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
25303540455055

Find the sample standard deviation s of the data: 32, 34, 23, 59, 46.

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

A quantity is normally distributed with mean 100 and standard deviation 15. What proportion P of values is above 70?

μ = 100σ = 15x = 70
P =
3.ProbabilityMedium
xp(x)10.220.2530.3540.0550.15μ = 2.7

For the distribution shown, find the variance Var(X) of X.

x: 1, 2, 3, 4, 5p(x): 0.2, 0.25, 0.35, 0.05, 0.15
Var(X) =
4.ProbabilityHard
xp(x)00.2510.0520.3530.340.05μ

X has the distribution shown, and Y = −2X + 1. Find the standard deviation σY of Y.

x: 0, 1, 2, 3, 4p(x): 0.25, 0.05, 0.35, 0.3, 0.05Y = −2X + 1
σY =
5.Normal distributionMedium
−3−2−10123zshaded area between −z and z = 0.5878

Drag z so that the shaded area between −z and z under the standard normal curve is 0.5878. Find z to two decimals.

A = 0.5878
z =
6.Normal distributionHard
μ = 178x = ?20%

The heights of adult men are normally distributed with mean 178 cm and standard deviation 7 cm. Find the value x below which 20% of them lie (the 20th percentile).

μ = 178σ = 7P = 20%
x = cm
7.Confidence intervalsMedium
n = 10, sd​ = 4.40LU = ?d̄ = 6.3E95%

The blood pressure of 10 patients is measured before and after a treatment. The differences d = before − after (mmHg) have mean 6.3 and standard deviation 4.4. Find the upper limit of the 95% confidence interval for the mean difference.

n = 10d̄ = 6.3sd = 4.495%t(9) = 2.262
upper limit = mmHg
8.Confidence intervalsHard
n = 25σ = ?Us = 295%

The diameters of 25 machined shafts have a sample standard deviation of 2 µm. Assuming the diameters are normally distributed, find the lower limit σ of the 95% confidence interval for the population standard deviation. With df = 24, 2.5% of the χ² distribution lies below 12.401 and 2.5% lies above 39.364.

n = 25s = 2 µmχ² = 12.401, 39.364
σ = µm
9.Hypothesis tests & confidence intervalsMedium
t −2.01102.011

A sample of 49 students has a mean score of 195.3 and a sample standard deviation of 15. Find the test statistic t for H₀: μ = 200.

μ₀ = 200x̄ = 195.3s = 15n = 49
t =
10.Hypothesis tests & confidence intervalsHard
z −2.57602.576

A sample of 36 measurements has mean 198. The population standard deviation is 4. Find the lower limit L of the 99% confidence interval for the mean.

x̄ = 198σ = 4n = 3699%
L =
11.Linear regression & correlationMedium
xy123222316417511xy241520

The table gives 5 paired observations of x = car age in years and y = price in k€. Their sums are given. Find Pearson's correlation coefficient r from these sums.

n = 5Σx = 15Σy = 89Σx² = 55Σy² = 1679Σxy = 238
r =
12.Linear regression & correlationHard
xy131228327424522620xy246202530ŷ = 21

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. For which value x does the line predict ŷ = 21?

n = 6ŷ = 21
x =

3Solutions

1.Descriptive statistics

Answer: s = 13.95

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 = 194/5 = 38.8

Squared deviations

Σ(x − x̄)² = (−6.8)² + (−4.8)² + (−15.8)² + 20.2² + 7.2² = 778.8

Standard deviation

s = √(778.8/4) = 13.95

Answer

s = 13.95
2.Descriptive statistics

Answer: P = 0.9772

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 = −2

Standard normal table

Φ(−2) = 0.02275
P = 1 − 0.02275 = 0.9772

Answer

P = 0.9772
3.Probability

Answer: Var(X) = 1.61

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.

Mean

μ = 1·0.2 + 2·0.25 + 3·0.35 + 4·0.05 + 5·0.15 = 2.7

Mean of the squares

E(X²) = 1·0.2 + 4·0.25 + 9·0.35 + 16·0.05 + 25·0.15 = 8.9

Variance

Var(X) = 8.9 − 2.7² = 1.61

Answer

Var(X) = 1.61
4.Probability

Answer: σY = 2.472

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 = 1.85, Var(X) = 1.527, σX = 1.236

Linear change

σY = |−2|·1.236 = 2.472

Answer

σY = 2.472

The mean does move: E(Y) = −2·1.85 + 1 = −2.7.

5.Normal distribution

Answer: z = 0.82

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 + 0.5878)/2 = 0.7939

Read the z-table backwards

Find 0.7939 inside the table and read off z.

z = 0.82

Answer

z = 0.82
6.Normal distribution

Answer: x = 172.1 cm

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) = 0.2 → z = −0.8416

Back to x

x = 178 + −0.8416·7 = 172.11 cm

Answer

x = 172.11 cm
7.Confidence intervals

Answer: upper limit = 9.447 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 = 9, t = 2.262

Margin of error

E = 2.262·4.4/√10 = 3.147

Interval

6.3 ± 3.147 → [3.153, 9.447]

Answer

upper limit = 9.447 mmHg
8.Confidence intervals

Answer: σ = 1.562 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 = 24·2² = 96

Limits for σ²

96/39.364 = 2.439
96/12.401 = 7.741

Limit for σ

σ = √2.439 = 1.562 µm

Answer

σ = 1.562 µm

95% interval for σ: [1.562, 2.782] µm.

9.Hypothesis tests & confidence intervals

Answer: t = −2.193

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/√49 = 15/7 = 2.143

Test statistic

t = (195.3 − 200)/2.143 = −2.193

Answer

t = −2.193
t0.025; df = 48 = 2.011

|t| > 2.011, so H₀ is rejected at the 5% level (two-sided).

10.Hypothesis tests & confidence intervals

Answer: L = 196.3

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 = 2.576·4/√36 = 1.717

Interval limit

L = 198 − 1.717 = 196.3

Answer

L = 196.3
11.Linear regression & correlation

Answer: r = −0.9419

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 = 238 − 15·89/5 = −29
Sxx = 55 − 15²/5 = 10
Syy = 1679 − 89²/5 = 94.8

Correlation coefficient

r = −29/√(10·94.8) = −0.9419

Answer

r = −0.9419
12.Linear regression & correlation

Answer: x = 5.496

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 = 152/6 = 25.33

Sums of squares and products

Sxx = Σ(x − x̄)² = 17.5
Sxy = Σ(x − x̄)(y − ȳ) = −38

Regression line

b = −38/17.5 = −2.171
a = 25.33 − (−2.171)·3.5 = 32.93
ŷ = 32.93 − 2.171·x

Solve for x

x = (21 − 32.93)/(−2.171) = 5.496

Answer

x = 5.496

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