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
50.575P = ?

The exam scores are normally distributed with mean 65 points and standard deviation 10 points. What fraction lies between 50.5 and 75 points? Give P as a decimal.

μ = 65σ = 10x1 = 50.5x2 = 75
P =
2.Normal distributionHard
μ = 178x = ?10%

The heights of adult men are normally distributed with mean 178 cm and standard deviation 7 cm. Find the value x below which 10% of them lie (the 10th percentile).

μ = 178σ = 7P = 10%
x = cm
3.Confidence intervalsMedium
n = 6, s = 9LU = ?x̄ = 40.5E99%

A sample of 6 items has mean 40.5 and standard deviation s = 9. The population is roughly normal. Find the upper limit of the 99% confidence interval for the mean.

n = 6x̄ = 40.5s = 999%t(5) = 4.032
upper limit =
4.Confidence intervalsHard
σ = 27, n = ?x̄ − Ex̄ + Ex̄E = 2.595%

The population standard deviation is 27. 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 2.5?

σ = 27E = 2.595%
n =
5.Hypothesis tests & confidence intervalsMedium
−1.961.96z = ?

Two production lines make the same part. Line 1: a sample of 30 parts has mean length 103 mm, and the line's known standard deviation is 7 mm. Line 2: a sample of 40 parts has mean 106.9 mm, known standard deviation 19 mm. Find the test statistic z for H₀: μ₁ = μ₂, taking the difference as x̄₁ − x̄₂ (line 1 minus line 2).

n₁ = 30x̄₁ = 103 mmσ₁ = 7 mmn₂ = 40x̄₂ = 106.9 mmσ₂ = 19 mm
z =
6.Hypothesis tests & confidence intervalsHard
σ = 5, n = ?x̄ − Ex̄ + Ex̄E = 1.599%

How large must a sample be so that the 99% confidence interval for a mean has a margin of error of at most 1.5? The standard deviation is 5. Find the smallest sample size n.

σ = 5E = 1.599%
n =
7.Linear regression & correlationMedium
xy125220317416510610xy24610152025

The table gives 6 paired observations of x = car age in years and y = price in k€. Find the coefficient of determination r² of the least-squares line.

n = 6
r² =
8.Linear regression & correlationHard
xy2334346368431052xy5104050x = 3

The table gives 5 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 (3, 35) is made. Find its residual e = y − ŷ.

n = 5(3, 35)
e =
9.Chi-square testsMedium
yesnoTotalmen324880women312960Total6377140

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
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 11 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 = 0.7678

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 = (50.5 − 65)/10 = −1.45
z₂ = (75 − 65)/10 = 1

Area between

P = Φ(1) − Φ(−1.45) = 0.8413 − 0.07353 = 0.7678

Answer

P = 0.7678
2.Normal distribution

Answer: x = 169 cm

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) = 0.1 → z = −1.282

Back to x

x = 178 + −1.282·7 = 169.03 cm

Answer

x = 169.03 cm
3.Confidence intervals

Answer: upper limit = 55.31

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 = 5, t = 4.032

Margin of error

E = 4.032·9/√6 = 14.81

Interval

40.5 ± 14.81 → [25.69, 55.31]

Answer

upper limit = 55.31
4.Confidence intervals

Answer: n = 449

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 = (1.96·27/2.5)² = 448.1

Round up

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

n = 449

Answer

n = 449
5.Hypothesis tests & confidence intervals

Answer: z = −1.195

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.

Standard error of the difference

SE = √(7²/30 + 19²/40) = √10.66 = 3.265

Test statistic

z = (103 − 106.9)/3.265 = −1.195

Answer

z = −1.195

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

6.Hypothesis tests & confidence intervals

Answer: n = 74

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).

Solve for n

n = (2.576·5/1.5)² = 73.73

Round up

n = 74

Answer

n = 74
7.Linear regression & correlation

Answer: r² = 0.9479

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 = 98/6 = 16.33

Sums of squares and products

Sxx = Σ(x − x̄)² = 17.5
Sxy = Σ(x − x̄)(y − ȳ) = −53
Syy = Σ(y − ȳ)² = 169.3

Coefficient of determination

r² = −53²/(17.5·169.3) = 0.9479

Answer

r² = 0.9479

About 94.79% of the variation in y is explained by the line.

8.Linear regression & correlation

Answer: e = 2.45

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 = 198/5 = 39.6

Sums of squares and products

Sxx = Σ(x − x̄)² = 40
Sxy = Σ(x − x̄)(y − ȳ) = 94

Regression line

b = 94/40 = 2.35
a = 39.6 − 2.35·6 = 25.5
ŷ = 25.5 + 2.35·x

Predicted value

ŷ = 25.5 + 2.35·3 = 32.55

Residual

e = 35 − 32.55 = 2.45

Answer

e = 2.45

The point lies above the line.

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

men: 80·63/140 = 36, 80·77/140 = 44
women: 60·63/140 = 27, 60·77/140 = 33

Chi-square statistic

χ² = (32 − 36)²/36 + (48 − 44)²/44 + (31 − 27)²/27 + (29 − 33)²/33
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: χ²crit = 18.31

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.

Degrees of freedom

df = (2 − 1)·(11 − 1) = 10

Table

χ²crit(df = 10, α = 0.05) = 18.307

Answer

χ²crit = 18.307

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

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