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
20253035404550

Find the sample standard deviation s of the data: 44, 16, 54, 17, 17, 49, 26.

n = 7
s =
2.Descriptive statisticsHard
z −2−1012

Exam scores are normally distributed with mean 100 and standard deviation 15. Find the z-score z of a score of 70.

μ = 100σ = 15x = 70
z =
3.ProbabilityMedium
P(at least one six) = ??1?2?3P(six) = 1/6 on each roll

A fair die is rolled 3 times. What is the probability P of getting at least one six?

n = 3
P =
4.ProbabilityHard
n = 7, p = 0.4: P(X ≤ 1) = ?k01234567

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

n = 7p = 0.4
P =
5.Normal distributionMedium
−3−2−10123zshaded area between −z and z = 0.9051

Drag z so that the shaded area between −z and z under the standard normal curve is 0.9051. Find z to two decimals.

A = 0.9051
z =
6.Normal distributionHard
4 s0X − Y?

Runner A's lap time X and runner B's lap time Y are independent and normal: X has mean 52 s and standard deviation 4 s, Y has mean 48 s and standard deviation 3 s. Find the probability P that X is greater than Y. Round z to two decimals.

μX = 52 s, σX = 4 sμY = 48 s, σY = 3 s
P =
7.Confidence intervalsMedium
μ₁ − μ₂0LUx̄₁ − x̄₂E = ?90%

The battery life of two phone models is compared. Model 1: 40 phones, mean 106.5 h, known standard deviation 17 h. Model 2: 40 phones, mean 118.5 h, known standard deviation 9 h. Find the margin of error E of the 90% confidence interval for μ₁ − μ₂.

n₁ = 40x̄₁ = 106.5 hσ₁ = 17 hn₂ = 40x̄₂ = 118.5 hσ₂ = 9 h90%
E = h
8.Confidence intervalsHard
n = 20σ = ?Us = 9.895%

The diameters of 20 machined shafts have a sample standard deviation of 9.8 µm. Assuming the diameters are normally distributed, find the lower 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 = 9.8 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 97.83 and a sample standard deviation of 5. Find the test statistic t for H₀: μ = 100.

μ₀ = 100x̄ = 97.83s = 5n = 36
t =
10.Hypothesis tests & confidence intervalsHard
σ = 4, n = ?x̄ − Ex̄ + Ex̄E = 399%

How large must a sample be so that the 99% confidence interval for a mean has a margin of error of at most 3? The standard deviation is 4. Find the smallest sample size n.

σ = 4E = 399%
n =
11.Linear regression & correlationMedium
xy239348451556661xy246405060

The table gives 5 paired observations of x = advertising spend in k€ and y = sales in k€. Find the coefficient of determination r² of the least-squares line.

n = 5
r² =
12.Linear regression & correlationHard
xy264136198231025xy5101020x = 5

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. A new observation (5, 11) is made. Find its residual e = y − ŷ.

n = 5(5, 11)
e =

3Solutions

1.Descriptive statistics

Answer: s = 16.63

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 = 223/7 = 31.86

Squared deviations

Σ(x − x̄)² = 12.14² + (−15.86)² + 22.14² + (−14.86)² + (−14.86)² + 17.14² + (−5.857)² = 1659

Standard deviation

s = √(1659/6) = 16.63

Answer

s = 16.63
2.Descriptive statistics

Answer: z = −2

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 = (70 − 100)/15 = −2

Answer

z = −2
3.Probability

Answer: P = 0.4213

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.

Complement

P(no six) = (5/6)3 = 0.5787
P = 1 − 0.5787 = 0.4213

Answer

P = 0.4213
4.Probability

Answer: P = 0.1586

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) = 0.67 = 0.02799
P(X = 1) = 7·0.4·0.66 = 0.1306

Add them

P = 0.02799 + 0.1306 = 0.1586

Answer

P = 0.1586
5.Normal distribution

Answer: z = 1.67

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 + 0.9051)/2 = 0.9525

Read the z-table backwards

Find 0.9525 inside the table and read off z.

z = 1.67

Answer

z = 1.67
6.Normal distribution

Answer: P = 0.7881

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 difference

μ = 52 − 48 = 4 s, σ = 5 s

Standardise

z = (0 − 4)/5 = −0.8

Probability

P = 1 − Φ(−0.8) = 0.7881

Answer

P = 0.7881
7.Confidence intervals

Answer: E = 5.003 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 = 106.5 − 118.5 = −12

Standard error

SE = √(17²/40 + 9²/40) = 3.041

Margin of error

E = 1.645·3.041 = 5.003

Answer

E = 5.003 h

The interval does not contain 0, so the means differ significantly.

8.Confidence intervals

Answer: σ = 7.453 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 = 19·9.8² = 1825

Limits for σ²

1825/32.852 = 55.54
1825/8.907 = 204.9

Limit for σ

σ = √55.54 = 7.453 µm

Answer

σ = 7.453 µm

95% interval for σ: [7.453, 14.31] µm.

9.Hypothesis tests & confidence intervals

Answer: t = −2.604

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 = 5/√36 = 5/6 = 0.8333

Test statistic

t = (97.83 − 100)/0.8333 = −2.604

Answer

t = −2.604
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 = 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).

Solve for n

n = (2.576·4/3)² = 11.8

Round up

n = 12

Answer

n = 12
11.Linear regression & correlation

Answer: r² = 0.9727

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 = 20/5 = 4
ȳ = 255/5 = 51

Sums of squares and products

Sxx = Σ(x − x̄)² = 10
Sxy = Σ(x − x̄)(y − ȳ) = 52
Syy = Σ(y − ȳ)² = 278

Coefficient of determination

r² = 52²/(10·278) = 0.9727

Answer

r² = 0.9727

About 97.27% of the variation in y is explained by the line.

12.Linear regression & correlation

Answer: e = −3.8

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 = 86/5 = 17.2

Sums of squares and products

Sxx = Σ(x − x̄)² = 40
Sxy = Σ(x − x̄)(y − ȳ) = 96

Regression line

b = 96/40 = 2.4
a = 17.2 − 2.4·6 = 2.8
ŷ = 2.8 + 2.4·x

Predicted value

ŷ = 2.8 + 2.4·5 = 14.8

Residual

e = 11 − 14.8 = −3.8

Answer

e = −3.8

The point lies below the line.

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