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
15202530354045

Find the sample standard deviation s of the data: 17, 45, 26, 26, 14.

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

A quantity is normally distributed with mean 170 and standard deviation 15. What proportion P of values is below 185?

μ = 170σ = 15x = 185
P =
3.ProbabilityMedium
xp(x)00.310.3520.0530.2540.05μ = 1.4

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

x: 0, 1, 2, 3, 4p(x): 0.3, 0.35, 0.05, 0.25, 0.05
Var(X) =
4.ProbabilityHard
xp(x)00.2520.3540.0560.35μ

X has the distribution shown, and Y = 0.5X + 1. Find the standard deviation σY of Y.

x: 0, 2, 4, 6p(x): 0.25, 0.35, 0.05, 0.35Y = 0.5X + 1
σY =
5.Normal distributionMedium
60 kg63.6 kgT = X₁ + … + X₅?

The masses of boxes are independent and normal with mean 12 kg and standard deviation 0.8 kg. Find the probability P that the total mass of 5 of them exceeds 63.6 kg. Round z to two decimals.

μ = 12 kgσ = 0.8 kgn = 5> 63.6 kg
P =
6.Normal distributionHard
235240245250255260265gtop 25%

The bag weights are normally distributed with mean 250 g and standard deviation 5 g. Drag the cut-off so the shaded top 25% is to the right of it. Find the cut-off x.

μ = 250 gσ = 5 g25%
x = g
7.Confidence intervalsMedium
n = 15, s = 17.5LU = ?x̄ = 23E90%

A sample of 15 items has mean 23 and standard deviation s = 17.5. The population is roughly normal. Find the upper limit of the 90% confidence interval for the mean.

n = 15x̄ = 23s = 17.590%t(14) = 1.761
upper limit =
8.Confidence intervalsHard
n = 25σ = ?Us = 10.995%

The diameters of 25 machined shafts have a sample standard deviation of 10.9 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 = 10.9 µmχ² = 12.401, 39.364
σ = µm
9.Hypothesis tests & confidence intervalsMedium
t −2.06402.064

A sample of 25 students has a mean score of 77.4 and a sample standard deviation of 10. Find the test statistic t for H₀: μ = 75.

μ₀ = 75x̄ = 77.4s = 10n = 25
t =
10.Hypothesis tests & confidence intervalsHard
μ₀x̄c

A one-sided z test of H₀: μ = 80 against H₁: μ < 80 is done at α = 0.05 with a sample of 64; the standard deviation is known to be 12. In reality the mean is 75.65. Find the power 1 − β of the test in percent.

μ₀ = 80μ = 75.65σ = 12n = 64α = 0.05
1 − β = %
11.Linear regression & correlationMedium
xy1877208722100241122611128132xy202580100120

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

n = 6
r =
12.Linear regression & correlationHard
xy364573783910211116xy5106080100120ŷ = 108

The table gives 5 paired observations of x = hours studied and y = exam score. Use the least-squares line ŷ = a + b·x. For which value x does the line predict ŷ = 108?

n = 5ŷ = 108
x =

3Solutions

1.Descriptive statistics

Answer: s = 12.1

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 = 128/5 = 25.6

Squared deviations

Σ(x − x̄)² = (−8.6)² + 19.4² + 0.4² + 0.4² + (−11.6)² = 585.2

Standard deviation

s = √(585.2/4) = 12.1

Answer

s = 12.1
2.Descriptive statistics

Answer: P = 0.8413

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 = 1

Standard normal table

Φ(1) = 0.8413
P = 0.8413

Answer

P = 0.8413
3.Probability

Answer: Var(X) = 1.64

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 = 0·0.3 + 1·0.35 + 2·0.05 + 3·0.25 + 4·0.05 = 1.4

Mean of the squares

E(X²) = 0·0.3 + 1·0.35 + 4·0.05 + 9·0.25 + 16·0.05 = 3.6

Variance

Var(X) = 3.6 − 1.4² = 1.64

Answer

Var(X) = 1.64
4.Probability

Answer: σY = 1.204

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 = 3, Var(X) = 5.8, σX = 2.408

Linear change

σY = |0.5|·2.408 = 1.204

Answer

σY = 1.204

The mean does move: E(Y) = 0.5·3 + 1 = 2.5.

5.Normal distribution

Answer: P = 0.02222

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 = 60 kg, σT = 0.8·√5 = 1.789 kg

Standardise

z = (63.6 − 60)/1.789 = 2.01

Probability

P = 1 − Φ(2.01) = 0.02222

Answer

P = 0.02222
6.Normal distribution

Answer: x = 253.4 g

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

Φ(z) = 0.75

z from the table

z ≈ 0.67

Back to the original units

x = 250 + 0.67·5 = 253.4 g

Answer

x = 253.4 g
7.Confidence intervals

Answer: upper limit = 30.96

Confidence interval for a mean: x̄ ± margin of error. With a known population σ the margin is z·σ/√n; with only the sample standard deviation s, use t with n − 1 degrees of freedom: t·s/√n. Critical values: z = 1.645 (90%), 1.96 (95%), 2.576 (99%). To get a margin E or smaller, you need n ≥ (z·σ/E)², rounded up.

Critical value

df = n − 1 = 14, t = 1.761

Margin of error

E = 1.761·17.5/√15 = 7.957

Interval

23 ± 7.957 → [15.04, 30.96]

Answer

upper limit = 30.96
8.Confidence intervals

Answer: σ = 8.511 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 = 24·10.9² = 2851

Limits for σ²

2851/39.364 = 72.44
2851/12.401 = 229.9

Limit for σ

σ = √72.44 = 8.511 µm

Answer

σ = 8.511 µm

95% interval for σ: [8.511, 15.16] µm.

9.Hypothesis tests & confidence intervals

Answer: t = 1.2

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/√25 = 10/5 = 2

Test statistic

t = (77.4 − 75)/2 = 1.2

Answer

t = 1.2
t0.025; df = 24 = 2.064

|t| < 2.064, so H₀ is not rejected at the 5% level (two-sided).

10.Hypothesis tests & confidence intervals

Answer: 1 − β = 89.53 %

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 = 80 − 1.645·1.5 = 77.53

z under the true mean

z = (77.53 − 75.65)/1.5 = 1.255

Power

power = P(Z < 1.255) = 89.53%
1 − β = 89.53%

Answer

1 − β = 89.53%
11.Linear regression & correlation

Answer: r = 0.9775

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 = 138/6 = 23
ȳ = 619/6 = 103.2

Sums of squares and products

Sxx = Σ(x − x̄)² = 70
Sxy = Σ(x − x̄)(y − ȳ) = 359
Syy = Σ(y − ȳ)² = 1927

Correlation coefficient

r = 359/√(70·1927) = 0.9775

Answer

r = 0.9775
12.Linear regression & correlation

Answer: x = 10.07

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 = 35/5 = 7
ȳ = 438/5 = 87.6

Sums of squares and products

Sxx = Σ(x − x̄)² = 40
Sxy = Σ(x − x̄)(y − ȳ) = 266

Regression line

b = 266/40 = 6.65
a = 87.6 − 6.65·7 = 41.05
ŷ = 41.05 + 6.65·x

Solve for x

x = (108 − 41.05)/6.65 = 10.07

Answer

x = 10.07

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