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
10152025303540455055

Find the sample standard deviation s of the data: 45, 10, 59, 18, 47.

n = 5
s =
2.Descriptive statisticsHard
z −2−1012

A quantity is normally distributed with mean 100 and standard deviation 8. What proportion P of values is below 116?

μ = 100σ = 8x = 116
P =
3.ProbabilityMedium
red: 7blue: 9

A bag holds 7 red and 9 blue balls. You draw two balls without putting the first one back. What is the probability P that both are red?

7 / 9
P =
4.ProbabilityHard
xp(x)10.220.2530.2540.3μ

X has the distribution shown, and Y = 3X + 5. Find the standard deviation σY of Y.

x: 1, 2, 3, 4p(x): 0.2, 0.25, 0.25, 0.3Y = 3X + 5
σY =
5.Normal distributionMedium
105 s113 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 113 s. Round z to two decimals.

μX = 60, σX = 5μY = 45, σY = 4X + Y > 113 s
P =
6.Normal distributionHard
557085100115130145pointstop 20%

The IQ scores are normally distributed with mean 100 points and standard deviation 15 points. Drag the cut-off so the shaded top 20% is to the right of it. Find the cut-off x.

μ = 100 pointsσ = 15 points20%
x = points
7.Confidence intervalsMedium
n = 11, s = 8.5LU = ?x̄ = 51.5E99%

A sample of 11 items has mean 51.5 and standard deviation s = 8.5. The population is roughly normal. Find the upper limit of the 99% confidence interval for the mean.

n = 11x̄ = 51.5s = 8.599%t(10) = 3.169
upper limit =
8.Confidence intervalsHard
n = 30Lσ = ?s = 5.495%

The diameters of 30 machined shafts have a sample standard deviation of 5.4 µm. Assuming the diameters are normally distributed, find the upper limit σ of the 95% confidence interval for the population standard deviation. With df = 29, 2.5% of the χ² distribution lies below 16.047 and 2.5% lies above 45.722.

n = 30s = 5.4 µmχ² = 16.047, 45.722
σ = µm
9.Hypothesis tests & confidence intervalsMedium
−2.0932.093t = ?df = 19

Two teaching methods are compared. Group 1: 15 students, mean score 68.5, standard deviation 9.5. Group 2: 6 students, mean score 64.3, standard deviation 6. 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 = 15x̄₁ = 68.5s₁ = 9.5n₂ = 6x̄₂ = 64.3s₂ = 6
t =
10.Hypothesis tests & confidence intervalsHard
μ₀x̄c

A one-sided z test of H₀: μ = 100 against H₁: μ > 100 is done at α = 0.10 with a sample of 100; the standard deviation is known to be 25. In reality the mean is 103. Find the probability of a Type II error β in percent.

μ₀ = 100μ = 103σ = 25n = 100α = 0.10
β = %
11.Linear regression & correlationMedium
xy15921810621126241412715630180xy15202530100150

The table gives 6 paired observations of x = temperature in °C and y = ice creams sold. Find the coefficient of determination r² of the least-squares line.

n = 6
r² =
12.Linear regression & correlationHard
xy13734956978199011105xy51050100ŷ = 85

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 = 85?

n = 6ŷ = 85
x =

3Solutions

1.Descriptive statistics

Answer: s = 20.8

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 = 179/5 = 35.8

Squared deviations

Σ(x − x̄)² = 9.2² + (−25.8)² + 23.2² + (−17.8)² + 11.2² = 1731

Standard deviation

s = √(1731/4) = 20.8

Answer

s = 20.8
2.Descriptive statistics

Answer: P = 0.9772

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 = 2

Standard normal table

Φ(2) = 0.9772
P = 0.9772

Answer

P = 0.9772
3.Probability

Answer: P = 0.175

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.

First and second draw

P = 7/16 · 6/15 = 42/240 = 0.175

Answer

P = 0.175
4.Probability

Answer: σY = 3.324

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 = 2.65, Var(X) = 1.228, σX = 1.108

Linear change

σY = |3|·1.108 = 3.324

Answer

σY = 3.324

The mean does move: E(Y) = 3·2.65 + 5 = 12.95.

5.Normal distribution

Answer: P = 0.1056

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 = (113 − 105)/6.403 = 1.25

Upper tail

P = 1 − Φ(1.25) = 1 − 0.8944 = 0.1056

Answer

P = 0.1056
6.Normal distribution

Answer: x = 112.6 points

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) = 0.8

z from the table

z ≈ 0.84

Back to the original units

x = 100 + 0.84·15 = 112.6 points

Answer

x = 112.6 points
7.Confidence intervals

Answer: upper limit = 59.62

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 = 10, t = 3.169

Margin of error

E = 3.169·8.5/√11 = 8.122

Interval

51.5 ± 8.122 → [43.38, 59.62]

Answer

upper limit = 59.62
8.Confidence intervals

Answer: σ = 7.259 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 = 29·5.4² = 845.6

Limits for σ²

845.6/45.722 = 18.5
845.6/16.047 = 52.7

Limit for σ

σ = √52.7 = 7.259 µm

Answer

σ = 7.259 µm

95% interval for σ: [4.301, 7.259] µm.

9.Hypothesis tests & confidence intervals

Answer: t = 0.9975

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 = (14·9.5² + 5·6²)/19 = 75.97
sp = 8.716

Standard error

SE = 8.716·√(1/15 + 1/6) = 4.21

Test statistic

t = (68.5 − 64.3)/4.21 = 0.9975

Answer

t = 0.9975

With df = 19, the two-sided 5% critical value is 2.093; |t| is smaller, so H₀ is not rejected.

10.Hypothesis tests & confidence intervals

Answer: β = 53.27 %

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 = 25/√100 = 2.5
x̄c = 100 + 1.282·2.5 = 103.2

z under the true mean

z = (103.2 − 103)/2.5 = 0.082

Type II error

power = P(Z > 0.082) = 46.73%
β = 100% − 46.73% = 53.27%

Answer

β = 53.27%
11.Linear regression & correlation

Answer: r² = 0.9942

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 = 135/6 = 22.5
ȳ = 801/6 = 133.5

Sums of squares and products

Sxx = Σ(x − x̄)² = 157.5
Sxy = Σ(x − x̄)(y − ȳ) = 907.5
Syy = Σ(y − ȳ)² = 5260

Coefficient of determination

r² = 907.5²/(157.5·5260) = 0.9942

Answer

r² = 0.9942

About 99.42% of the variation in y is explained by the line.

12.Linear regression & correlation

Answer: x = 7.94

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 = 36/6 = 6
ȳ = 431/6 = 71.83

Sums of squares and products

Sxx = Σ(x − x̄)² = 70
Sxy = Σ(x − x̄)(y − ȳ) = 475

Regression line

b = 475/70 = 6.786
a = 71.83 − 6.786·6 = 31.12
ŷ = 31.12 + 6.786·x

Solve for x

x = (85 − 31.12)/6.786 = 7.94

Answer

x = 7.94

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