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
−3−2−10123zshaded area to the right of z = 0.9162

Drag z so that the shaded area under the standard normal curve, to the right of z, is 0.9162. Find z to two decimals.

A = 0.9162
z =
2.Normal distributionHard
235240245250255260265gbottom 15%

The bag weights are normally distributed with mean 250 g and standard deviation 5 g. Drag the cut-off so the shaded bottom 15% is to the left of it. Find the cut-off x.

μ = 250 gσ = 5 g15%
x = g
3.Confidence intervalsMedium
μ₁ − μ₂0LU = ?x̄₁ − x̄₂E99%

The battery life of two phone models is compared. Model 1: 80 phones, mean 106 h, known standard deviation 5 h. Model 2: 60 phones, mean 92 h, known standard deviation 9 h. Find the upper limit of the 99% confidence interval for μ₁ − μ₂.

n₁ = 80x̄₁ = 106 hσ₁ = 5 hn₂ = 60x̄₂ = 92 hσ₂ = 9 h99%
upper limit = h
4.Confidence intervalsHard
σ = 20, n = ?x̄ − Ex̄ + Ex̄E = 499%

The population standard deviation is 20. 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 4?

σ = 20E = 499%
n =
5.Hypothesis tests & confidence intervalsMedium
t −1.98401.984

A sample of 100 students has a mean score of 197.4 and a sample standard deviation of 12. Find the test statistic t for H₀: μ = 200.

μ₀ = 200x̄ = 197.4s = 12n = 100
t =
6.Hypothesis tests & confidence intervalsHard
z −1.9601.96

A sample of 81 measurements has mean 120.7. The population standard deviation is 8. Find the upper limit U of the 95% confidence interval for the mean.

x̄ = 120.7σ = 8n = 8195%
U =
7.Linear regression & correlationMedium
xy284126178231028xy510102030

The table gives 5 paired observations of x = load in kN and y = extension in mm. Find Pearson's correlation coefficient r.

n = 5
r =
8.Linear regression & correlationHard
xy23544264684810531253xy5104050ŷ = 53

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. For which value x does the line predict ŷ = 53?

n = 6ŷ = 53
x =
9.Chi-square testsMedium
improvednot improvedTotalnew drug233760placebo404080Total6377140

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 = 1402 × 2
χ² =
10.Chi-square testsHard

A chi-square goodness-of-fit test compares observed counts in 5 categories with fully specified expected proportions (no parameters are estimated from the data). How many degrees of freedom df does the test have?

k = 5
df =

3Solutions

1.Normal distribution

Answer: z = −1.38

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 − 0.9162 = 0.0838

Read the z-table backwards

Find 0.0838 inside the table and read off z.

z = −1.38

Answer

z = −1.38
2.Normal distribution

Answer: x = 244.8 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) = 0.15

z from the table

z ≈ −1.04

Back to the original units

x = 250 − 1.04·5 = 244.8 g

Answer

x = 244.8 g
3.Confidence intervals

Answer: upper limit = 17.32 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 = 106 − 92 = 14

Standard error

SE = √(5²/80 + 9²/60) = 1.289

Margin of error

E = 2.576·1.289 = 3.321

Interval

14 ± 3.321 → [10.68, 17.32]

Answer

upper limit = 17.32 h

The interval does not contain 0, so the means differ significantly.

4.Confidence intervals

Answer: n = 166

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 = (2.576·20/4)² = 165.9

Round up

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

n = 166

Answer

n = 166
5.Hypothesis tests & confidence intervals

Answer: t = −2.167

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/√100 = 12/10 = 1.2

Test statistic

t = (197.4 − 200)/1.2 = −2.167

Answer

t = −2.167
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: U = 122.4

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 = 1.96·8/√81 = 1.742

Interval limit

U = 120.7 + 1.742 = 122.4

Answer

U = 122.4
7.Linear regression & correlation

Answer: r = 0.9979

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 = 88/5 = 17.6

Sums of squares and products

Sxx = Σ(x − x̄)² = 40
Sxy = Σ(x − x̄)(y − ȳ) = 102
Syy = Σ(y − ȳ)² = 261.2

Correlation coefficient

r = 102/√(40·261.2) = 0.9979

Answer

r = 0.9979
8.Linear regression & correlation

Answer: x = 10.83

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 = 277/6 = 46.17

Sums of squares and products

Sxx = Σ(x − x̄)² = 70
Sxy = Σ(x − x̄)(y − ȳ) = 125

Regression line

b = 125/70 = 1.786
a = 46.17 − 1.786·7 = 33.67
ŷ = 33.67 + 1.786·x

Solve for x

x = (53 − 33.67)/1.786 = 10.83

Answer

x = 10.83
9.Chi-square tests

Answer: χ² = 1.886

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

new drug: 60·63/140 = 27, 60·77/140 = 33
placebo: 80·63/140 = 36, 80·77/140 = 44

Chi-square statistic

χ² = (23 − 27)²/27 + (37 − 33)²/33 + (40 − 36)²/36 + (40 − 44)²/44
χ² = 1.886

Answer

χ² = 1.886

df = 1. At α = 0.05 the critical value is 3.841, so H₀ (independence) is not rejected.

10.Chi-square tests

Answer: df = 4

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 = 5 − 1 = 4

Answer

df = 4

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