Statistics 2: Statistical Inference: cram pack

Formula sheets, the key method steps for every topic and 10 practice problems, with full solutions at the back.

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1Formulas and methods

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

Chi-square tests

Test statistic
χ² = Σ(O − E)²/E
Goodness of fit
E = n·p, df = k − 1 − (estimated parameters)
Independence
E = row total × column total / grand total, df = (r − 1)(c − 1)
Critical values (α = 0.05)
df 1: 3.841, 2: 5.991, 3: 7.815, 4: 9.488
Decision
reject H0 if χ² > χ²crit

2Practice problems

1.Normal distributionMedium
32 min29.5 minX + Y?

X is normal with mean 12 min and standard deviation 1.5 min; Y is independent and normal with mean 20 min and standard deviation 2 min. Find the probability P that X + Y is more than 29.5 min. Round z to two decimals.

μX = 12, σX = 1.5μY = 20, σY = 2X + Y > 29.5 min
P =
2.Normal distributionHard
10 cm0X − Y?

The heights X of men and Y of women in a group are independent and normal: X has mean 180 cm and standard deviation 10 cm, Y has mean 170 cm and standard deviation 12 cm. Find the probability P that X is greater than Y. Round z to two decimals.

μX = 180 cm, σX = 10 cmμY = 170 cm, σY = 12 cm
P =
3.Confidence intervalsMedium
n = 13, s = 15LUx̄ = 65E = ?95%

A sample of 13 items has mean 65 and standard deviation s = 15. The population is roughly normal. Find the margin of error E of a 95% confidence interval for the mean.

n = 13x̄ = 65s = 1595%t(12) = 2.179
E =
4.Confidence intervalsHard
σ = 23, n = ?x̄ − Ex̄ + Ex̄E = 595%

The population standard deviation is 23. How large a sample n do you need so that a 95% confidence interval for the mean has a margin of error of at most 5?

σ = 23E = 595%
n =
5.Hypothesis tests & confidence intervalsMedium
−2.1792.179t = ?df = 12

Two teaching methods are compared. Group 1: 8 students, mean score 96, standard deviation 5. Group 2: 6 students, mean score 88.7, standard deviation 5. Assume equal population variances and use the pooled two-sample t test. Find the test statistic t for H₀: μ₁ = μ₂, taking the difference as x̄₁ − x̄₂ (group 1 minus group 2).

n₁ = 8x̄₁ = 96s₁ = 5n₂ = 6x̄₂ = 88.7s₂ = 5
t =
6.Hypothesis tests & confidence intervalsHard
H₀H₁μ₀x̄cμ₀ + 4α = 0.01βσ = 21, power 80%, n = ?

A one-sided z test at α = 0.01 must detect a shift of the mean of 4 units with a power of 80%. The standard deviation is 21. Find the minimum sample size n.

α = 0.01power = 80%δ = 4σ = 21zα = 2.326zβ = 0.842
n =
7.Linear regression & correlationMedium
xy294186238281036xy510102030

The table gives 5 paired observations of x = load in kN and y = extension in mm. Find Pearson's correlation coefficient r.

n = 5
r =
8.Linear regression & correlationHard
xy1544186021702478279430102xy15202530406080100ŷ = 63

The table gives 6 paired observations of x = temperature in °C and y = ice creams sold. Use the least-squares line ŷ = a + b·x. For which value x does the line predict ŷ = 63?

n = 6ŷ = 63
x =
9.Chi-square testsMedium
no defectminormajorTotalday shift11141540night shift19162560Total303040100

The contingency table shows observed counts. Compute the chi-square statistic χ² for the test of independence between the row and column variables. Do not use a continuity correction.

N = 1002 × 3
χ² =
10.Chi-square testsHard

A chi-square goodness-of-fit test compares observed counts in 7 categories with fully specified expected proportions (no parameters are estimated from the data). How many degrees of freedom df does the test have?

k = 7
df =

3Solutions

1.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 sum

μ = 32 min, σ = √(1.5² + 2²) = 2.5 min

Standardise

z = (29.5 − 32)/2.5 = −1

Upper tail

P = 1 − Φ(−1) = 1 − 0.1587 = 0.8413

Answer

P = 0.8413
2.Normal distribution

Answer: P = 0.7389

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 = 180 − 170 = 10 cm, σ = 15.62 cm

Standardise

z = (0 − 10)/15.62 = −0.64

Probability

P = 1 − Φ(−0.64) = 0.7389

Answer

P = 0.7389
3.Confidence intervals

Answer: E = 9.065

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 = 12, t = 2.179

Margin of error

E = 2.179·15/√13 = 9.065

Answer

E = 9.065
4.Confidence intervals

Answer: n = 82

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.

Solve for n

n = (1.96·23/5)² = 81.29

Round up

A sample size is a whole number, and rounding down would make the margin too large.

n = 82

Answer

n = 82
5.Hypothesis tests & confidence intervals

Answer: t = 2.703

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.

Pooled variance

sp² = (7·5² + 5·5²)/12 = 25
sp = 5

Standard error

SE = 5·√(1/8 + 1/6) = 2.7

Test statistic

t = (96 − 88.7)/2.7 = 2.703

Answer

t = 2.703

With df = 12, the two-sided 5% critical value is 2.179; |t| is larger, so H₀ is rejected.

6.Hypothesis tests & confidence intervals

Answer: n = 277

Sample size for a test: to detect a true shift δ with power 1 − β in a one-sided z test at level α, the standard error must satisfy δ = (z_α + z_β)·σ/√n, so n = ((z_α + z_β)·σ/δ)², rounded up.

Solve for n

n = ((2.326 + 0.842)·21/4)² = 276.6

Round up

n = 277

Answer

n = 277
7.Linear regression & correlation

Answer: r = 0.9937

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 = 114/5 = 22.8

Sums of squares and products

Sxx = Σ(x − x̄)² = 40
Sxy = Σ(x − x̄)(y − ȳ) = 128
Syy = Σ(y − ȳ)² = 414.8

Correlation coefficient

r = 128/√(40·414.8) = 0.9937

Answer

r = 0.9937
8.Linear regression & correlation

Answer: x = 19.44

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 = 135/6 = 22.5
ȳ = 448/6 = 74.67

Sums of squares and products

Sxx = Σ(x − x̄)² = 157.5
Sxy = Σ(x − x̄)(y − ȳ) = 600

Regression line

b = 600/157.5 = 3.81
a = 74.67 − 3.81·22.5 = −11.05
ŷ = −11.05 + 3.81·x

Solve for x

x = (63 − (−11.05))/3.81 = 19.44

Answer

x = 19.44
9.Chi-square tests

Answer: χ² = 0.7986

Chi-square tests: compare observed counts O with the counts E expected under H₀ using χ² = Σ(O − E)²/E. Goodness of fit: E = n·p for each category, df = k − 1 (minus one more for each parameter estimated from the data). Independence in an r × c table: E = row total × column total / grand total, df = (r − 1)(c − 1). Reject H₀ when χ² exceeds the critical value for the chosen α.

Expected counts

day shift: 40·30/100 = 12, 40·30/100 = 12, 40·40/100 = 16
night shift: 60·30/100 = 18, 60·30/100 = 18, 60·40/100 = 24

Chi-square statistic

χ² = (11 − 12)²/12 + (14 − 12)²/12 + (15 − 16)²/16 + (19 − 18)²/18 + (16 − 18)²/18 + (25 − 24)²/24
χ² = 0.7986

Answer

χ² = 0.7986

df = 2. At α = 0.05 the critical value is 5.991, so H₀ (independence) is not rejected.

10.Chi-square tests

Answer: df = 6

Chi-square tests: compare observed counts O with the counts E expected under H₀ using χ² = Σ(O − E)²/E. Goodness of fit: E = n·p for each category, df = k − 1 (minus one more for each parameter estimated from the data). Independence in an r × c table: E = row total × column total / grand total, df = (r − 1)(c − 1). Reject H₀ when χ² exceeds the critical value for the chosen α.

Degrees of freedom

df = 7 − 1 = 6

Answer

df = 6

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