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
15202530354045505560

Find the sample standard deviation s of the data: 43, 60, 14, 30, 39.

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

Exam scores are normally distributed with mean 500 and standard deviation 25. Find the z-score z of a score of 470.

μ = 500σ = 25x = 470
z =
3.ProbabilityMedium
P(at least one six) = ??1?2P(six) = 1/6 on each roll

A fair die is rolled 2 times. What is the probability P of getting at least one six?

n = 2
P =
4.ProbabilityHard
n = 12, p = 0.5: P(X ≤ 1) = ?k0123456789101112

12 components are tested, and each one fails with probability 0.5, independently. What is the probability P that at most one fails?

n = 12p = 0.5
P =
5.Normal distributionMedium
248.5251P = ?

The bag weights are normally distributed with mean 250 g and standard deviation 5 g. What fraction lies between 248.5 and 251 g? Give P as a decimal.

μ = 250σ = 5x1 = 248.5x2 = 251
P =
6.Normal distributionHard
10 g0X − Y?

The mass X of a product from line 1 and Y from line 2 are independent and normal: X has mean 500 g and standard deviation 8 g, Y has mean 490 g and standard deviation 6 g. Find the probability P that X is greater than Y. Round z to two decimals.

μX = 500 g, σX = 8 gμY = 490 g, σY = 6 g
P =
7.Confidence intervalsMedium
μ₁ − μ₂0L = ?Ux̄₁ − x̄₂E99%

The battery life of two phone models is compared. Model 1: 40 phones, mean 53.5 h, known standard deviation 16 h. Model 2: 30 phones, mean 64.5 h, known standard deviation 17 h. Find the lower limit of the 99% confidence interval for μ₁ − μ₂.

n₁ = 40x̄₁ = 53.5 hσ₁ = 16 hn₂ = 30x̄₂ = 64.5 hσ₂ = 17 h99%
lower limit = h
8.Confidence intervalsHard
n = 16σ = ?Us = 2.695%

The diameters of 16 machined shafts have a sample standard deviation of 2.6 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 = 2.6 µmχ² = 6.262, 27.488
σ = µm
9.Hypothesis tests & confidence intervalsMedium
t −2.0302.03

A sample of 36 students has a mean score of 206 and a sample standard deviation of 15. Find the test statistic t for H₀: μ = 200.

μ₀ = 200x̄ = 206s = 15n = 36
t =
10.Hypothesis tests & confidence intervalsHard
−1.961.96z = ?

In a trial, 65 of 100 patients on treatment A recover, and 68 of 120 patients on treatment B. Using the pooled proportion, find the test statistic z for H₀: p₁ = p₂.

x₁ = 65n₁ = 100x₂ = 68n₂ = 120
z =
11.Linear regression & correlationMedium
xy1540185321602472279430104xy15202530406080100

The table gives 6 paired observations of x = temperature in °C and y = ice creams sold. Their sums are given. Find Pearson's correlation coefficient r from these sums.

n = 6Σx = 135Σy = 423Σx² = 3195Σy² = 32845Σxy = 10200
r =
12.Linear regression & correlationHard
xy12121931641651169xy246101520x = 2.5

The table gives 6 paired observations of x = car age in years and y = price in k€. Use the least-squares line ŷ = a + b·x. A new observation (2.5, 16) is made. Find its residual e = y − ŷ.

n = 6(2.5, 16)
e =

3Solutions

1.Descriptive statistics

Answer: s = 16.93

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 = 186/5 = 37.2

Squared deviations

Σ(x − x̄)² = 5.8² + 22.8² + (−23.2)² + (−7.2)² + 1.8² = 1147

Standard deviation

s = √(1147/4) = 16.93

Answer

s = 16.93
2.Descriptive statistics

Answer: z = −1.2

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 = (470 − 500)/25 = −1.2

Answer

z = −1.2
3.Probability

Answer: P = 0.3056

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)2 = 0.6944
P = 1 − 0.6944 = 0.3056

Answer

P = 0.3056
4.Probability

Answer: P = 0.003174

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) = 0.512 = 0.0002441
P(X = 1) = 12·0.5·0.511 = 0.00293

Add them

P = 0.0002441 + 0.00293 = 0.003174

Answer

P = 0.003174
5.Normal distribution

Answer: P = 0.1972

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-scores

z₁ = (248.5 − 250)/5 = −0.3
z₂ = (251 − 250)/5 = 0.2

Area between

P = Φ(0.2) − Φ(−0.3) = 0.5793 − 0.3821 = 0.1972

Answer

P = 0.1972
6.Normal distribution

Answer: P = 0.8413

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 = 500 − 490 = 10 g, σ = 10 g

Standardise

z = (0 − 10)/10 = −1

Probability

P = 1 − Φ(−1) = 0.8413

Answer

P = 0.8413
7.Confidence intervals

Answer: lower limit = −21.31 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 = 53.5 − 64.5 = −11

Standard error

SE = √(16²/40 + 17²/30) = 4.004

Margin of error

E = 2.576·4.004 = 10.31

Interval

−11 ± 10.31 → [−21.31, −0.6853]

Answer

lower limit = −21.31 h

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

8.Confidence intervals

Answer: σ = 1.921 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 = 15·2.6² = 101.4

Limits for σ²

101.4/27.488 = 3.689
101.4/6.262 = 16.19

Limit for σ

σ = √3.689 = 1.921 µm

Answer

σ = 1.921 µm

95% interval for σ: [1.921, 4.024] µm.

9.Hypothesis tests & confidence intervals

Answer: t = 2.4

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 = 15/√36 = 15/6 = 2.5

Test statistic

t = (206 − 200)/2.5 = 2.4

Answer

t = 2.4
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: z = 1.259

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 = 65/100 = 0.65
p̂₂ = 68/120 = 0.5667
p̂ = 133/220 = 0.6045

Standard error under H₀

SE = √(0.6045·0.3955·(1/100 + 1/120)) = 0.0662

Test statistic

z = (0.65 − 0.5667)/0.0662 = 1.259

Answer

z = 1.259

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

11.Linear regression & correlation

Answer: r = 0.989

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.

Centred sums

Sxy = 10200 − 135·423/6 = 682.5
Sxx = 3195 − 135²/6 = 157.5
Syy = 32845 − 423²/6 = 3024

Correlation coefficient

r = 682.5/√(157.5·3024) = 0.989

Answer

r = 0.989
12.Linear regression & correlation

Answer: e = −1.733

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 = 21/6 = 3.5
ȳ = 92/6 = 15.33

Sums of squares and products

Sxx = Σ(x − x̄)² = 17.5
Sxy = Σ(x − x̄)(y − ȳ) = −42

Regression line

b = −42/17.5 = −2.4
a = 15.33 − (−2.4)·3.5 = 23.73
ŷ = 23.73 − 2.4·x

Predicted value

ŷ = 23.73 + (−2.4)·2.5 = 17.73

Residual

e = 16 − 17.73 = −1.733

Answer

e = −1.733

The point lies below the line.

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