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
175.9188.85P = ?

The heights of adult men are normally distributed with mean 178 cm and standard deviation 7 cm. What fraction lies between 175.9 and 188.85 cm? Give P as a decimal.

μ = 178σ = 7x1 = 175.9x2 = 188.85
P =
2.Normal distributionHard
4 s0X − Y?

Runner A's lap time X and runner B's lap time Y are independent and normal: X has mean 52 s and standard deviation 4 s, Y has mean 48 s and standard deviation 3 s. Find the probability P that X is greater than Y. Round z to two decimals.

μX = 52 s, σX = 4 sμY = 48 s, σY = 3 s
P =
3.Confidence intervalsMedium
n = 10, s = 19L = ?Ux̄ = 93E90%

A sample of 10 items has mean 93 and standard deviation s = 19. The population is roughly normal. Find the lower limit of the 90% confidence interval for the mean.

n = 10x̄ = 93s = 1990%t(9) = 1.833
lower limit =
4.Confidence intervalsHard
p ≈ 0.6, n = ?p̂ − Ep̂ + Ep̂E = 0.0395%

A pilot study suggests that about 0.6 of products are defective. How many products n must be inspected so that a 95% confidence interval for the proportion has a margin of error of at most 0.03?

p ≈ 0.6E = 0.0395%
n =
5.Hypothesis tests & confidence intervalsMedium
−2.5712.571t = ?df = 5

6 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 −8.8 and the standard deviation of the differences is 9.8. Find the paired test statistic t for H₀: μd = 0.

n = 6d̄ = −8.8sd = 9.8
t =
6.Hypothesis tests & confidence intervalsHard
σ = 8, 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 8. Find the smallest sample size n.

σ = 8E = 195%
n =
7.Linear regression & correlationMedium
xy127221319419518611xy24610152025

The table gives 6 paired observations of x = car age in years and y = price in k€. Find Pearson's correlation coefficient r.

n = 6
r =
8.Linear regression & correlationHard
xy148250356454556666xy2465060x = 3.5

The table gives 6 paired observations of x = hours studied and y = exam score. Use the least-squares line ŷ = a + b·x. A new observation (3.5, 58) is made. Find its residual e = y − ŷ.

n = 6(3.5, 58)
e =
9.Chi-square testsMedium
lowmediumhighTotalurban20142862rural34173586Total543163148

The contingency table shows observed counts. For a chi-square test of independence, find the expected count E for the cell “rural, low”.

row total = 86column total = 54N = 148
E =
10.Chi-square testsHard

A chi-square goodness-of-fit test compares observed counts in 10 categories with fully specified expected proportions (no parameters are estimated from the data). How many degrees of freedom df does the test have?

k = 10
df =

3Solutions

1.Normal distribution

Answer: P = 0.5573

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 = (175.9 − 178)/7 = −0.3
z₂ = (188.85 − 178)/7 = 1.55

Area between

P = Φ(1.55) − Φ(−0.3) = 0.9394 − 0.3821 = 0.5573

Answer

P = 0.5573
2.Normal distribution

Answer: P = 0.7881

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 = 52 − 48 = 4 s, σ = 5 s

Standardise

z = (0 − 4)/5 = −0.8

Probability

P = 1 − Φ(−0.8) = 0.7881

Answer

P = 0.7881
3.Confidence intervals

Answer: lower limit = 81.99

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 = 9, t = 1.833

Margin of error

E = 1.833·19/√10 = 11.01

Interval

93 ± 11.01 → [81.99, 104]

Answer

lower limit = 81.99
4.Confidence intervals

Answer: n = 1025

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 = 1.96²·0.6·0.4/0.03² = 1024

Round up

n = 1025

Answer

n = 1025
5.Hypothesis tests & confidence intervals

Answer: t = −2.2

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 of the mean difference

SE = 9.8/√6 = 4.001

Test statistic

t = −8.8/4.001 = −2.2

Answer

t = −2.2

With df = 5, the two-sided 5% critical value is 2.571; |t| is smaller, so H₀ is not rejected.

6.Hypothesis tests & confidence intervals

Answer: n = 246

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 = (1.96·8/1)² = 245.9

Round up

n = 246

Answer

n = 246
7.Linear regression & correlation

Answer: r = −0.923

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 = 21/6 = 3.5
ȳ = 115/6 = 19.17

Sums of squares and products

Sxx = Σ(x − x̄)² = 17.5
Sxy = Σ(x − x̄)(y − ȳ) = −44.5
Syy = Σ(y − ȳ)² = 132.8

Correlation coefficient

r = −44.5/√(17.5·132.8) = −0.923

Answer

r = −0.923
8.Linear regression & correlation

Answer: e = 3

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 = 21/6 = 3.5
ȳ = 330/6 = 55

Sums of squares and products

Sxx = Σ(x − x̄)² = 17.5
Sxy = Σ(x − x̄)(y − ȳ) = 53

Regression line

b = 53/17.5 = 3.029
a = 55 − 3.029·3.5 = 44.4
ŷ = 44.4 + 3.029·x

Predicted value

ŷ = 44.4 + 3.029·3.5 = 55

Residual

e = 58 − 55 = 3

Answer

e = 3

The point lies above the line.

9.Chi-square tests

Answer: E = 31.38

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 count

E = 86·54/148 = 31.38

Answer

E = 31.38
10.Chi-square tests

Answer: df = 9

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 = 10 − 1 = 9

Answer

df = 9

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