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
The IQ scores are normally distributed with mean 100 points and standard deviation 15 points. What fraction lies between 100.75 and 103.75 points? Give P as a decimal.
μ = 100σ = 15x1 = 100.75x2 = 103.75
P =
2.Normal distributionHard
The bag weights are normally distributed with mean 250 g and standard deviation 5 g. Find the value x below which 25% of them lie (the 25th percentile).
μ = 250σ = 5P = 25%
x = g
3.Confidence intervalsMedium
The battery life of two phone models is compared. Model 1: 60 phones, mean 46.5 h, known standard deviation 20 h. Model 2: 40 phones, mean 32 h, known standard deviation 15 h. Find the upper limit of the 99% confidence interval for μ₁ − μ₂.
The population standard deviation is 27. How large a sample n do you need so that a 90% confidence interval for the mean has a margin of error of at most 1?
σ = 27E = 190%
n =
5.Hypothesis tests & confidence intervalsMedium
A sample of 25 students has a mean score of 504.2 and a sample standard deviation of 10. Find the test statistic t for H₀: μ = 500.
μ₀ = 500x̄ = 504.2s = 10n = 25
t =
6.Hypothesis tests & confidence intervalsHard
A sample of 100 measurements has mean 201. The population standard deviation is 12. Find the upper limit U of the 95% confidence interval for the mean.
x̄ = 201σ = 12n = 10095%
U =
7.Linear regression & correlationMedium
The table gives 5 paired observations of x = car age in years and y = price in k€. Find Pearson's correlation coefficient r.
n = 5
r =
8.Linear regression & correlationHard
The table gives 6 paired observations of x = hours studied and y = exam score. Use the least-squares line ŷ = a + b·x. For which value x does the line predict ŷ = 51?
n = 6ŷ = 51
x =
9.Chi-square testsMedium
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 × 3
χ² =
10.Chi-square testsHard
A chi-square test of independence is carried out on a contingency table with 2 rows and 6 columns (not counting the totals). Using the table, find the critical value χ²crit at significance level α = 0.01.
α = 0.01
χ²crit =
3Solutions
1.Normal distribution
Answer: P = 0.07877
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 = (100.75 − 100)/15 = 0.05
z₂ = (103.75 − 100)/15 = 0.25
Area between
P = Φ(0.25) − Φ(0.05) = 0.5987 − 0.5199 = 0.07877
Answer
P = 0.07877
2.Normal distribution
Answer: x = 246.6 g
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) = 0.25 → z = −0.6745
Back to x
x = 250 + −0.6745·5 = 246.63 g
Answer
x = 246.63 g
3.Confidence intervals
Answer: upper limit = 23.53 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 = 46.5 − 32 = 14.5
Standard error
SE = √(20²/60 + 15²/40) = 3.506
Margin of error
E = 2.576·3.506 = 9.031
Interval
14.5 ± 9.031 → [5.469, 23.53]
Answer
upper limit = 23.53 h
The interval does not contain 0, so the means differ significantly.
4.Confidence intervals
Answer: n = 1973
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 = (1.645·27/1)² = 1973
Round up
A sample size is a whole number, and rounding down would make the margin too large.
n = 1973
Answer
n = 1973
5.Hypothesis tests & confidence intervals
Answer: t = 2.1
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/√25 = 10/5 = 2
Test statistic
t = (504.2 − 500)/2 = 2.1
Answer
t = 2.1
t0.025; df = 24 = 2.064
|t| > 2.064, so H₀ is rejected at the 5% level (two-sided).
6.Hypothesis tests & confidence intervals
Answer: U = 203.4
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 = 1.96·12/√100 = 2.352
Interval limit
U = 201 + 2.352 = 203.4
Answer
U = 203.4
7.Linear regression & correlation
Answer: r = −0.9771
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 = 99/5 = 19.8
Sums of squares and products
Sxx = Σ(x − x̄)² = 10
Sxy = Σ(x − x̄)(y − ȳ) = −26
Syy = Σ(y − ȳ)² = 70.8
Correlation coefficient
r = −26/√(10·70.8) = −0.9771
Answer
r = −0.9771
8.Linear regression & correlation
Answer: x = 3.07
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 = 434/6 = 72.33
Sums of squares and products
Sxx = Σ(x − x̄)² = 70
Sxy = Σ(x − x̄)(y − ȳ) = 380
Regression line
b = 380/70 = 5.429
a = 72.33 − 5.429·7 = 34.33
ŷ = 34.33 + 5.429·x
Solve for x
x = (51 − 34.33)/5.429 = 3.07
Answer
x = 3.07
9.Chi-square tests
Answer: χ² = 5.533
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.
df = 2. At α = 0.05 the critical value is 5.991, so H₀ (independence) is not rejected.
10.Chi-square tests
Answer: χ²crit = 15.09
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)·(6 − 1) = 5
Table
χ²crit(df = 5, α = 0.01) = 15.086
Answer
χ²crit = 15.086
Reject H₀ if the computed χ² is larger than this value.