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 = 0.7330

Drag z so that the shaded area between −z and z under the standard normal curve is 0.7330. Find z to two decimals.

A = 0.7330
z =
2.Normal distributionHard
0.04 mm0D − d?

Hole diameters D are normal with mean 25.04 mm and standard deviation 0.015 mm. Shaft diameters d are independent and normal with mean 25 mm and standard deviation 0.02 mm. A randomly chosen shaft fits when D − d > 0. Find the probability P that it fits. Round z to two decimals.

hole: μ = 25.04 mm, σ = 0.015 mmshaft: μ = 25 mm, σ = 0.02 mm
P =
3.Confidence intervalsMedium
n = 12, s = 16.5LUx̄ = 36E = ?95%

A sample of 12 items has mean 36 and standard deviation s = 16.5. The population is roughly normal. Find the margin of error E of a 95% confidence interval for the mean.

n = 12x̄ = 36s = 16.595%t(11) = 2.201
E =
4.Confidence intervalsHard
σ = 30, n = ?x̄ − Ex̄ + Ex̄E = 1.590%

The population standard deviation is 30. How large a sample n do you need so that a 90% confidence interval for the mean has a margin of error of at most 1.5?

σ = 30E = 1.590%
n =
5.Hypothesis tests & confidence intervalsMedium
−2.2622.262t = ?df = 9

10 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 −1.57 and the standard deviation of the differences is 6.2. Find the paired test statistic t for H₀: μd = 0.

n = 10d̄ = −1.57sd = 6.2
t =
6.Hypothesis tests & confidence intervalsHard
σ = 5, n = ?x̄ − Ex̄ + Ex̄E = 2.599%

How large must a sample be so that the 99% confidence interval for a mean has a margin of error of at most 2.5? The standard deviation is 5. Find the smallest sample size n.

σ = 5E = 2.599%
n =
7.Linear regression & correlationMedium
xy123222329427528xy2425

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

n = 5
r =
8.Linear regression & correlationHard
xy189520104221252413826143xy2025100120140x = 19

The table gives 5 paired observations of x = temperature in °C and y = ice creams sold. Use the least-squares line ŷ = a + b·x. A new observation (19, 98) is made. Find its residual e = y − ŷ.

n = 5(19, 98)
e =
9.Chi-square testsMedium
lowmediumhighTotalurban253342100rural352738100Total606080200

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
dfα = 0.05α = 0.0113.8416.63525.9919.21037.81511.34549.48813.277511.07015.086612.59216.812714.06718.475815.50720.090916.91921.6661018.30723.209

A chi-square goodness-of-fit test compares observed counts in 11 categories with fully specified expected proportions (no parameters are estimated from the data). Using the table, find the critical value χ²crit at significance level α = 0.05.

α = 0.05
χ²crit =

3Solutions

1.Normal distribution

Answer: z = 1.11

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 + 0.7330)/2 = 0.8665

Read the z-table backwards

Find 0.8665 inside the table and read off z.

z = 1.11

Answer

z = 1.11
2.Normal distribution

Answer: P = 0.9452

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 = 25.04 − 25 = 0.04 mm, σ = 0.025 mm

Standardise

z = (0 − 0.04)/0.025 = −1.6

Probability

P = 1 − Φ(−1.6) = 0.9452

Answer

P = 0.9452
3.Confidence intervals

Answer: E = 10.48

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 = 11, t = 2.201

Margin of error

E = 2.201·16.5/√12 = 10.48

Answer

E = 10.48
4.Confidence intervals

Answer: n = 1083

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 = (1.645·30/1.5)² = 1082

Round up

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

n = 1083

Answer

n = 1083
5.Hypothesis tests & confidence intervals

Answer: t = −0.8008

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 = 6.2/√10 = 1.961

Test statistic

t = −1.57/1.961 = −0.8008

Answer

t = −0.8008

With df = 9, the two-sided 5% critical value is 2.262; |t| is smaller, so H₀ is not rejected.

6.Hypothesis tests & confidence intervals

Answer: n = 27

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 = (2.576·5/2.5)² = 26.54

Round up

n = 27

Answer

n = 27
7.Linear regression & correlation

Answer: r = 0.7615

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 = 129/5 = 25.8

Sums of squares and products

Sxx = Σ(x − x̄)² = 10
Sxy = Σ(x − x̄)(y − ȳ) = 15
Syy = Σ(y − ȳ)² = 38.8

Correlation coefficient

r = 15/√(10·38.8) = 0.7615

Answer

r = 0.7615
8.Linear regression & correlation

Answer: e = −3.5

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 = 110/5 = 22
ȳ = 605/5 = 121

Sums of squares and products

Sxx = Σ(x − x̄)² = 40
Sxy = Σ(x − x̄)(y − ȳ) = 260

Regression line

b = 260/40 = 6.5
a = 121 − 6.5·22 = −22
ŷ = −22 + 6.5·x

Predicted value

ŷ = −22 + 6.5·19 = 101.5

Residual

e = 98 − 101.5 = −3.5

Answer

e = −3.5

The point lies below the line.

9.Chi-square tests

Answer: χ² = 2.467

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: 100·60/200 = 30, 100·60/200 = 30, 100·80/200 = 40
rural: 100·60/200 = 30, 100·60/200 = 30, 100·80/200 = 40

Chi-square statistic

χ² = (25 − 30)²/30 + (33 − 30)²/30 + (42 − 40)²/40 + (35 − 30)²/30 + (27 − 30)²/30 + (38 − 40)²/40
χ² = 2.467

Answer

χ² = 2.467

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

10.Chi-square tests

Answer: χ²crit = 18.31

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 = 11 − 1 = 10

Table

χ²crit(df = 10, α = 0.05) = 18.307

Answer

χ²crit = 18.307

Reject H₀ if the computed χ² is larger than this value.

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