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
22.738P = ?

The delivery times are normally distributed with mean 32 min and standard deviation 6 min. What fraction lies between 22.7 and 38 min? Give P as a decimal.

μ = 32σ = 6x1 = 22.7x2 = 38
P =
2.Normal distributionHard
235240245250255260265gtop 80%

The bag weights are normally distributed with mean 250 g and standard deviation 5 g. Drag the cut-off so the shaded top 80% is to the right of it. Find the cut-off x.

μ = 250 gσ = 5 g80%
x = g
3.Confidence intervalsMedium
n = 20, s = 11.5LU = ?x̄ = 39E90%

A sample of 20 items has mean 39 and standard deviation s = 11.5. The population is roughly normal. Find the upper limit of the 90% confidence interval for the mean.

n = 20x̄ = 39s = 11.590%t(19) = 1.729
upper limit =
4.Confidence intervalsHard
μ₁ − μ₂, t(19) = 1.7290LU = ?x̄₁ − x̄₂E90%

Two fertilisers are tested. Plot group 1: 16 plots, mean yield 28.5 kg, standard deviation 2.5 kg. Plot group 2: 5 plots, mean 34.5 kg, standard deviation 10 kg. Assume equal variances (pooled t, df = n₁ + n₂ − 2). Find the upper limit of the 90% confidence interval for μ₁ − μ₂.

n₁ = 16x̄₁ = 28.5s₁ = 2.5n₂ = 5x̄₂ = 34.5s₂ = 1090%t(19) = 1.729
upper limit = kg
5.Hypothesis tests & confidence intervalsMedium
−2.2012.201t = ?df = 11

12 employees take a typing test before and after a training course. For each person the difference d = after − before (words per minute) is recorded. The mean difference is 8.57 and the standard deviation of the differences is 9.9. Find the paired test statistic t for H₀: μd = 0.

n = 12d̄ = 8.57sd = 9.9
t =
6.Hypothesis tests & confidence intervalsHard
z −1.9601.96

A sample of 36 measurements has mean 101.5. The population standard deviation is 4. Find the lower limit L of the 95% confidence interval for the mean.

x̄ = 101.5σ = 4n = 3695%
L =
7.Linear regression & correlationMedium
xy125224326422521620xy24620222426

The table gives 6 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 = 6Σx = 21Σy = 138Σx² = 91Σy² = 3202Σxy = 464
r =
8.Linear regression & correlationHard
xy130223321421516xy2415202530ŷ = 24

The table gives 5 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 = 24?

n = 5ŷ = 24
x =
9.Chi-square testsMedium
lowmediumhighTotalurban473340120rural13274080Total606080200

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 = 2002 × 3
χ² =
10.Chi-square testsHard
dfα = 0.05α = 0.0113.8416.63525.9919.21037.81511.34549.48813.277511.07015.086612.59216.812714.06718.475815.50720.090916.91921.6661018.30723.209

A chi-square test of independence is carried out on a contingency table with 3 rows and 4 columns (not counting the totals). Using the table, find the critical value χ²crit at significance level α = 0.01.

α = 0.01
χ²crit =

3Solutions

1.Normal distribution

Answer: P = 0.7808

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-scores

z₁ = (22.7 − 32)/6 = −1.55
z₂ = (38 − 32)/6 = 1

Area between

P = Φ(1) − Φ(−1.55) = 0.8413 − 0.06057 = 0.7808

Answer

P = 0.7808
2.Normal distribution

Answer: x = 245.8 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) = 0.2

z from the table

z ≈ −0.84

Back to the original units

x = 250 − 0.84·5 = 245.8 g

Answer

x = 245.8 g
3.Confidence intervals

Answer: upper limit = 43.45

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 = 19, t = 1.729

Margin of error

E = 1.729·11.5/√20 = 4.446

Interval

39 ± 4.446 → [34.55, 43.45]

Answer

upper limit = 43.45
4.Confidence intervals

Answer: upper limit = −1.484 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 = (15·2.5² + 4·10²)/19 = 25.99
sp = 5.098

Margin of error

E = 1.729·5.098·√(1/16 + 1/5) = 4.516

Interval

−6 ± 4.516 → [−10.52, −1.484]

Answer

upper limit = −1.484 kg
5.Hypothesis tests & confidence intervals

Answer: t = 2.999

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 of the mean difference

SE = 9.9/√12 = 2.858

Test statistic

t = 8.57/2.858 = 2.999

Answer

t = 2.999

With df = 11, the two-sided 5% critical value is 2.201; |t| is larger, so H₀ is rejected.

6.Hypothesis tests & confidence intervals

Answer: L = 100.2

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).

Margin of error

E = 1.96·4/√36 = 1.307

Interval limit

L = 101.5 − 1.307 = 100.2

Answer

L = 100.2
7.Linear regression & correlation

Answer: r = −0.8583

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 = 464 − 21·138/6 = −19
Sxx = 91 − 21²/6 = 17.5
Syy = 3202 − 138²/6 = 28

Correlation coefficient

r = −19/√(17.5·28) = −0.8583

Answer

r = −0.8583
8.Linear regression & correlation

Answer: x = 2.4

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 = 15/5 = 3
ȳ = 111/5 = 22.2

Sums of squares and products

Sxx = Σ(x − x̄)² = 10
Sxy = Σ(x − x̄)(y − ȳ) = −30

Regression line

b = −30/10 = −3
a = 22.2 − (−3)·3 = 31.2
ŷ = 31.2 − 3·x

Solve for x

x = (24 − 31.2)/(−3) = 2.4

Answer

x = 2.4
9.Chi-square tests

Answer: χ² = 12.36

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

urban: 120·60/200 = 36, 120·60/200 = 36, 120·80/200 = 48
rural: 80·60/200 = 24, 80·60/200 = 24, 80·80/200 = 32

Chi-square statistic

χ² = (47 − 36)²/36 + (33 − 36)²/36 + (40 − 48)²/48 + (13 − 24)²/24 + (27 − 24)²/24 + (40 − 32)²/32
χ² = 12.36

Answer

χ² = 12.36

df = 2. At α = 0.05 the critical value is 5.991, so H₀ (independence) is rejected.

10.Chi-square tests

Answer: χ²crit = 16.81

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 = (3 − 1)·(4 − 1) = 6

Table

χ²crit(df = 6, α = 0.01) = 16.812

Answer

χ²crit = 16.812

Reject H₀ if the computed χ² is larger than this value.

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