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
32 min34 minX + Y?

X is normal with mean 12 min and standard deviation 1.5 min; Y is independent and normal with mean 20 min and standard deviation 2 min. Find the probability P that X + Y is more than 34 min. Round z to two decimals.

μX = 12, σX = 1.5μY = 20, σY = 2X + Y > 34 min
P =
2.Normal distributionHard
10 g0X − Y?

The mass X of a product from line 1 and Y from line 2 are independent and normal: X has mean 500 g and standard deviation 8 g, Y has mean 490 g and standard deviation 6 g. Find the probability P that X is greater than Y. Round z to two decimals.

μX = 500 g, σX = 8 gμY = 490 g, σY = 6 g
P =
3.Confidence intervalsMedium
x̄E₁ = 5.995%, n = 25x̄E₂ = ?90%, n = 25

A 95% confidence interval for a mean, based on 25 observations and a known σ, has a margin of error of 5.9. What is the margin of error E₂ of a 90% interval from the same data?

95%n₁ = 25E₁ = 5.990%
E₂ =
4.Confidence intervalsHard
μ₁ − μ₂, t(20) = 2.0860LUx̄₁ − x̄₂E = ?95%

Two fertilisers are tested. Plot group 1: 6 plots, mean yield 19 kg, standard deviation 3.5 kg. Plot group 2: 16 plots, mean 9.5 kg, standard deviation 2.5 kg. Assume equal variances (pooled t, df = n₁ + n₂ − 2). Find the margin of error E of the 95% confidence interval for μ₁ − μ₂.

n₁ = 6x̄₁ = 19s₁ = 3.5n₂ = 16x̄₂ = 9.5s₂ = 2.595%t(20) = 2.086
E = kg
5.Hypothesis tests & confidence intervalsMedium
t −2.01102.011

A sample of 49 students has a mean score of 98.51 and a sample standard deviation of 4. Find the test statistic t for H₀: μ = 100.

μ₀ = 100x̄ = 98.51s = 4n = 49
t =
6.Hypothesis tests & confidence intervalsHard
z −2.57602.576

A sample of 100 measurements has mean 202.6. The population standard deviation is 15. Find the upper limit U of the 99% confidence interval for the mean.

x̄ = 202.6σ = 15n = 10099%
U =
7.Linear regression & correlationMedium
xy163119402243255628723182xy15202530406080

The table gives 6 paired observations of x = temperature in °C and y = ice creams sold. Find Pearson's correlation coefficient r.

n = 6
r =
8.Linear regression & correlationHard
xy12221831641458xy24101520x = 2.5

The table gives 5 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 = 5x = 2.5
ŷ =
9.Chi-square testsMedium
improvednot improvedTotalnew drug273360placebo451560Total7248120

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 = 1202 × 2
χ² =
10.Chi-square testsHard

A chi-square goodness-of-fit test checks whether counts in 9 categories follow a Poisson distribution whose mean is estimated from the same data. How many degrees of freedom df does the test have?

k = 9
df =

3Solutions

1.Normal distribution

Answer: P = 0.2119

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 sum

μ = 32 min, σ = √(1.5² + 2²) = 2.5 min

Standardise

z = (34 − 32)/2.5 = 0.8

Upper tail

P = 1 − Φ(0.8) = 1 − 0.7881 = 0.2119

Answer

P = 0.2119
2.Normal distribution

Answer: P = 0.8413

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 = 500 − 490 = 10 g, σ = 10 g

Standardise

z = (0 − 10)/10 = −1

Probability

P = 1 − Φ(−1) = 0.8413

Answer

P = 0.8413
3.Confidence intervals

Answer: E₂ = 4.952

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 = (1.645/1.96)·√(25/25) = 0.8393

New margin

E₂ = 5.9·0.8393 = 4.952

Answer

E₂ = 4.952
4.Confidence intervals

Answer: E = 2.78 kg

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.

Pooled standard deviation

sp² = (5·3.5² + 15·2.5²)/20 = 7.75
sp = 2.784

Margin of error

E = 2.086·2.784·√(1/6 + 1/16) = 2.78

Answer

E = 2.78 kg
5.Hypothesis tests & confidence intervals

Answer: t = −2.607

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 = 4/√49 = 4/7 = 0.5714

Test statistic

t = (98.51 − 100)/0.5714 = −2.607

Answer

t = −2.607
t0.025; df = 48 = 2.011

|t| > 2.011, so H₀ is rejected at the 5% level (two-sided).

6.Hypothesis tests & confidence intervals

Answer: U = 206.5

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).

Margin of error

E = 2.576·15/√100 = 3.864

Interval limit

U = 202.6 + 3.864 = 206.5

Answer

U = 206.5
7.Linear regression & correlation

Answer: r = 0.9832

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 = 141/6 = 23.5
ȳ = 324/6 = 54

Sums of squares and products

Sxx = Σ(x − x̄)² = 157.5
Sxy = Σ(x − x̄)(y − ȳ) = 546
Syy = Σ(y − ȳ)² = 1958

Correlation coefficient

r = 546/√(157.5·1958) = 0.9832

Answer

r = 0.9832
8.Linear regression & correlation

Answer: ŷ = 17.2

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 = 78/5 = 15.6

Sums of squares and products

Sxx = Σ(x − x̄)² = 10
Sxy = Σ(x − x̄)(y − ȳ) = −32

Regression line

b = −32/10 = −3.2
a = 15.6 − (−3.2)·3 = 25.2
ŷ = 25.2 − 3.2·x

Prediction

ŷ = 25.2 + (−3.2)·2.5 = 17.2

Answer

ŷ = 17.2
9.Chi-square tests

Answer: χ² = 11.25

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

new drug: 60·72/120 = 36, 60·48/120 = 24
placebo: 60·72/120 = 36, 60·48/120 = 24

Chi-square statistic

χ² = (27 − 36)²/36 + (33 − 24)²/24 + (45 − 36)²/36 + (15 − 24)²/24
χ² = 11.25

Answer

χ² = 11.25

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

10.Chi-square tests

Answer: df = 7

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 = 9 − 1 − 1 = 7

Answer

df = 7

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