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 = 0.0694

Drag z so that the shaded area under the standard normal curve, to the right of z, is 0.0694. Find z to two decimals.

A = 0.0694
z =
2.Normal distributionHard
10 cm0X − Y?

The heights X of men and Y of women in a group are independent and normal: X has mean 180 cm and standard deviation 10 cm, Y has mean 170 cm and standard deviation 12 cm. Find the probability P that X is greater than Y. Round z to two decimals.

μX = 180 cm, σX = 10 cmμY = 170 cm, σY = 12 cm
P =
3.Confidence intervalsMedium
n = 9, s = 19LU = ?x̄ = 65E90%

A sample of 9 items has mean 65 and standard deviation s = 19. The population is roughly normal. Find the upper limit of the 90% confidence interval for the mean.

n = 9x̄ = 65s = 1990%t(8) = 1.86
upper limit =
4.Confidence intervalsHard
σ = 30, n = ?x̄ − Ex̄ + Ex̄E = 490%

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 4?

σ = 30E = 490%
n =
5.Hypothesis tests & confidence intervalsMedium
t −2.01102.011

A sample of 49 students has a mean score of 99.49 and a sample standard deviation of 4. Find the test statistic t for H₀: μ = 100.

μ₀ = 100x̄ = 99.49s = 4n = 49
t =
6.Hypothesis tests & confidence intervalsHard
H₀H₁μ₀x̄cμ₀ + 4α = 0.01βσ = 13, power 80%, n = ?

A one-sided z test at α = 0.01 must detect a shift of the mean of 4 units with a power of 80%. The standard deviation is 13. Find the minimum sample size n.

α = 0.01power = 80%δ = 4σ = 13zα = 2.326zβ = 0.842
n =
7.Linear regression & correlationMedium
xy1473178720982311226134xy15202580100120140

The table gives 5 paired observations of x = temperature in °C and y = ice creams sold. Find Pearson's correlation coefficient r.

n = 5
r =
8.Linear regression & correlationHard
xy23745666787610851299xy510406080100x = 11

The table gives 6 paired observations of x = advertising spend in k€ and y = sales in k€. Use the least-squares line ŷ = a + b·x. A new observation (11, 88) is made. Find its residual e = y − ŷ.

n = 6(11, 88)
e =
9.Chi-square testsMedium
lowmediumhighTotalurban382844110rural20332578Total586169188

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

row total = 78column total = 61N = 188
E =
10.Chi-square testsHard

A chi-square test of independence is carried out on a contingency table with 2 rows and 2 columns (not counting the totals). How many degrees of freedom df does the test have?

2 × 2
df =

3Solutions

1.Normal distribution

Answer: z = 1.48

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 − 0.0694 = 0.9306

Read the z-table backwards

Find 0.9306 inside the table and read off z.

z = 1.48

Answer

z = 1.48
2.Normal distribution

Answer: P = 0.7389

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 = 180 − 170 = 10 cm, σ = 15.62 cm

Standardise

z = (0 − 10)/15.62 = −0.64

Probability

P = 1 − Φ(−0.64) = 0.7389

Answer

P = 0.7389
3.Confidence intervals

Answer: upper limit = 76.78

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 = 8, t = 1.86

Margin of error

E = 1.86·19/√9 = 11.78

Interval

65 ± 11.78 → [53.22, 76.78]

Answer

upper limit = 76.78
4.Confidence intervals

Answer: n = 153

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 = (1.645·30/4)² = 152.2

Round up

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

n = 153

Answer

n = 153
5.Hypothesis tests & confidence intervals

Answer: t = −0.8925

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 = (99.49 − 100)/0.5714 = −0.8925

Answer

t = −0.8925
t0.025; df = 48 = 2.011

|t| < 2.011, so H₀ is not rejected at the 5% level (two-sided).

6.Hypothesis tests & confidence intervals

Answer: n = 107

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 = ((2.326 + 0.842)·13/4)² = 106

Round up

n = 107

Answer

n = 107
7.Linear regression & correlation

Answer: r = 0.9913

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 = 100/5 = 20
ȳ = 504/5 = 100.8

Sums of squares and products

Sxx = Σ(x − x̄)² = 90
Sxy = Σ(x − x̄)(y − ȳ) = 441
Syy = Σ(y − ȳ)² = 2199

Correlation coefficient

r = 441/√(90·2199) = 0.9913

Answer

r = 0.9913
8.Linear regression & correlation

Answer: e = −5.2

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 = 42/6 = 7
ȳ = 420/6 = 70

Sums of squares and products

Sxx = Σ(x − x̄)² = 70
Sxy = Σ(x − x̄)(y − ȳ) = 406

Regression line

b = 406/70 = 5.8
a = 70 − 5.8·7 = 29.4
ŷ = 29.4 + 5.8·x

Predicted value

ŷ = 29.4 + 5.8·11 = 93.2

Residual

e = 88 − 93.2 = −5.2

Answer

e = −5.2

The point lies below the line.

9.Chi-square tests

Answer: E = 25.31

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.

Expected count

E = 78·61/188 = 25.31

Answer

E = 25.31
10.Chi-square tests

Answer: df = 1

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 = (2 − 1)·(2 − 1) = 1

Answer

df = 1

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