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
167.15190.6P = ?

The heights of adult men are normally distributed with mean 178 cm and standard deviation 7 cm. What fraction lies between 167.15 and 190.6 cm? Give P as a decimal.

μ = 178σ = 7x1 = 167.15x2 = 190.6
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
μ₁ − μ₂0LU = ?x̄₁ − x̄₂E99%

The battery life of two phone models is compared. Model 1: 40 phones, mean 123 h, known standard deviation 9 h. Model 2: 100 phones, mean 125.5 h, known standard deviation 10 h. Find the upper limit of the 99% confidence interval for μ₁ − μ₂.

n₁ = 40x̄₁ = 123 hσ₁ = 9 hn₂ = 100x̄₂ = 125.5 hσ₂ = 10 h99%
upper limit = h
4.Confidence intervalsHard
σ = 24, n = ?x̄ − Ex̄ + Ex̄E = 0.599%

The population standard deviation is 24. 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 0.5?

σ = 24E = 0.599%
n =
5.Hypothesis tests & confidence intervalsMedium
t −1.98401.984

A sample of 100 students has a mean score of 48.96 and a sample standard deviation of 4. Find the test statistic t for H₀: μ = 50.

μ₀ = 50x̄ = 48.96s = 4n = 100
t =
6.Hypothesis tests & confidence intervalsHard
−1.961.96z = ?

In a trial, 83 of 150 patients on treatment A recover, and 201 of 300 patients on treatment B. Using the pooled proportion, find the test statistic z for H₀: p₁ = p₂.

x₁ = 83n₁ = 150x₂ = 201n₂ = 300
z =
7.Linear regression & correlationMedium
xy294136168211025xy51010152025

The table gives 5 paired observations of x = load in kN and y = extension in mm. Find the coefficient of determination r² of the least-squares line.

n = 5
r² =
8.Linear regression & correlationHard
xy21041462582910391246xy5102040x = 11

The table gives 6 paired observations of x = load in kN and y = extension in mm. Use the least-squares line ŷ = a + b·x. A new observation (11, 40) is made. Find its residual e = y − ŷ.

n = 6(11, 40)
e =
9.Chi-square testsMedium
busbikecarTotalstudents19114272staff13134167Total322483139

The contingency table shows observed counts. For a chi-square test of independence, find the expected count E for the cell “staff, bus”.

row total = 67column total = 32N = 139
E =
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 checks whether counts in 4 categories follow a Poisson distribution whose mean is estimated from the same data. Using the table, find the critical value χ²crit at significance level α = 0.05.

α = 0.05
χ²crit =

3Solutions

1.Normal distribution

Answer: P = 0.9035

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-scores

z₁ = (167.15 − 178)/7 = −1.55
z₂ = (190.6 − 178)/7 = 1.8

Area between

P = Φ(1.8) − Φ(−1.55) = 0.9641 − 0.06057 = 0.9035

Answer

P = 0.9035
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: upper limit = 1.98 h

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.

Point estimate

x̄₁ − x̄₂ = 123 − 125.5 = −2.5

Standard error

SE = √(9²/40 + 10²/100) = 1.739

Margin of error

E = 2.576·1.739 = 4.48

Interval

−2.5 ± 4.48 → [−6.98, 1.98]

Answer

upper limit = 1.98 h

The interval contains 0, so the data do not show a significant difference.

4.Confidence intervals

Answer: n = 15289

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 = (2.576·24/0.5)² = 15289

Round up

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

n = 15289

Answer

n = 15289
5.Hypothesis tests & confidence intervals

Answer: t = −2.6

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/√100 = 4/10 = 0.4

Test statistic

t = (48.96 − 50)/0.4 = −2.6

Answer

t = −2.6
t0.025; df = 99 = 1.984

|t| > 1.984, so H₀ is rejected at the 5% level (two-sided).

6.Hypothesis tests & confidence intervals

Answer: z = −2.418

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 = 83/150 = 0.5533
p̂₂ = 201/300 = 0.67
p̂ = 284/450 = 0.6311

Standard error under H₀

SE = √(0.6311·0.3689·(1/150 + 1/300)) = 0.04825

Test statistic

z = (0.5533 − 0.67)/0.04825 = −2.418

Answer

z = −2.418

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

7.Linear regression & correlation

Answer: r² = 0.995

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 = 30/5 = 6
ȳ = 84/5 = 16.8

Sums of squares and products

Sxx = Σ(x − x̄)² = 40
Sxy = Σ(x − x̄)(y − ȳ) = 80
Syy = Σ(y − ȳ)² = 160.8

Coefficient of determination

r² = 80²/(40·160.8) = 0.995

Answer

r² = 0.995

About 99.5% of the variation in y is explained by the line.

8.Linear regression & correlation

Answer: e = −1.967

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 = 42/6 = 7
ȳ = 163/6 = 27.17

Sums of squares and products

Sxx = Σ(x − x̄)² = 70
Sxy = Σ(x − x̄)(y − ȳ) = 259

Regression line

b = 259/70 = 3.7
a = 27.17 − 3.7·7 = 1.267
ŷ = 1.267 + 3.7·x

Predicted value

ŷ = 1.267 + 3.7·11 = 41.97

Residual

e = 40 − 41.97 = −1.967

Answer

e = −1.967

The point lies below the line.

9.Chi-square tests

Answer: E = 15.42

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 count

E = 67·32/139 = 15.42

Answer

E = 15.42
10.Chi-square tests

Answer: χ²crit = 5.991

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 = 4 − 1 − 1 = 2

Table

χ²crit(df = 2, α = 0.05) = 5.991

Answer

χ²crit = 5.991

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

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