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 to the right of z = 0.8315

Drag z so that the shaded area under the standard normal curve, to the right of z, is 0.8315. Find z to two decimals.

A = 0.8315
z =
2.Normal distributionHard
μ = 500x = ?10%

The battery lifetimes are normally distributed with mean 500 h and standard deviation 40 h. Find the value x below which 10% of them lie (the 10th percentile).

μ = 500σ = 40P = 10%
x = h
3.Confidence intervalsMedium
x̄E₁ = 9.999%, n = 50x̄E₂ = ?90%, n = 50

A 99% confidence interval for a mean, based on 50 observations and a known σ, has a margin of error of 9.9. What is the margin of error E₂ of a 90% interval from the same data?

99%n₁ = 50E₁ = 9.990%
E₂ =
4.Confidence intervalsHard
σ = 39, n = ?x̄ − Ex̄ + Ex̄E = 299%

The population standard deviation is 39. How large a sample n do you need so that a 99% confidence interval for the mean has a margin of error of at most 2?

σ = 39E = 299%
n =
5.Hypothesis tests & confidence intervalsMedium
−2.0452.045t = ?df = 29

Two teaching methods are compared. Group 1: 14 students, mean score 28.5, standard deviation 2. Group 2: 17 students, mean score 27.8, standard deviation 2. 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 = 14x̄₁ = 28.5s₁ = 2n₂ = 17x̄₂ = 27.8s₂ = 2
t =
6.Hypothesis tests & confidence intervalsHard
σ = 10, n = ?x̄ − Ex̄ + Ex̄E = 2.595%

How large must a sample be so that the 95% confidence interval for a mean has a margin of error of at most 2.5? The standard deviation is 10. Find the smallest sample size n.

σ = 10E = 2.595%
n =
7.Linear regression & correlationMedium
xy130236344447548xy24304050

The table gives 5 paired observations of x = advertising spend in k€ and y = sales in k€. Their sums are given. Find Pearson's correlation coefficient r from these sums.

n = 5Σx = 15Σy = 205Σx² = 55Σy² = 8645Σxy = 662
r =
8.Linear regression & correlationHard
xy128225324424524623xy246242628x = 2.5

The table gives 6 paired observations of x = car age in years and y = price in k€. Use the least-squares line ŷ = a + b·x. Predict ŷ for x = 2.5.

n = 6x = 2.5
ŷ =
9.Chi-square testsMedium
no defectminormajorTotalday shift9284380night shift15362980Total246472160

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 = 1602 × 3
χ² =
10.Chi-square testsHard
yesnoTotalage 18–3078280age 31–50251540age 51+251540Total12832160dfα = 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 = 0.05 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 = 1603 × 2α = 0.05
χ² =

3Solutions

1.Normal distribution

Answer: z = −0.96

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 − 0.8315 = 0.1685

Read the z-table backwards

Find 0.1685 inside the table and read off z.

z = −0.96

Answer

z = −0.96
2.Normal distribution

Answer: x = 448.7 h

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) = 0.1 → z = −1.282

Back to x

x = 500 + −1.282·40 = 448.74 h

Answer

x = 448.74 h
3.Confidence intervals

Answer: E₂ = 6.322

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 = (1.645/2.576)·√(50/50) = 0.6386

New margin

E₂ = 9.9·0.6386 = 6.322

Answer

E₂ = 6.322
4.Confidence intervals

Answer: n = 2524

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 = (2.576·39/2)² = 2523

Round up

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

n = 2524

Answer

n = 2524
5.Hypothesis tests & confidence intervals

Answer: t = 0.9698

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 = (13·2² + 16·2²)/29 = 4
sp = 2

Standard error

SE = 2·√(1/14 + 1/17) = 0.7218

Test statistic

t = (28.5 − 27.8)/0.7218 = 0.9698

Answer

t = 0.9698

With df = 29, the two-sided 5% critical value is 2.045; |t| is smaller, so H₀ is not rejected.

6.Hypothesis tests & confidence intervals

Answer: n = 62

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 = (1.96·10/2.5)² = 61.47

Round up

n = 62

Answer

n = 62
7.Linear regression & correlation

Answer: r = 0.9594

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 = 662 − 15·205/5 = 47
Sxx = 55 − 15²/5 = 10
Syy = 8645 − 205²/5 = 240

Correlation coefficient

r = 47/√(10·240) = 0.9594

Answer

r = 0.9594
8.Linear regression & correlation

Answer: ŷ = 25.47

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 = 148/6 = 24.67

Sums of squares and products

Sxx = Σ(x − x̄)² = 17.5
Sxy = Σ(x − x̄)(y − ȳ) = −14

Regression line

b = −14/17.5 = −0.8
a = 24.67 − (−0.8)·3.5 = 27.47
ŷ = 27.47 − 0.8·x

Prediction

ŷ = 27.47 + (−0.8)·2.5 = 25.47

Answer

ŷ = 25.47
9.Chi-square tests

Answer: χ² = 5.222

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: 80·24/160 = 12, 80·64/160 = 32, 80·72/160 = 36
night shift: 80·24/160 = 12, 80·64/160 = 32, 80·72/160 = 36

Chi-square statistic

χ² = (9 − 12)²/12 + (28 − 32)²/32 + (43 − 36)²/36 + (15 − 12)²/12 + (36 − 32)²/32 + (29 − 36)²/36
χ² = 5.222

Answer

χ² = 5.222

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

10.Chi-square tests

Answer: χ² = 30.63

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

age 18–30: 64, 16
age 31–50: 32, 8
age 51+: 32, 8

Chi-square statistic

χ² = (78 − 64)²/64 + (2 − 16)²/16 + (25 − 32)²/32 + (15 − 8)²/8 + (25 − 32)²/32 + (15 − 8)²/8
χ² = 30.63

Decision

df = 2, χ²crit = 5.991
30.63 > 5.991

Answer

χ² = 30.63

χ² is above the critical value, so H₀ is rejected: the variables appear to be related.

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