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
−3−2−10123zshaded area between −z and z = 0.4448

Drag z so that the shaded area between −z and z under the standard normal curve is 0.4448. Find z to two decimals.

A = 0.4448
z =
2.Normal distributionHard
157164171178185192199cmbottom 60%

The heights of adult men are normally distributed with mean 178 cm and standard deviation 7 cm. Drag the cut-off so the shaded bottom 60% is to the left of it. Find the cut-off x.

μ = 178 cmσ = 7 cm60%
x = cm
3.Confidence intervalsMedium
x̄E₁ = 8.595%, n = 64x̄E₂ = ?95%, n = 256

A 95% confidence interval for a mean, based on 64 observations and a known σ, has a margin of error of 8.5. What is the margin of error E₂ if the same study uses 256 observations instead?

95%n₁ = 64E₁ = 8.5n₂ = 256
E₂ =
4.Confidence intervalsHard
μ₁ − μ₂, t(19) = 1.7290LUx̄₁ − x̄₂E = ?90%

Two fertilisers are tested. Plot group 1: 10 plots, mean yield 23 kg, standard deviation 3.5 kg. Plot group 2: 11 plots, mean 22 kg, standard deviation 1.5 kg. Assume equal variances (pooled t, df = n₁ + n₂ − 2). Find the margin of error E of the 90% confidence interval for μ₁ − μ₂.

n₁ = 10x̄₁ = 23s₁ = 3.5n₂ = 11x̄₂ = 22s₂ = 1.590%t(19) = 1.729
E = kg
5.Hypothesis tests & confidence intervalsMedium
t −1.98401.984

A sample of 100 students has a mean score of 119.4 and a sample standard deviation of 4. Find the test statistic t for H₀: μ = 120.

μ₀ = 120x̄ = 119.4s = 4n = 100
t =
6.Hypothesis tests & confidence intervalsHard
σ = 20, n = ?x̄ − Ex̄ + Ex̄E = 195%

How large must a sample be so that the 95% confidence interval for a mean has a margin of error of at most 1? The standard deviation is 20. Find the smallest sample size n.

σ = 20E = 195%
n =
7.Linear regression & correlationMedium
xy16461950226225722878xy202550607080

The table gives 5 paired observations of x = temperature in °C and y = ice creams sold. Find the coefficient of determination r² of the least-squares line.

n = 5
r² =
8.Linear regression & correlationHard
xy2245688131018xy51051015ŷ = 3

The table gives 5 paired observations of x = load in kN and y = extension in mm. Use the least-squares line ŷ = a + b·x. For which value x does the line predict ŷ = 3?

n = 5ŷ = 3
x =
9.Chi-square testsMedium
coughno coughTotalsmokers186280non-smokers243660Total4298140

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 = 1402 × 2
χ² =
10.Chi-square testsHard

A chi-square goodness-of-fit test checks whether counts in 8 categories follow a Poisson distribution whose mean is estimated from the same data. How many degrees of freedom df does the test have?

k = 8
df =

3Solutions

1.Normal distribution

Answer: z = 0.59

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 + 0.4448)/2 = 0.7224

Read the z-table backwards

Find 0.7224 inside the table and read off z.

z = 0.59

Answer

z = 0.59
2.Normal distribution

Answer: x = 179.8 cm

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) = 0.6

z from the table

z ≈ 0.25

Back to the original units

x = 178 + 0.25·7 = 179.8 cm

Answer

x = 179.8 cm
3.Confidence intervals

Answer: E₂ = 4.25

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.

Ratio of margins

E₂/E₁ = (1.96/1.96)·√(64/256) = 0.5

New margin

E₂ = 8.5·0.5 = 4.25

Answer

E₂ = 4.25
4.Confidence intervals

Answer: E = 1.997 kg

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.

Pooled standard deviation

sp² = (9·3.5² + 10·1.5²)/19 = 6.987
sp = 2.643

Margin of error

E = 1.729·2.643·√(1/10 + 1/11) = 1.997

Answer

E = 1.997 kg
5.Hypothesis tests & confidence intervals

Answer: t = −1.5

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 = 4/√100 = 4/10 = 0.4

Test statistic

t = (119.4 − 120)/0.4 = −1.5

Answer

t = −1.5
t0.025; df = 99 = 1.984

|t| < 1.984, so H₀ is not rejected at the 5% level (two-sided).

6.Hypothesis tests & confidence intervals

Answer: n = 1537

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).

Solve for n

n = (1.96·20/1)² = 1537

Round up

n = 1537

Answer

n = 1537
7.Linear regression & correlation

Answer: r² = 0.9793

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 = 110/5 = 22
ȳ = 308/5 = 61.6

Sums of squares and products

Sxx = Σ(x − x̄)² = 90
Sxy = Σ(x − x̄)(y − ȳ) = 258
Syy = Σ(y − ȳ)² = 755.2

Coefficient of determination

r² = 258²/(90·755.2) = 0.9793

Answer

r² = 0.9793

About 97.93% of the variation in y is explained by the line.

8.Linear regression & correlation

Answer: x = 2.9

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 = 46/5 = 9.2

Sums of squares and products

Sxx = Σ(x − x̄)² = 40
Sxy = Σ(x − x̄)(y − ȳ) = 80

Regression line

b = 80/40 = 2
a = 9.2 − 2·6 = −2.8
ŷ = −2.8 + 2·x

Solve for x

x = (3 − (−2.8))/2 = 2.9

Answer

x = 2.9
9.Chi-square tests

Answer: χ² = 5

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

smokers: 80·42/140 = 24, 80·98/140 = 56
non-smokers: 60·42/140 = 18, 60·98/140 = 42

Chi-square statistic

χ² = (18 − 24)²/24 + (62 − 56)²/56 + (24 − 18)²/18 + (36 − 42)²/42
χ² = 5

Answer

χ² = 5

df = 1. At α = 0.05 the critical value is 3.841, so H₀ (independence) is 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 = 8 − 1 − 1 = 6

Answer

df = 6

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