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.

New problemsBack to the course

In the print window, choose “Save as PDF” to keep a copy on your device.

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 s114.6 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 114.6 s. Round z to two decimals.

μX = 60, σX = 5μY = 45, σY = 4X + Y > 114.6 s
P =
2.Normal distributionHard
235240245250255260265gbottom 30%

The bag weights are normally distributed with mean 250 g and standard deviation 5 g. Drag the cut-off so the shaded bottom 30% is to the left of it. Find the cut-off x.

μ = 250 gσ = 5 g30%
x = g
3.Confidence intervalsMedium
x̄E₁ = 10.895%, n = 36x̄E₂ = ?95%, n = 9

A 95% confidence interval for a mean, based on 36 observations and a known σ, has a margin of error of 10.8. What is the margin of error E₂ if the same study uses 9 observations instead?

95%n₁ = 36E₁ = 10.8n₂ = 9
E₂ =
4.Confidence intervalsHard
n = 11Lσ = ?s = 7.695%

The diameters of 11 machined shafts have a sample standard deviation of 7.6 µm. Assuming the diameters are normally distributed, find the upper limit σ of the 95% confidence interval for the population standard deviation. With df = 10, 2.5% of the χ² distribution lies below 3.247 and 2.5% lies above 20.483.

n = 11s = 7.6 µmχ² = 3.247, 20.483
σ = µm
5.Hypothesis tests & confidence intervalsMedium
t −2.13102.131

A sample of 16 students has a mean score of 94.5 and a sample standard deviation of 10. Find the test statistic t for H₀: μ = 100.

μ₀ = 100x̄ = 94.5s = 10n = 16
t =
6.Hypothesis tests & confidence intervalsHard
H₀H₁μ₀x̄cμ₀ + 3α = 0.01βσ = 12, power 90%, n = ?

A one-sided z test at α = 0.01 must detect a shift of the mean of 3 units with a power of 90%. The standard deviation is 12. Find the minimum sample size n.

α = 0.01power = 90%δ = 3σ = 12zα = 2.326zβ = 1.282
n =
7.Linear regression & correlationMedium
xy141242345454555xy2440455055

The table gives 5 paired observations of x = advertising spend in k€ and y = sales in k€. Find Pearson's correlation coefficient r.

n = 5
r =
8.Linear regression & correlationHard
xy2544576738771083xy510607080x = 9

The table gives 5 paired observations of x = hours studied and y = exam score. Use the least-squares line ŷ = a + b·x. Predict ŷ for x = 9.

n = 5x = 9
ŷ =
9.Chi-square testsMedium
lowmediumhighTotalurban673221120rural532839120Total1206060240

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 = 2402 × 3
χ² =
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 test of independence is carried out on a contingency table with 2 rows and 4 columns (not counting the totals). Using the table, find the critical value χ²crit at significance level α = 0.05.

α = 0.05
χ²crit =

3Solutions

1.Normal distribution

Answer: P = 0.06681

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 = (114.6 − 105)/6.403 = 1.5

Upper tail

P = 1 − Φ(1.5) = 1 − 0.9332 = 0.06681

Answer

P = 0.06681
2.Normal distribution

Answer: x = 247.4 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) = 0.3

z from the table

z ≈ −0.52

Back to the original units

x = 250 − 0.52·5 = 247.4 g

Answer

x = 247.4 g
3.Confidence intervals

Answer: E₂ = 21.6

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 = (1.96/1.96)·√(36/9) = 2

New margin

E₂ = 10.8·2 = 21.6

Answer

E₂ = 21.6
4.Confidence intervals

Answer: σ = 13.34 µm

Interval for a variance: (n − 1)s²/σ² has a χ² distribution with n − 1 degrees of freedom. The 95% interval for σ² runs from (n − 1)s² divided by the upper χ² value (2.5% above it) to (n − 1)s² divided by the lower χ² value (2.5% below it); the interval for σ is the square root of both limits. It is not symmetric around s.

Sum of squares

(n − 1)s² = 10·7.6² = 577.6

Limits for σ²

577.6/20.483 = 28.2
577.6/3.247 = 177.9

Limit for σ

σ = √177.9 = 13.34 µm

Answer

σ = 13.34 µm

95% interval for σ: [5.31, 13.34] µm.

5.Hypothesis tests & confidence intervals

Answer: t = −2.2

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 = 10/√16 = 10/4 = 2.5

Test statistic

t = (94.5 − 100)/2.5 = −2.2

Answer

t = −2.2
t0.025; df = 15 = 2.131

|t| > 2.131, so H₀ is rejected at the 5% level (two-sided).

6.Hypothesis tests & confidence intervals

Answer: n = 209

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 + 1.282)·12/3)² = 208.3

Round up

n = 209

Answer

n = 209
7.Linear regression & correlation

Answer: r = 0.9502

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 = 237/5 = 47.4

Sums of squares and products

Sxx = Σ(x − x̄)² = 10
Sxy = Σ(x − x̄)(y − ȳ) = 40
Syy = Σ(y − ȳ)² = 177.2

Correlation coefficient

r = 40/√(10·177.2) = 0.9502

Answer

r = 0.9502
8.Linear regression & correlation

Answer: ŷ = 80.5

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 = 344/5 = 68.8

Sums of squares and products

Sxx = Σ(x − x̄)² = 40
Sxy = Σ(x − x̄)(y − ȳ) = 156

Regression line

b = 156/40 = 3.9
a = 68.8 − 3.9·6 = 45.4
ŷ = 45.4 + 3.9·x

Prediction

ŷ = 45.4 + 3.9·9 = 80.5

Answer

ŷ = 80.5
9.Chi-square tests

Answer: χ² = 7.3

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

urban: 120·120/240 = 60, 120·60/240 = 30, 120·60/240 = 30
rural: 120·120/240 = 60, 120·60/240 = 30, 120·60/240 = 30

Chi-square statistic

χ² = (67 − 60)²/60 + (32 − 30)²/30 + (21 − 30)²/30 + (53 − 60)²/60 + (28 − 30)²/30 + (39 − 30)²/30
χ² = 7.3

Answer

χ² = 7.3

df = 2. At α = 0.05 the critical value is 5.991, so H₀ (independence) is rejected.

10.Chi-square tests

Answer: χ²crit = 7.815

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)·(4 − 1) = 3

Table

χ²crit(df = 3, α = 0.05) = 7.815

Answer

χ²crit = 7.815

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

ExamProblems · examproblems.com/courses/statistics-2-inference