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
2025303540455055

Find the sample standard deviation s of the data: 52, 48, 20, 57, 48.

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

A quantity is normally distributed with mean 70 and standard deviation 8. What proportion P of values is below 74.8?

μ = 70σ = 8x = 74.8
P =
3.ProbabilityMedium
n = 10k = 5N = ?

In how many ways N can a committee of 5 people be chosen from a group of 10? The order does not matter.

n = 10k = 5
N =
4.ProbabilityHard
xp(x)00.310.3520.35μ

X has the distribution shown, and Y = 5X + 10. Find the standard deviation σY of Y.

x: 0, 1, 2p(x): 0.3, 0.35, 0.35Y = 5X + 10
σY =
5.Normal distributionMedium
−3−2−10123zshaded area between −z and z = 0.3401

Drag z so that the shaded area between −z and z under the standard normal curve is 0.3401. Find z to two decimals.

A = 0.3401
z =
6.Normal distributionHard
557085100115130145pointstop 60%

The IQ scores are normally distributed with mean 100 points and standard deviation 15 points. Drag the cut-off so the shaded top 60% is to the right of it. Find the cut-off x.

μ = 100 pointsσ = 15 points60%
x = points
7.Confidence intervalsMedium
μ₁ − μ₂0LU = ?x̄₁ − x̄₂E99%

The battery life of two phone models is compared. Model 1: 100 phones, mean 71.5 h, known standard deviation 8 h. Model 2: 30 phones, mean 77 h, known standard deviation 13 h. Find the upper limit of the 99% confidence interval for μ₁ − μ₂.

n₁ = 100x̄₁ = 71.5 hσ₁ = 8 hn₂ = 30x̄₂ = 77 hσ₂ = 13 h99%
upper limit = h
8.Confidence intervalsHard
p = 0.5, n = ?p̂ − Ep̂ + Ep̂E = 0.0190%

Nothing is known in advance about the proportion of defective products. How many products n must be inspected so that a 90% confidence interval for the proportion has a margin of error of at most 0.01?

p = 0.5E = 0.0190%
n =
9.Hypothesis tests & confidence intervalsMedium
t −2.06402.064

A sample of 25 students has a mean score of 114.7 and a sample standard deviation of 12. Find the test statistic t for H₀: μ = 120.

μ₀ = 120x̄ = 114.7s = 12n = 25
t =
10.Hypothesis tests & confidence intervalsHard
z −1.9601.96

A sample of 49 measurements has mean 75.9. The population standard deviation is 8. Find the lower limit L of the 95% confidence interval for the mean.

x̄ = 75.9σ = 8n = 4995%
L =
11.Linear regression & correlationMedium
xy141252359462569xy2440506070

The table gives 5 paired observations of x = hours studied and y = exam score. Find Pearson's correlation coefficient r.

n = 5
r =
12.Linear regression & correlationHard
xy294156258331038xy51010203040ŷ = 20

The table gives 5 paired observations of x = load in kN and y = extension in mm. Use the least-squares line ŷ = a + b·x. For which value x does the line predict ŷ = 20?

n = 5ŷ = 20
x =

3Solutions

1.Descriptive statistics

Answer: s = 14.46

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 = 225/5 = 45

Squared deviations

Σ(x − x̄)² = 7² + 3² + (−25)² + 12² + 3² = 836

Standard deviation

s = √(836/4) = 14.46

Answer

s = 14.46
2.Descriptive statistics

Answer: P = 0.7257

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 = 0.6

Standard normal table

Φ(0.6) = 0.7257
P = 0.7257

Answer

P = 0.7257
3.Probability

Answer: N = 252

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(10, 5) = (10·9·8·7·6)/(5·4·3·2·1) = 252

Answer

N = 252
4.Probability

Answer: σY = 4.023

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 = 1.05, Var(X) = 0.6475, σX = 0.8047

Linear change

σY = |5|·0.8047 = 4.023

Answer

σY = 4.023

The mean does move: E(Y) = 5·1.05 + 10 = 15.25.

5.Normal distribution

Answer: z = 0.44

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 + 0.3401)/2 = 0.6700

Read the z-table backwards

Find 0.6700 inside the table and read off z.

z = 0.44

Answer

z = 0.44
6.Normal distribution

Answer: x = 96.3 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) = 0.4

z from the table

z ≈ −0.25

Back to the original units

x = 100 − 0.25·15 = 96.3 points

Answer

x = 96.3 points
7.Confidence intervals

Answer: upper limit = 0.952 h

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.

Point estimate

x̄₁ − x̄₂ = 71.5 − 77 = −5.5

Standard error

SE = √(8²/100 + 13²/30) = 2.505

Margin of error

E = 2.576·2.505 = 6.452

Interval

−5.5 ± 6.452 → [−11.95, 0.952]

Answer

upper limit = 0.952 h

The interval contains 0, so the data do not show a significant difference.

8.Confidence intervals

Answer: n = 6766

Confidence interval for a proportion: p̂ ± z·√(p̂(1 − p̂)/n), where p̂ = x/n. To reach a margin E you need n ≥ z²·p(1 − p)/E², rounded up; without a prior estimate use p = 0.5, which gives the largest n.

Solve for n

n = 1.645²·0.5·0.5/0.01² = 6765

Round up

n = 6766

Answer

n = 6766
9.Hypothesis tests & confidence intervals

Answer: t = −2.208

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 = (114.7 − 120)/2.4 = −2.208

Answer

t = −2.208
t0.025; df = 24 = 2.064

|t| > 2.064, so H₀ is rejected at the 5% level (two-sided).

10.Hypothesis tests & confidence intervals

Answer: L = 73.66

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 = 1.96·8/√49 = 2.24

Interval limit

L = 75.9 − 2.24 = 73.66

Answer

L = 73.66
11.Linear regression & correlation

Answer: r = 0.9804

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 = 15/5 = 3
ȳ = 283/5 = 56.6

Sums of squares and products

Sxx = Σ(x − x̄)² = 10
Sxy = Σ(x − x̄)(y − ȳ) = 66
Syy = Σ(y − ȳ)² = 453.2

Correlation coefficient

r = 66/√(10·453.2) = 0.9804

Answer

r = 0.9804
12.Linear regression & correlation

Answer: x = 4.947

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 = 120/5 = 24

Sums of squares and products

Sxx = Σ(x − x̄)² = 40
Sxy = Σ(x − x̄)(y − ȳ) = 152

Regression line

b = 152/40 = 3.8
a = 24 − 3.8·6 = 1.2
ŷ = 1.2 + 3.8·x

Solve for x

x = (20 − 1.2)/3.8 = 4.947

Answer

x = 4.947

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