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
101520253035404550

Find the sample standard deviation s of the data: 23, 10, 54, 22, 40.

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

Exam scores are normally distributed with mean 60 and standard deviation 20. Find the z-score z of a score of 85.

μ = 60σ = 20x = 85
z =
3.ProbabilityMedium
xp(x)10.120.330.340.3μ = 2.8

For the distribution shown, find the standard deviation σ of X.

x: 1, 2, 3, 4p(x): 0.1, 0.3, 0.3, 0.3
σ =
4.ProbabilityHard
xp(x)00.1510.320.3530.2μ

X has the distribution shown, and Y = 0.5X + 20. Find the standard deviation σY of Y.

x: 0, 1, 2, 3p(x): 0.15, 0.3, 0.35, 0.2Y = 0.5X + 20
σY =
5.Normal distributionMedium
−3−2−10123zshaded area between −z and z = 0.4039

Drag z so that the shaded area between −z and z under the standard normal curve is 0.4039. Find z to two decimals.

A = 0.4039
z =
6.Normal distributionHard
0.04 mm0D − d?

Hole diameters D are normal with mean 25.04 mm and standard deviation 0.015 mm. Shaft diameters d are independent and normal with mean 25 mm and standard deviation 0.015 mm. A randomly chosen shaft fits when D − d > 0. Find the probability P that it fits. Round z to two decimals.

hole: μ = 25.04 mm, σ = 0.015 mmshaft: μ = 25 mm, σ = 0.015 mm
P =
7.Confidence intervalsMedium
μ₁ − μ₂0LUx̄₁ − x̄₂E = ?95%

The battery life of two phone models is compared. Model 1: 80 phones, mean 80.5 h, known standard deviation 6 h. Model 2: 40 phones, mean 76.5 h, known standard deviation 17 h. Find the margin of error E of the 95% confidence interval for μ₁ − μ₂.

n₁ = 80x̄₁ = 80.5 hσ₁ = 6 hn₂ = 40x̄₂ = 76.5 hσ₂ = 17 h95%
E = h
8.Confidence intervalsHard
n = 11Lσ = ?s = 3.695%

The diameters of 11 machined shafts have a sample standard deviation of 3.6 µm. Assuming the diameters are normally distributed, find the upper limit σ of the 95% confidence interval for the population standard deviation. With df = 10, 2.5% of the χ² distribution lies below 3.247 and 2.5% lies above 20.483.

n = 11s = 3.6 µmχ² = 3.247, 20.483
σ = µm
9.Hypothesis tests & confidence intervalsMedium
t −1.9901.99

A sample of 81 students has a mean score of 51.56 and a sample standard deviation of 20. Find the test statistic t for H₀: μ = 50.

μ₀ = 50x̄ = 51.56s = 20n = 81
t =
10.Hypothesis tests & confidence intervalsHard
μ₀x̄c

A one-sided z test of H₀: μ = 100 against H₁: μ > 100 is done at α = 0.05 with a sample of 64; the standard deviation is known to be 12. In reality the mean is 103.75. Find the power 1 − β of the test in percent.

μ₀ = 100μ = 103.75σ = 12n = 64α = 0.05
1 − β = %
11.Linear regression & correlationMedium
xy123221323419519617xy246182022

The table gives 6 paired observations of x = car age in years and y = price in k€. Find the coefficient of determination r² of the least-squares line.

n = 6
r² =
12.Linear regression & correlationHard
xy130225327424521618xy246202530x = 5.5

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 (5.5, 16) is made. Find its residual e = y − ŷ.

n = 6(5.5, 16)
e =

3Solutions

1.Descriptive statistics

Answer: s = 17.24

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 = 149/5 = 29.8

Squared deviations

Σ(x − x̄)² = (−6.8)² + (−19.8)² + 24.2² + (−7.8)² + 10.2² = 1189

Standard deviation

s = √(1189/4) = 17.24

Answer

s = 17.24
2.Descriptive statistics

Answer: z = 1.25

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 = (85 − 60)/20 = 1.25

Answer

z = 1.25
3.Probability

Answer: σ = 0.9798

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 = 1·0.1 + 2·0.3 + 3·0.3 + 4·0.3 = 2.8

Mean of the squares

E(X²) = 1·0.1 + 4·0.3 + 9·0.3 + 16·0.3 = 8.8

Variance

Var(X) = 8.8 − 2.8² = 0.96
σ = √0.96 = 0.9798

Answer

σ = 0.9798
4.Probability

Answer: σY = 0.4848

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 = 1.6, Var(X) = 0.94, σX = 0.9695

Linear change

σY = |0.5|·0.9695 = 0.4848

Answer

σY = 0.4848

The mean does move: E(Y) = 0.5·1.6 + 20 = 20.8.

5.Normal distribution

Answer: z = 0.53

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 + 0.4039)/2 = 0.7019

Read the z-table backwards

Find 0.7019 inside the table and read off z.

z = 0.53

Answer

z = 0.53
6.Normal distribution

Answer: P = 0.9706

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 = 25.04 − 25 = 0.04 mm, σ = 0.02121 mm

Standardise

z = (0 − 0.04)/0.02121 = −1.89

Probability

P = 1 − Φ(−1.89) = 0.9706

Answer

P = 0.9706
7.Confidence intervals

Answer: E = 5.43 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 = 80.5 − 76.5 = 4

Standard error

SE = √(6²/80 + 17²/40) = 2.77

Margin of error

E = 1.96·2.77 = 5.43

Answer

E = 5.43 h

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

8.Confidence intervals

Answer: σ = 6.318 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 = 10·3.6² = 129.6

Limits for σ²

129.6/20.483 = 6.327
129.6/3.247 = 39.91

Limit for σ

σ = √39.91 = 6.318 µm

Answer

σ = 6.318 µm

95% interval for σ: [2.515, 6.318] µm.

9.Hypothesis tests & confidence intervals

Answer: t = 0.702

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 = 20/√81 = 20/9 = 2.222

Test statistic

t = (51.56 − 50)/2.222 = 0.702

Answer

t = 0.702
t0.025; df = 80 = 1.99

|t| < 1.99, so H₀ is not rejected at the 5% level (two-sided).

10.Hypothesis tests & confidence intervals

Answer: 1 − β = 80.37 %

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 = 12/√64 = 1.5
x̄c = 100 + 1.645·1.5 = 102.5

z under the true mean

z = (102.5 − 103.75)/1.5 = −0.855

Power

power = P(Z > −0.855) = 80.37%
1 − β = 80.37%

Answer

1 − β = 80.37%
11.Linear regression & correlation

Answer: r² = 0.7792

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 = 122/6 = 20.33

Sums of squares and products

Sxx = Σ(x − x̄)² = 17.5
Sxy = Σ(x − x̄)(y − ȳ) = −20
Syy = Σ(y − ȳ)² = 29.33

Coefficient of determination

r² = −20²/(17.5·29.33) = 0.7792

Answer

r² = 0.7792

About 77.92% of the variation in y is explained by the line.

12.Linear regression & correlation

Answer: e = −3.881

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 = 145/6 = 24.17

Sums of squares and products

Sxx = Σ(x − x̄)² = 17.5
Sxy = Σ(x − x̄)(y − ȳ) = −37.5

Regression line

b = −37.5/17.5 = −2.143
a = 24.17 − (−2.143)·3.5 = 31.67
ŷ = 31.67 − 2.143·x

Predicted value

ŷ = 31.67 + (−2.143)·5.5 = 19.88

Residual

e = 16 − 19.88 = −3.881

Answer

e = −3.881

The point lies below the line.

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