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 = 0.9615

Drag z so that the shaded area between −z and z under the standard normal curve is 0.9615. Find z to two decimals.

A = 0.9615
z =
2.Normal distributionHard
157164171178185192199cmbottom 10%

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 10% is to the left of it. Find the cut-off x.

μ = 178 cmσ = 7 cm10%
x = cm
3.Confidence intervalsMedium
n = 30, s = 18LUx̄ = 39E = ?90%

A sample of 30 items has mean 39 and standard deviation s = 18. The population is roughly normal. Find the margin of error E of a 90% confidence interval for the mean.

n = 30x̄ = 39s = 1890%t(29) = 1.699
E =
4.Confidence intervalsHard
p ≈ 0.7, n = ?p̂ − Ep̂ + Ep̂E = 0.02599%

A pilot study suggests that about 0.7 of products are defective. How many products n must be inspected so that a 99% confidence interval for the proportion has a margin of error of at most 0.025?

p ≈ 0.7E = 0.02599%
n =
5.Hypothesis tests & confidence intervalsMedium
−1.961.96z = ?

Two production lines make the same part. Line 1: a sample of 60 parts has mean length 110.5 mm, and the line's known standard deviation is 16 mm. Line 2: a sample of 100 parts has mean 104.2 mm, known standard deviation 9 mm. Find the test statistic z for H₀: μ₁ = μ₂, taking the difference as x̄₁ − x̄₂ (line 1 minus line 2).

n₁ = 60x̄₁ = 110.5 mmσ₁ = 16 mmn₂ = 100x̄₂ = 104.2 mmσ₂ = 9 mm
z =
6.Hypothesis tests & confidence intervalsHard
−1.961.96z = ?

In a trial, 105 of 150 patients on treatment A recover, and 146 of 200 patients on treatment B. Using the pooled proportion, find the test statistic z for H₀: p₁ = p₂.

x₁ = 105n₁ = 150x₂ = 146n₂ = 200
z =
7.Linear regression & correlationMedium
xy150244355451560xy2445505560

The table gives 5 paired observations of x = hours studied and y = exam score. Find Pearson's correlation coefficient r.

n = 5
r =
8.Linear regression & correlationHard
xy145257356466566xy245060x = 2.5

The table gives 5 paired observations of x = hours studied and y = exam score. Use the least-squares line ŷ = a + b·x. Predict ŷ for x = 2.5.

n = 5x = 2.5
ŷ =
9.Chi-square testsMedium
coughno coughTotalsmokers433780non-smokers295180Total7288160

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 × 2
χ² =
10.Chi-square testsHard
passfailTotalline 1641680line 2691180line 373340Total14060200dfα = 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 = 2003 × 2α = 0.01
χ² =

3Solutions

1.Normal distribution

Answer: z = 2.07

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 + 0.9615)/2 = 0.9808

Read the z-table backwards

Find 0.9808 inside the table and read off z.

z = 2.07

Answer

z = 2.07
2.Normal distribution

Answer: x = 169 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) = 0.1

z from the table

z ≈ −1.28

Back to the original units

x = 178 − 1.28·7 = 169 cm

Answer

x = 169 cm
3.Confidence intervals

Answer: E = 5.583

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 = 29, t = 1.699

Margin of error

E = 1.699·18/√30 = 5.583

Answer

E = 5.583
4.Confidence intervals

Answer: n = 2230

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 = 2.576²·0.7·0.3/0.025² = 2230

Round up

n = 2230

Answer

n = 2230
5.Hypothesis tests & confidence intervals

Answer: z = 2.796

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.

Standard error of the difference

SE = √(16²/60 + 9²/100) = √5.077 = 2.253

Test statistic

z = (110.5 − 104.2)/2.253 = 2.796

Answer

z = 2.796

|z| > 1.96, so H₀ is rejected at the 5% level (two-sided).

6.Hypothesis tests & confidence intervals

Answer: z = −0.6167

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.

Sample proportions

p̂₁ = 105/150 = 0.7
p̂₂ = 146/200 = 0.73
p̂ = 251/350 = 0.7171

Standard error under H₀

SE = √(0.7171·0.2829·(1/150 + 1/200)) = 0.04865

Test statistic

z = (0.7 − 0.73)/0.04865 = −0.6167

Answer

z = −0.6167

|z| < 1.96, so H₀ is not rejected at the 5% level (two-sided).

7.Linear regression & correlation

Answer: r = 0.7165

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 = 260/5 = 52

Sums of squares and products

Sxx = Σ(x − x̄)² = 10
Sxy = Σ(x − x̄)(y − ȳ) = 27
Syy = Σ(y − ȳ)² = 142

Correlation coefficient

r = 27/√(10·142) = 0.7165

Answer

r = 0.7165
8.Linear regression & correlation

Answer: ŷ = 55.45

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 = 290/5 = 58

Sums of squares and products

Sxx = Σ(x − x̄)² = 10
Sxy = Σ(x − x̄)(y − ȳ) = 51

Regression line

b = 51/10 = 5.1
a = 58 − 5.1·3 = 42.7
ŷ = 42.7 + 5.1·x

Prediction

ŷ = 42.7 + 5.1·2.5 = 55.45

Answer

ŷ = 55.45
9.Chi-square tests

Answer: χ² = 4.949

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: 80·72/160 = 36, 80·88/160 = 44
non-smokers: 80·72/160 = 36, 80·88/160 = 44

Chi-square statistic

χ² = (43 − 36)²/36 + (37 − 44)²/44 + (29 − 36)²/36 + (51 − 44)²/44
χ² = 4.949

Answer

χ² = 4.949

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

10.Chi-square tests

Answer: χ² = 66.37

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

line 1: 56, 24
line 2: 56, 24
line 3: 28, 12

Chi-square statistic

χ² = (64 − 56)²/56 + (16 − 24)²/24 + (69 − 56)²/56 + (11 − 24)²/24 + (7 − 28)²/28 + (33 − 12)²/12
χ² = 66.37

Decision

df = 2, χ²crit = 9.210
66.37 > 9.210

Answer

χ² = 66.37

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

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