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
750 g769.7 gX + Y?

X is normal with mean 250 g and standard deviation 4 g; Y is independent and normal with mean 500 g and standard deviation 9 g. Find the probability P that X + Y is more than 769.7 g. Round z to two decimals.

μX = 250, σX = 4μY = 500, σY = 9X + Y > 769.7 g
P =
2.Normal distributionHard
μ = 250x = ?80%

The bag weights are normally distributed with mean 250 g and standard deviation 5 g. Find the value x below which 80% of them lie (the 80th percentile).

μ = 250σ = 5P = 80%
x = g
3.Confidence intervalsMedium
μ₁ − μ₂0L = ?Ux̄₁ − x̄₂E99%

The battery life of two phone models is compared. Model 1: 100 phones, mean 94.5 h, known standard deviation 4 h. Model 2: 40 phones, mean 81 h, known standard deviation 19 h. Find the lower limit of the 99% confidence interval for μ₁ − μ₂.

n₁ = 100x̄₁ = 94.5 hσ₁ = 4 hn₂ = 40x̄₂ = 81 hσ₂ = 19 h99%
lower limit = h
4.Confidence intervalsHard
p = 0.5, n = ?p̂ − Ep̂ + Ep̂E = 0.0390%

Nothing is known in advance about the proportion of defective products. How many products n must be inspected so that a 90% confidence interval for the proportion has a margin of error of at most 0.03?

p = 0.5E = 0.0390%
n =
5.Hypothesis tests & confidence intervalsMedium
t −2.13102.131

A sample of 16 students has a mean score of 116.6 and a sample standard deviation of 15. Find the test statistic t for H₀: μ = 120.

μ₀ = 120x̄ = 116.6s = 15n = 16
t =
6.Hypothesis tests & confidence intervalsHard
z −1.9601.96

A sample of 16 measurements has mean 43.5. The population standard deviation is 20. Find the lower limit L of the 95% confidence interval for the mean.

x̄ = 43.5σ = 20n = 1695%
L =
7.Linear regression & correlationMedium
xy143346555764965xy55060

The table gives 5 paired observations of x = hours studied and y = exam score. Their sums are given. Find Pearson's correlation coefficient r from these sums.

n = 5Σx = 25Σy = 273Σx² = 165Σy² = 15311Σxy = 1489
r =
8.Linear regression & correlationHard
xy254106188221030xy510102030ŷ = 26

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 = 26?

n = 5ŷ = 26
x =
9.Chi-square testsMedium
lowmediumhighTotalurban222355100rural183745100Total4060100200

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
yesnoTotalplant A202040plant B285280plant C243660Total72108180dfα = 0.05α = 0.0113.8416.63525.9919.21037.81511.34549.48813.277511.07015.086612.59216.812714.06718.475815.50720.090916.91921.6661018.30723.209

The table shows observed counts. Test at α = 0.01 whether the two variables are independent (no continuity correction). Compute the test statistic χ², then compare it with the critical value from the table.

N = 1803 × 2α = 0.01
χ² =

3Solutions

1.Normal distribution

Answer: P = 0.02275

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 = 750 g, σ = √(4² + 9²) = 9.849 g

Standardise

z = (769.7 − 750)/9.849 = 2

Upper tail

P = 1 − Φ(2) = 1 − 0.9772 = 0.02275

Answer

P = 0.02275
2.Normal distribution

Answer: x = 254.2 g

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 for the percentile

Φ(z) = 0.8 → z = 0.8416

Back to x

x = 250 + 0.8416·5 = 254.21 g

Answer

x = 254.21 g
3.Confidence intervals

Answer: lower limit = 5.693 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 = 94.5 − 81 = 13.5

Standard error

SE = √(4²/100 + 19²/40) = 3.031

Margin of error

E = 2.576·3.031 = 7.807

Interval

13.5 ± 7.807 → [5.693, 21.31]

Answer

lower limit = 5.693 h

The interval does not contain 0, so the means differ significantly.

4.Confidence intervals

Answer: n = 752

Confidence interval for a proportion: p̂ ± z·√(p̂(1 − p̂)/n), where p̂ = x/n. To reach a margin E you need n ≥ z²·p(1 − p)/E², rounded up; without a prior estimate use p = 0.5, which gives the largest n.

Solve for n

n = 1.645²·0.5·0.5/0.03² = 751.7

Round up

n = 752

Answer

n = 752
5.Hypothesis tests & confidence intervals

Answer: t = −0.9067

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 = 15/√16 = 15/4 = 3.75

Test statistic

t = (116.6 − 120)/3.75 = −0.9067

Answer

t = −0.9067
t0.025; df = 15 = 2.131

|t| < 2.131, so H₀ is not rejected at the 5% level (two-sided).

6.Hypothesis tests & confidence intervals

Answer: L = 33.7

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 = 1.96·20/√16 = 9.8

Interval limit

L = 43.5 − 9.8 = 33.7

Answer

L = 33.7
7.Linear regression & correlation

Answer: r = 0.974

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 = 1489 − 25·273/5 = 124
Sxx = 165 − 25²/5 = 40
Syy = 15311 − 273²/5 = 405.2

Correlation coefficient

r = 124/√(40·405.2) = 0.974

Answer

r = 0.974
8.Linear regression & correlation

Answer: x = 8.903

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 = 85/5 = 17

Sums of squares and products

Sxx = Σ(x − x̄)² = 40
Sxy = Σ(x − x̄)(y − ȳ) = 124

Regression line

b = 124/40 = 3.1
a = 17 − 3.1·6 = −1.6
ŷ = −1.6 + 3.1·x

Solve for x

x = (26 − (−1.6))/3.1 = 8.903

Answer

x = 8.903
9.Chi-square tests

Answer: χ² = 4.667

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: 100·40/200 = 20, 100·60/200 = 30, 100·100/200 = 50
rural: 100·40/200 = 20, 100·60/200 = 30, 100·100/200 = 50

Chi-square statistic

χ² = (22 − 20)²/20 + (23 − 30)²/30 + (55 − 50)²/50 + (18 − 20)²/20 + (37 − 30)²/30 + (45 − 50)²/50
χ² = 4.667

Answer

χ² = 4.667

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

10.Chi-square tests

Answer: χ² = 2.5

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

plant A: 16, 24
plant B: 32, 48
plant C: 24, 36

Chi-square statistic

χ² = (20 − 16)²/16 + (20 − 24)²/24 + (28 − 32)²/32 + (52 − 48)²/48 + (24 − 24)²/24 + (36 − 36)²/36
χ² = 2.5

Decision

df = 2, χ²crit = 9.210
2.5 < 9.210

Answer

χ² = 2.5

χ² is below the critical value, so H₀ is not rejected: there is no significant evidence of a relationship.

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