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: 13, 46, 48, 51, 33.

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

Exam scores are normally distributed with mean 70 and standard deviation 20. Find the z-score z of a score of 60.

μ = 70σ = 20x = 60
z =
3.ProbabilityMedium
n = 12k = 4N = ?

In how many ways N can a committee of 4 people be chosen from a group of 12? The order does not matter.

n = 12k = 4
N =
4.ProbabilityHard
n = 7, p = 0.25: P(X ≤ 1) = ?k01234567

7 components are tested, and each one fails with probability 0.25, independently. What is the probability P that at most one fails?

n = 7p = 0.25
P =
5.Normal distributionMedium
250 kg251.6 kgT = X₁ + … + X₁₀?

The masses of bags of cement are independent and normal with mean 25 kg and standard deviation 0.5 kg. Find the probability P that the total mass of 10 of them exceeds 251.6 kg. Round z to two decimals.

μ = 25 kgσ = 0.5 kgn = 10> 251.6 kg
P =
6.Normal distributionHard
μ = 500x = ?90%

The battery lifetimes are normally distributed with mean 500 h and standard deviation 40 h. Find the value x below which 90% of them lie (the 90th percentile).

μ = 500σ = 40P = 90%
x = h
7.Confidence intervalsMedium
μ₁ − μ₂0L = ?Ux̄₁ − x̄₂E95%

The battery life of two phone models is compared. Model 1: 100 phones, mean 58.5 h, known standard deviation 15 h. Model 2: 50 phones, mean 63 h, known standard deviation 8 h. Find the lower limit of the 95% confidence interval for μ₁ − μ₂.

n₁ = 100x̄₁ = 58.5 hσ₁ = 15 hn₂ = 50x̄₂ = 63 hσ₂ = 8 h95%
lower limit = h
8.Confidence intervalsHard
n = 16Lσ = ?s = 9.295%

The diameters of 16 machined shafts have a sample standard deviation of 9.2 µm. Assuming the diameters are normally distributed, find the upper 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 = 9.2 µmχ² = 6.262, 27.488
σ = µm
9.Hypothesis tests & confidence intervalsMedium
t −1.99801.998

A sample of 64 students has a mean score of 502.1 and a sample standard deviation of 10. Find the test statistic t for H₀: μ = 500.

μ₀ = 500x̄ = 502.1s = 10n = 64
t =
10.Hypothesis tests & confidence intervalsHard
μ₀x̄c

A one-sided z test of H₀: μ = 100 against H₁: μ > 100 is done at α = 0.01 with a sample of 25; the standard deviation is known to be 10. In reality the mean is 106.8. Find the probability of a Type II error β in percent.

μ₀ = 100μ = 106.8σ = 10n = 25α = 0.01
β = %
11.Linear regression & correlationMedium
xy2464506588681075xy510506070

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
xy127224323419519xy242025ŷ = 23

The table gives 5 paired observations of x = car age in years and y = price in k€. Use the least-squares line ŷ = a + b·x. For which value x does the line predict ŷ = 23?

n = 5ŷ = 23
x =

3Solutions

1.Descriptive statistics

Answer: s = 15.67

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 = 191/5 = 38.2

Squared deviations

Σ(x − x̄)² = (−25.2)² + 7.8² + 9.8² + 12.8² + (−5.2)² = 982.8

Standard deviation

s = √(982.8/4) = 15.67

Answer

s = 15.67
2.Descriptive statistics

Answer: z = −0.5

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 = (60 − 70)/20 = −0.5

Answer

z = −0.5
3.Probability

Answer: N = 495

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(12, 4) = (12·11·10·9)/(4·3·2·1) = 495

Answer

N = 495
4.Probability

Answer: P = 0.4449

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) = 0.757 = 0.1335
P(X = 1) = 7·0.25·0.756 = 0.3115

Add them

P = 0.1335 + 0.3115 = 0.4449

Answer

P = 0.4449
5.Normal distribution

Answer: P = 0.1562

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 total

μT = 250 kg, σT = 0.5·√10 = 1.581 kg

Standardise

z = (251.6 − 250)/1.581 = 1.01

Probability

P = 1 − Φ(1.01) = 0.1562

Answer

P = 0.1562
6.Normal distribution

Answer: x = 551.3 h

Normal distribution: convert a value x to a z-score, z = (x − μ)/σ, which counts standard deviations from the mean. A z-table (or Φ(z)) gives the area to the left of z, which is P(X < x). The area to the right is 1 − Φ(z), and the area between two values is the difference of their two areas.

z for the percentile

Φ(z) = 0.9 → z = 1.282

Back to x

x = 500 + 1.282·40 = 551.26 h

Answer

x = 551.26 h
7.Confidence intervals

Answer: lower limit = −8.183 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 = 58.5 − 63 = −4.5

Standard error

SE = √(15²/100 + 8²/50) = 1.879

Margin of error

E = 1.96·1.879 = 3.683

Interval

−4.5 ± 3.683 → [−8.183, −0.8175]

Answer

lower limit = −8.183 h

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

8.Confidence intervals

Answer: σ = 14.24 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 = 15·9.2² = 1270

Limits for σ²

1270/27.488 = 46.19
1270/6.262 = 202.7

Limit for σ

σ = √202.7 = 14.24 µm

Answer

σ = 14.24 µm

95% interval for σ: [6.796, 14.24] µm.

9.Hypothesis tests & confidence intervals

Answer: t = 1.68

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 = 10/√64 = 10/8 = 1.25

Test statistic

t = (502.1 − 500)/1.25 = 1.68

Answer

t = 1.68
t0.025; df = 63 = 1.998

|t| < 1.998, so H₀ is not rejected at the 5% level (two-sided).

10.Hypothesis tests & confidence intervals

Answer: β = 14.14 %

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 = 10/√25 = 2
x̄c = 100 + 2.326·2 = 104.7

z under the true mean

z = (104.7 − 106.8)/2 = −1.074

Type II error

power = P(Z > −1.074) = 85.86%
β = 100% − 85.86% = 14.14%

Answer

β = 14.14%
11.Linear regression & correlation

Answer: r = 0.9918

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 = 297/5 = 59.4

Sums of squares and products

Sxx = Σ(x − x̄)² = 40
Sxy = Σ(x − x̄)(y − ȳ) = 152
Syy = Σ(y − ȳ)² = 587.2

Correlation coefficient

r = 152/√(40·587.2) = 0.9918

Answer

r = 0.9918
12.Linear regression & correlation

Answer: x = 2.714

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 = 112/5 = 22.4

Sums of squares and products

Sxx = Σ(x − x̄)² = 10
Sxy = Σ(x − x̄)(y − ȳ) = −21

Regression line

b = −21/10 = −2.1
a = 22.4 − (−2.1)·3 = 28.7
ŷ = 28.7 − 2.1·x

Solve for x

x = (23 − 28.7)/(−2.1) = 2.714

Answer

x = 2.714

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