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
105 s108.2 sX + Y?

X is normal with mean 60 s and standard deviation 5 s; Y is independent and normal with mean 45 s and standard deviation 4 s. Find the probability P that X + Y is more than 108.2 s. Round z to two decimals.

μX = 60, σX = 5μY = 45, σY = 4X + Y > 108.2 s
P =
2.Normal distributionHard
235240245250255260265gtop 90%

The bag weights are normally distributed with mean 250 g and standard deviation 5 g. Drag the cut-off so the shaded top 90% is to the right of it. Find the cut-off x.

μ = 250 gσ = 5 g90%
x = g
3.Confidence intervalsMedium
n = 11, s = 17LUx̄ = 62.5E = ?90%

A sample of 11 items has mean 62.5 and standard deviation s = 17. The population is roughly normal. Find the margin of error E of a 90% confidence interval for the mean.

n = 11x̄ = 62.5s = 1790%t(10) = 1.812
E =
4.Confidence intervalsHard
σ = 20, n = ?x̄ − Ex̄ + Ex̄E = 1.595%

The population standard deviation is 20. 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 1.5?

σ = 20E = 1.595%
n =
5.Hypothesis tests & confidence intervalsMedium
−2.1312.131t = ?df = 15

Two teaching methods are compared. Group 1: 12 students, mean score 57.5, standard deviation 3.5. Group 2: 5 students, mean score 61.8, standard deviation 4. Assume equal population variances and use the pooled two-sample t test. Find the test statistic t for H₀: μ₁ = μ₂, taking the difference as x̄₁ − x̄₂ (group 1 minus group 2).

n₁ = 12x̄₁ = 57.5s₁ = 3.5n₂ = 5x̄₂ = 61.8s₂ = 4
t =
6.Hypothesis tests & confidence intervalsHard
μ₀x̄c

A one-sided z test of H₀: μ = 50 against H₁: μ < 50 is done at α = 0.10 with a sample of 100; the standard deviation is known to be 30. In reality the mean is 42.8. Find the power 1 − β of the test in percent.

μ₀ = 50μ = 42.8σ = 30n = 100α = 0.10
1 − β = %
7.Linear regression & correlationMedium
xy21342162983610421249xy5102040

The table gives 6 paired observations of x = load in kN and y = extension in mm. Their sums are given. Find Pearson's correlation coefficient r from these sums.

n = 6Σx = 42Σy = 190Σx² = 364Σy² = 6912Σxy = 1580
r =
8.Linear regression & correlationHard
xy17922010423126261422916332184xy202530100150x = 24.5

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 = 24.5.

n = 6x = 24.5
ŷ =
9.Chi-square testsMedium
busbikecarTotalstudents4291970staff36411188Total785030158

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

row total = 70column total = 50N = 158
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 compares observed counts in 8 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 = 0.01.

α = 0.01
χ²crit =

3Solutions

1.Normal distribution

Answer: P = 0.3085

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 = 105 s, σ = √(5² + 4²) = 6.403 s

Standardise

z = (108.2 − 105)/6.403 = 0.5

Upper tail

P = 1 − Φ(0.5) = 1 − 0.6915 = 0.3085

Answer

P = 0.3085
2.Normal distribution

Answer: x = 243.6 g

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 = 250 − 1.28·5 = 243.6 g

Answer

x = 243.6 g
3.Confidence intervals

Answer: E = 9.288

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 = 10, t = 1.812

Margin of error

E = 1.812·17/√11 = 9.288

Answer

E = 9.288
4.Confidence intervals

Answer: n = 683

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 = (1.96·20/1.5)² = 683

Round up

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

n = 683

Answer

n = 683
5.Hypothesis tests & confidence intervals

Answer: t = −2.219

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.

Pooled variance

sp² = (11·3.5² + 4·4²)/15 = 13.25
sp = 3.64

Standard error

SE = 3.64·√(1/12 + 1/5) = 1.938

Test statistic

t = (57.5 − 61.8)/1.938 = −2.219

Answer

t = −2.219

With df = 15, the two-sided 5% critical value is 2.131; |t| is larger, so H₀ is rejected.

6.Hypothesis tests & confidence intervals

Answer: 1 − β = 86.82 %

Power of a test: power = 1 − β is the probability of rejecting H₀ when it is false. Find the cutoff x̄c from μ₀ and z_α, then compute the probability of landing beyond x̄c using the true mean: z = (x̄c − μ)/(σ/√n).

Critical sample mean

SE = 30/√100 = 3
x̄c = 50 − 1.282·3 = 46.15

z under the true mean

z = (46.15 − 42.8)/3 = 1.118

Power

power = P(Z < 1.118) = 86.82%
1 − β = 86.82%

Answer

1 − β = 86.82%
7.Linear regression & correlation

Answer: r = 0.9986

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 = 1580 − 42·190/6 = 250
Sxx = 364 − 42²/6 = 70
Syy = 6912 − 190²/6 = 895.3

Correlation coefficient

r = 250/√(70·895.3) = 0.9986

Answer

r = 0.9986
8.Linear regression & correlation

Answer: ŷ = 135.2

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 = 147/6 = 24.5
ȳ = 811/6 = 135.2

Sums of squares and products

Sxx = Σ(x − x̄)² = 157.5
Sxy = Σ(x − x̄)(y − ȳ) = 979.5

Regression line

b = 979.5/157.5 = 6.219
a = 135.2 − 6.219·24.5 = −17.2
ŷ = −17.2 + 6.219·x

Prediction

ŷ = −17.2 + 6.219·24.5 = 135.2

Answer

ŷ = 135.2
9.Chi-square tests

Answer: E = 22.15

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 = 70·50/158 = 22.15

Answer

E = 22.15
10.Chi-square tests

Answer: χ²crit = 18.48

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 = 8 − 1 = 7

Table

χ²crit(df = 7, α = 0.01) = 18.475

Answer

χ²crit = 18.475

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

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