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: 25, 34, 13, 50, 22.

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 108?

μ = 100σ = 8x = 108
P =
3.ProbabilityMedium
n = 14k = 4N = ?

In how many ways N can a committee of 4 people be chosen from a group of 14? The order does not matter.

n = 14k = 4
N =
4.ProbabilityHard
n = 5, p = 0.4: P(X ≤ 1) = ?k012345

5 components are tested, and each one fails with probability 0.4, independently. What is the probability P that at most one fails?

n = 5p = 0.4
P =
5.Normal distributionMedium
105 s117.8 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 117.8 s. Round z to two decimals.

μX = 60, σX = 5μY = 45, σY = 4X + Y > 117.8 s
P =
6.Normal distributionHard
557085100115130145pointstop 5%

The IQ scores are normally distributed with mean 100 points and standard deviation 15 points. Drag the cut-off so the shaded top 5% is to the right of it. Find the cut-off x.

μ = 100 pointsσ = 15 points5%
x = points
7.Confidence intervalsMedium
n = 15, sd​ = 20LU = ?d̄ = 5.8E90%

The blood pressure of 15 patients is measured before and after a treatment. The differences d = before − after (mmHg) have mean 5.8 and standard deviation 2. Find the upper limit of the 90% confidence interval for the mean difference.

n = 15d̄ = 5.8sd = 290%t(14) = 1.761
upper limit = mmHg
8.Confidence intervalsHard
n = 16σ = ?Us = 7.795%

The diameters of 16 machined shafts have a sample standard deviation of 7.7 µm. Assuming the diameters are normally distributed, find the lower limit σ of the 95% confidence interval for the population standard deviation. With df = 15, 2.5% of the χ² distribution lies below 6.262 and 2.5% lies above 27.488.

n = 16s = 7.7 µmχ² = 6.262, 27.488
σ = µm
9.Hypothesis tests & confidence intervalsMedium
t −2.06402.064

A sample of 25 students has a mean score of 201.7 and a sample standard deviation of 12. Find the test statistic t for H₀: μ = 200.

μ₀ = 200x̄ = 201.7s = 12n = 25
t =
10.Hypothesis tests & confidence intervalsHard
−1.961.96z = ?

In a trial, 125 of 250 patients on treatment A recover, and 83 of 150 patients on treatment B. Using the pooled proportion, find the test statistic z for H₀: p₁ = p₂.

x₁ = 125n₁ = 250x₂ = 83n₂ = 150
z =
11.Linear regression & correlationMedium
xy14341649186120552262xy1520405060

The table gives 5 paired observations of x = temperature in °C and y = ice creams sold. Find Pearson's correlation coefficient r.

n = 5
r =
12.Linear regression & correlationHard
xy1673198722101251132811831143xy15202530100150x = 20.5

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 (20.5, 90) is made. Find its residual e = y − ŷ.

n = 6(20.5, 90)
e =

3Solutions

1.Descriptive statistics

Answer: s = 14.02

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 = 144/5 = 28.8

Squared deviations

Σ(x − x̄)² = (−3.8)² + 5.2² + (−15.8)² + 21.2² + (−6.8)² = 786.8

Standard deviation

s = √(786.8/4) = 14.02

Answer

s = 14.02
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) = 0.8413
P = 0.8413

Answer

P = 0.8413
3.Probability

Answer: N = 1001

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, 4) = (14·13·12·11)/(4·3·2·1) = 1001

Answer

N = 1001
4.Probability

Answer: P = 0.337

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) = 0.65 = 0.07776
P(X = 1) = 5·0.4·0.64 = 0.2592

Add them

P = 0.07776 + 0.2592 = 0.337

Answer

P = 0.337
5.Normal distribution

Answer: P = 0.02275

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 = (117.8 − 105)/6.403 = 2

Upper tail

P = 1 − Φ(2) = 1 − 0.9772 = 0.02275

Answer

P = 0.02275
6.Normal distribution

Answer: x = 124.6 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) = 0.95

z from the table

z ≈ 1.64

Back to the original units

x = 100 + 1.64·15 = 124.6 points

Answer

x = 124.6 points
7.Confidence intervals

Answer: upper limit = 6.709 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 = 14, t = 1.761

Margin of error

E = 1.761·2/√15 = 0.9094

Interval

5.8 ± 0.9094 → [4.891, 6.709]

Answer

upper limit = 6.709 mmHg
8.Confidence intervals

Answer: σ = 5.688 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 = 15·7.7² = 889.4

Limits for σ²

889.4/27.488 = 32.35
889.4/6.262 = 142

Limit for σ

σ = √32.35 = 5.688 µm

Answer

σ = 5.688 µm

95% interval for σ: [5.688, 11.92] µm.

9.Hypothesis tests & confidence intervals

Answer: t = 0.7083

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 = 12/√25 = 12/5 = 2.4

Test statistic

t = (201.7 − 200)/2.4 = 0.7083

Answer

t = 0.7083
t0.025; df = 24 = 2.064

|t| < 2.064, so H₀ is not rejected at the 5% level (two-sided).

10.Hypothesis tests & confidence intervals

Answer: z = −1.034

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.

Sample proportions

p̂₁ = 125/250 = 0.5
p̂₂ = 83/150 = 0.5533
p̂ = 208/400 = 0.52

Standard error under H₀

SE = √(0.52·0.48·(1/250 + 1/150)) = 0.0516

Test statistic

z = (0.5 − 0.5533)/0.0516 = −1.034

Answer

z = −1.034

|z| < 1.96, so H₀ is not rejected at the 5% level (two-sided).

11.Linear regression & correlation

Answer: r = 0.8575

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 = 90/5 = 18
ȳ = 261/5 = 52.2

Sums of squares and products

Sxx = Σ(x − x̄)² = 40
Sxy = Σ(x − x̄)(y − ȳ) = 124
Syy = Σ(y − ȳ)² = 522.8

Correlation coefficient

r = 124/√(40·522.8) = 0.8575

Answer

r = 0.8575
12.Linear regression & correlation

Answer: e = −2.833

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 = 141/6 = 23.5
ȳ = 635/6 = 105.8

Sums of squares and products

Sxx = Σ(x − x̄)² = 157.5
Sxy = Σ(x − x̄)(y − ȳ) = 682.5

Regression line

b = 682.5/157.5 = 4.333
a = 105.8 − 4.333·23.5 = 4
ŷ = 4 + 4.333·x

Predicted value

ŷ = 4 + 4.333·20.5 = 92.83

Residual

e = 90 − 92.83 = −2.833

Answer

e = −2.833

The point lies below the line.

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