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 = 0.1841

Drag z so that the shaded area under the standard normal curve, to the right of z, is 0.1841. Find z to two decimals.

A = 0.1841
z =
2.Normal distributionHard
μ = 500x = ?20%

The battery lifetimes are normally distributed with mean 500 h and standard deviation 40 h. Find the value x below which 20% of them lie (the 20th percentile).

μ = 500σ = 40P = 20%
x = h
3.Confidence intervalsMedium
n = 20, sd​ = 7.30L = ?Ud̄ = 9.7E90%

The blood pressure of 20 patients is measured before and after a treatment. The differences d = before − after (mmHg) have mean 9.7 and standard deviation 7.3. Find the lower limit of the 90% confidence interval for the mean difference.

n = 20d̄ = 9.7sd = 7.390%t(19) = 1.729
lower limit = mmHg
4.Confidence intervalsHard
σ = 8, n = ?x̄ − Ex̄ + Ex̄E = 195%

The population standard deviation is 8. How large a sample n do you need so that a 95% confidence interval for the mean has a margin of error of at most 1?

σ = 8E = 195%
n =
5.Hypothesis tests & confidence intervalsMedium
t −2.13102.131

A sample of 16 students has a mean score of 112.3 and a sample standard deviation of 10. Find the test statistic t for H₀: μ = 120.

μ₀ = 120x̄ = 112.3s = 10n = 16
t =
6.Hypothesis tests & confidence intervalsHard
H₀H₁μ₀x̄cμ₀ + 4α = 0.05βσ = 29, power 95%, n = ?

A one-sided z test at α = 0.05 must detect a shift of the mean of 4 units with a power of 95%. The standard deviation is 29. Find the minimum sample size n.

α = 0.05power = 95%δ = 4σ = 29zα = 1.645zβ = 1.645
n =
7.Linear regression & correlationMedium
xy3415497579711181xy510406080

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 = 35Σy = 299Σx² = 285Σy² = 18933Σxy = 2297
r =
8.Linear regression & correlationHard
xy18852111024115271323014933173xy202530100150x = 22.5

The table gives 6 paired observations of x = temperature in °C and y = ice creams sold. Use the least-squares line ŷ = a + b·x. Predict ŷ for x = 22.5.

n = 6x = 22.5
ŷ =
9.Chi-square testsMedium
yesnoTotalmen7624100women204060Total9664160

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
yesnoTotalline 1162440line 2154560line 3112940Total4298140dfα = 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 = 1403 × 2α = 0.05
χ² =

3Solutions

1.Normal distribution

Answer: z = 0.9

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 − 0.1841 = 0.8159

Read the z-table backwards

Find 0.8159 inside the table and read off z.

z = 0.90

Answer

z = 0.90
2.Normal distribution

Answer: x = 466.3 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) = 0.2 → z = −0.8416

Back to x

x = 500 + −0.8416·40 = 466.34 h

Answer

x = 466.34 h
3.Confidence intervals

Answer: lower limit = 6.878 mmHg

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.

Critical value

df = 19, t = 1.729

Margin of error

E = 1.729·7.3/√20 = 2.822

Interval

9.7 ± 2.822 → [6.878, 12.52]

Answer

lower limit = 6.878 mmHg
4.Confidence intervals

Answer: n = 246

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 = (1.96·8/1)² = 245.9

Round up

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

n = 246

Answer

n = 246
5.Hypothesis tests & confidence intervals

Answer: t = −3.08

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 = 10/√16 = 10/4 = 2.5

Test statistic

t = (112.3 − 120)/2.5 = −3.08

Answer

t = −3.08
t0.025; df = 15 = 2.131

|t| > 2.131, so H₀ is rejected at the 5% level (two-sided).

6.Hypothesis tests & confidence intervals

Answer: n = 569

Sample size for a test: to detect a true shift δ with power 1 − β in a one-sided z test at level α, the standard error must satisfy δ = (z_α + z_β)·σ/√n, so n = ((z_α + z_β)·σ/δ)², rounded up.

Solve for n

n = ((1.645 + 1.645)·29/4)² = 568.9

Round up

n = 569

Answer

n = 569
7.Linear regression & correlation

Answer: r = 0.9941

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 = 2297 − 35·299/5 = 204
Sxx = 285 − 35²/5 = 40
Syy = 18933 − 299²/5 = 1053

Correlation coefficient

r = 204/√(40·1053) = 0.9941

Answer

r = 0.9941
8.Linear regression & correlation

Answer: ŷ = 110.9

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 = 153/6 = 25.5
ȳ = 764/6 = 127.3

Sums of squares and products

Sxx = Σ(x − x̄)² = 157.5
Sxy = Σ(x − x̄)(y − ȳ) = 861

Regression line

b = 861/157.5 = 5.467
a = 127.3 − 5.467·25.5 = −12.07
ŷ = −12.07 + 5.467·x

Prediction

ŷ = −12.07 + 5.467·22.5 = 110.9

Answer

ŷ = 110.9
9.Chi-square tests

Answer: χ² = 28.44

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

men: 100·96/160 = 60, 100·64/160 = 40
women: 60·96/160 = 36, 60·64/160 = 24

Chi-square statistic

χ² = (76 − 60)²/60 + (24 − 40)²/40 + (20 − 36)²/36 + (40 − 24)²/24
χ² = 28.44

Answer

χ² = 28.44

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

10.Chi-square tests

Answer: χ² = 2.738

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: 12, 28
line 2: 18, 42
line 3: 12, 28

Chi-square statistic

χ² = (16 − 12)²/12 + (24 − 28)²/28 + (15 − 18)²/18 + (45 − 42)²/42 + (11 − 12)²/12 + (29 − 28)²/28
χ² = 2.738

Decision

df = 2, χ²crit = 5.991
2.738 < 5.991

Answer

χ² = 2.738

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

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