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
248.5259P = ?

The bag weights are normally distributed with mean 250 g and standard deviation 5 g. What fraction lies between 248.5 and 259 g? Give P as a decimal.

μ = 250σ = 5x1 = 248.5x2 = 259
P =
2.Normal distributionHard
157164171178185192199cmtop 15%

The heights of adult men are normally distributed with mean 178 cm and standard deviation 7 cm. Drag the cut-off so the shaded top 15% is to the right of it. Find the cut-off x.

μ = 178 cmσ = 7 cm15%
x = cm
3.Confidence intervalsMedium
n = 16, s = 1.5L = ?Ux̄ = 16.5E99%

A sample of 16 items has mean 16.5 and standard deviation s = 1.5. The population is roughly normal. Find the lower limit of the 99% confidence interval for the mean.

n = 16x̄ = 16.5s = 1.599%t(15) = 2.947
lower limit =
4.Confidence intervalsHard
σ = 20, n = ?x̄ − Ex̄ + Ex̄E = 495%

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 4?

σ = 20E = 495%
n =
5.Hypothesis tests & confidence intervalsMedium
t −2.01102.011

A sample of 49 students has a mean score of 105 and a sample standard deviation of 12. Find the test statistic t for H₀: μ = 100.

μ₀ = 100x̄ = 105s = 12n = 49
t =
6.Hypothesis tests & confidence intervalsHard
z −1.9601.96

A sample of 100 measurements has mean 98.8. The population standard deviation is 15. Find the lower limit L of the 95% confidence interval for the mean.

x̄ = 98.8σ = 15n = 10095%
L =
7.Linear regression & correlationMedium
xy126225325422521618xy2462025

The table gives 6 paired observations of x = car age in years and y = price in k€. Find Pearson's correlation coefficient r.

n = 6
r =
8.Linear regression & correlationHard
xy143256354459567xy245060x = 2.5

The table gives 5 paired observations of x = hours studied and y = exam score. Use the least-squares line ŷ = a + b·x. A new observation (2.5, 55) is made. Find its residual e = y − ŷ.

n = 5(2.5, 55)
e =
9.Chi-square testsMedium
lowmediumhighTotalurban42182686rural12151946Total543345132

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

row total = 86column total = 45N = 132
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 7 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 = 0.582

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 = (248.5 − 250)/5 = −0.3
z₂ = (259 − 250)/5 = 1.8

Area between

P = Φ(1.8) − Φ(−0.3) = 0.9641 − 0.3821 = 0.582

Answer

P = 0.582
2.Normal distribution

Answer: x = 185.3 cm

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) = 0.85

z from the table

z ≈ 1.04

Back to the original units

x = 178 + 1.04·7 = 185.3 cm

Answer

x = 185.3 cm
3.Confidence intervals

Answer: lower limit = 15.39

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 = 15, t = 2.947

Margin of error

E = 2.947·1.5/√16 = 1.105

Interval

16.5 ± 1.105 → [15.39, 17.61]

Answer

lower limit = 15.39
4.Confidence intervals

Answer: n = 97

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 = (1.96·20/4)² = 96.04

Round up

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

n = 97

Answer

n = 97
5.Hypothesis tests & confidence intervals

Answer: t = 2.917

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 = 12/√49 = 12/7 = 1.714

Test statistic

t = (105 − 100)/1.714 = 2.917

Answer

t = 2.917
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: L = 95.86

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 = 1.96·15/√100 = 2.94

Interval limit

L = 98.8 − 2.94 = 95.86

Answer

L = 95.86
7.Linear regression & correlation

Answer: r = −0.9606

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 = 21/6 = 3.5
ȳ = 137/6 = 22.83

Sums of squares and products

Sxx = Σ(x − x̄)² = 17.5
Sxy = Σ(x − x̄)(y − ȳ) = −27.5
Syy = Σ(y − ȳ)² = 46.83

Correlation coefficient

r = −27.5/√(17.5·46.83) = −0.9606

Answer

r = −0.9606
8.Linear regression & correlation

Answer: e = 1.75

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 = 279/5 = 55.8

Sums of squares and products

Sxx = Σ(x − x̄)² = 10
Sxy = Σ(x − x̄)(y − ȳ) = 51

Regression line

b = 51/10 = 5.1
a = 55.8 − 5.1·3 = 40.5
ŷ = 40.5 + 5.1·x

Predicted value

ŷ = 40.5 + 5.1·2.5 = 53.25

Residual

e = 55 − 53.25 = 1.75

Answer

e = 1.75

The point lies above the line.

9.Chi-square tests

Answer: E = 29.32

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 = 86·45/132 = 29.32

Answer

E = 29.32
10.Chi-square tests

Answer: χ²crit = 16.81

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 = 7 − 1 = 6

Table

χ²crit(df = 6, α = 0.01) = 16.812

Answer

χ²crit = 16.812

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

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