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 between −z and z = 0.3035

Drag z so that the shaded area between −z and z under the standard normal curve is 0.3035. Find z to two decimals.

A = 0.3035
z =
2.Normal distributionHard
μ = 32x = ?97.5%

The delivery times are normally distributed with mean 32 min and standard deviation 6 min. Find the value x below which 97.5% of them lie (the 97.5th percentile).

μ = 32σ = 6P = 97.5%
x = min
3.Confidence intervalsMedium
μ₁ − μ₂0L = ?Ux̄₁ − x̄₂E90%

The battery life of two phone models is compared. Model 1: 50 phones, mean 127.5 h, known standard deviation 14 h. Model 2: 40 phones, mean 142 h, known standard deviation 4 h. Find the lower limit of the 90% confidence interval for μ₁ − μ₂.

n₁ = 50x̄₁ = 127.5 hσ₁ = 14 hn₂ = 40x̄₂ = 142 hσ₂ = 4 h90%
lower limit = h
4.Confidence intervalsHard
p ≈ 0.4, n = ?p̂ − Ep̂ + Ep̂E = 0.0495%

A pilot study suggests that about 0.4 of products are defective. How many products n must be inspected so that a 95% confidence interval for the proportion has a margin of error of at most 0.04?

p ≈ 0.4E = 0.0495%
n =
5.Hypothesis tests & confidence intervalsMedium
t −2.01102.011

A sample of 49 students has a mean score of 102.7 and a sample standard deviation of 8. Find the test statistic t for H₀: μ = 100.

μ₀ = 100x̄ = 102.7s = 8n = 49
t =
6.Hypothesis tests & confidence intervalsHard
−1.961.96z = ?

In a trial, 36 of 120 patients on treatment A recover, and 39 of 150 patients on treatment B. Using the pooled proportion, find the test statistic z for H₀: p₁ = p₂.

x₁ = 36n₁ = 120x₂ = 39n₂ = 150
z =
7.Linear regression & correlationMedium
xy2344386418461052xy5104050

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

n = 5Σx = 30Σy = 211Σx² = 220Σy² = 9101Σxy = 1354
r =
8.Linear regression & correlationHard
xy359461567670771xy46606570x = 5.5

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 = 5.5.

n = 5x = 5.5
ŷ =
9.Chi-square testsMedium
yesnoTotalmen441660women342660Total7842120

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 = 1202 × 2
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 goodness-of-fit test checks whether counts in 6 categories follow a Poisson distribution whose mean is estimated from the same data. Using the table, find the critical value χ²crit at significance level α = 0.05.

α = 0.05
χ²crit =

3Solutions

1.Normal distribution

Answer: z = 0.39

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 + 0.3035)/2 = 0.6517

Read the z-table backwards

Find 0.6517 inside the table and read off z.

z = 0.39

Answer

z = 0.39
2.Normal distribution

Answer: x = 43.76 min

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 for the percentile

Φ(z) = 0.975 → z = 1.96

Back to x

x = 32 + 1.96·6 = 43.76 min

Answer

x = 43.76 min
3.Confidence intervals

Answer: lower limit = −17.92 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 = 127.5 − 142 = −14.5

Standard error

SE = √(14²/50 + 4²/40) = 2.078

Margin of error

E = 1.645·2.078 = 3.419

Interval

−14.5 ± 3.419 → [−17.92, −11.08]

Answer

lower limit = −17.92 h

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

4.Confidence intervals

Answer: n = 577

Confidence interval for a proportion: p̂ ± z·√(p̂(1 − p̂)/n), where p̂ = x/n. To reach a margin E you need n ≥ z²·p(1 − p)/E², rounded up; without a prior estimate use p = 0.5, which gives the largest n.

Solve for n

n = 1.96²·0.4·0.6/0.04² = 576.2

Round up

n = 577

Answer

n = 577
5.Hypothesis tests & confidence intervals

Answer: t = 2.363

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 = 8/√49 = 8/7 = 1.143

Test statistic

t = (102.7 − 100)/1.143 = 2.363

Answer

t = 2.363
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: z = 0.7292

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.

Sample proportions

p̂₁ = 36/120 = 0.3
p̂₂ = 39/150 = 0.26
p̂ = 75/270 = 0.2778

Standard error under H₀

SE = √(0.2778·0.7222·(1/120 + 1/150)) = 0.05486

Test statistic

z = (0.3 − 0.26)/0.05486 = 0.7292

Answer

z = 0.7292

|z| < 1.96, so H₀ is not rejected at the 5% level (two-sided).

7.Linear regression & correlation

Answer: r = 0.9918

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 = 1354 − 30·211/5 = 88
Sxx = 220 − 30²/5 = 40
Syy = 9101 − 211²/5 = 196.8

Correlation coefficient

r = 88/√(40·196.8) = 0.9918

Answer

r = 0.9918
8.Linear regression & correlation

Answer: ŷ = 67.25

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 = 25/5 = 5
ȳ = 328/5 = 65.6

Sums of squares and products

Sxx = Σ(x − x̄)² = 10
Sxy = Σ(x − x̄)(y − ȳ) = 33

Regression line

b = 33/10 = 3.3
a = 65.6 − 3.3·5 = 49.1
ŷ = 49.1 + 3.3·x

Prediction

ŷ = 49.1 + 3.3·5.5 = 67.25

Answer

ŷ = 67.25
9.Chi-square tests

Answer: χ² = 3.663

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

men: 60·78/120 = 39, 60·42/120 = 21
women: 60·78/120 = 39, 60·42/120 = 21

Chi-square statistic

χ² = (44 − 39)²/39 + (16 − 21)²/21 + (34 − 39)²/39 + (26 − 21)²/21
χ² = 3.663

Answer

χ² = 3.663

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

10.Chi-square tests

Answer: χ²crit = 9.488

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 = 6 − 1 − 1 = 4

Table

χ²crit(df = 4, α = 0.05) = 9.488

Answer

χ²crit = 9.488

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

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