviation is known to be 15. A random sample of 100 observations has a mean of 48.38. Test H₀: μ = 50 against H₁: μ ≠ 50. Find the p-value p.

μ₀ = 50σ = 15n = 100x̄ = 48.38H₁: μ ≠ μ₀
p =
10.Hypothesis tests & confidence intervalsHard
z −2.57602.576

A sample of 25 measurements has mean 108.8. The population standard deviation is 20. Find the upper limit U of the 99% confidence interval for the mean.

x̄ = 108.8σ = 20n = 2599%
U =
11.Linear regression & correlationMedium
xy294126198251030xy510102030

The table gives 5 paired observations of x = load in kN and y = extension in mm. Find the coefficient of determination r² of the least-squares line.

n = 5
r² =
12.Linear regression & correlationHard
xy349455563669778883xy4686080ŷ = 73

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 = 73?

n = 6ŷ = 73
x =

3Solutions

1.Descriptive statistics

Answer: s = 14.68

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 = 244/7 = 34.86

Squared deviations

Σ(x − x̄)² = 3.143² + (−20.86)² + 9.143² + 17.14² + 14.14² + (−8.857)² + (−13.86)² = 1293

Standard deviation

s = √(1293/6) = 14.68

Answer

s = 14.68
2.Descriptive statistics

Answer: P = 0.06681

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 = −1.5

Standard normal table

Φ(−1.5) = 0.06681
P = 0.06681

Answer

P = 0.06681
3.Probability

Answer: σ = 1.195

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 = 0·0.35 + 1·0.2 + 2·0.2 + 3·0.25 = 1.35

Mean of the squares

E(X²) = 0·0.35 + 1·0.2 + 4·0.2 + 9·0.25 = 3.25

Variance

Var(X) = 3.25 − 1.35² = 1.427
σ = √1.427 = 1.195

Answer

σ = 1.195
4.Probability

Answer: P = 0.5767

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) = 0.87 = 0.2097
P(X = 1) = 7·0.2·0.86 = 0.367

Add them

P = 0.2097 + 0.367 = 0.5767

Answer

P = 0.5767
5.Normal distribution

Answer: P = 0.6915

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 = 750 g, σ = √(4² + 9²) = 9.849 g

Standardise

z = (745.1 − 750)/9.849 = −0.5

Upper tail

P = 1 − Φ(−0.5) = 1 − 0.3085 = 0.6915

Answer

P = 0.6915
6.Normal distribution

Answer: x = 521 h

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

Φ(z) = 0.7

z from the table

z ≈ 0.52

Back to the original units

x = 500 + 0.52·40 = 521 h

Answer

x = 521 h
7.Confidence intervals

Answer: lower limit = 110.3

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 = 7, t = 2.365

Margin of error

E = 2.365·2/√8 = 1.672

Interval

112 ± 1.672 → [110.3, 113.7]

Answer

lower limit = 110.3
8.Confidence intervals

Answer: σ = 18.43 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 = 10·10.5² = 1103

Limits for σ²

1103/20.483 = 53.83
1103/3.247 = 339.5

Limit for σ

σ = √339.5 = 18.43 µm

Answer

σ = 18.43 µm

95% interval for σ: [7.337, 18.43] µm.

9.Hypothesis tests & confidence intervals

Answer: p = 0.2801

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 = 15/√100 = 1.5
z = (48.38 − 50)/1.5 = −1.08

Tail area

Φ(1.08) = 0.8599
1 − Φ(1.08) = 0.1401

p-value

p = 2·0.1401 = 0.2801

Answer

p = 0.2801

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

10.Hypothesis tests & confidence intervals

Answer: U = 119.1

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).

Margin of error

E = 2.576·20/√25 = 10.3

Interval limit

U = 108.8 + 10.3 = 119.1

Answer

U = 119.1
11.Linear regression & correlation

Answer: r² = 0.9886

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 = 95/5 = 19

Sums of squares and products

Sxx = Σ(x − x̄)² = 40
Sxy = Σ(x − x̄)(y − ȳ) = 110
Syy = Σ(y − ȳ)² = 306

Coefficient of determination

r² = 110²/(40·306) = 0.9886

Answer

r² = 0.9886

About 98.86% of the variation in y is explained by the line.

12.Linear regression & correlation

Answer: x = 6.476

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 = 33/6 = 5.5
ȳ = 397/6 = 66.17

Sums of squares and products

Sxx = Σ(x − x̄)² = 17.5
Sxy = Σ(x − x̄)(y − ȳ) = 122.5

Regression line

b = 122.5/17.5 = 7
a = 66.17 − 7·5.5 = 27.67
ŷ = 27.67 + 7·x

Solve for x

x = (73 − 27.67)/7 = 6.476

Answer

x = 6.476

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