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 exam scores are normally distributed with mean 65 points and standard deviation 10 points. What fraction lies between 50.5 and 75.5 points? Give P as a decimal.
μ = 65σ = 10x1 = 50.5x2 = 75.5
P =
2.Normal distributionHard
The IQ scores are normally distributed with mean 100 points and standard deviation 15 points. Find the value x below which 99% of them lie (the 99th percentile).
μ = 100σ = 15P = 99%
x = points
3.Confidence intervalsMedium
A 99% confidence interval for a mean, based on 64 observations and a known σ, has a margin of error of 5.2. What is the margin of error E₂ of a 90% interval based on 16 observations?
99%n₁ = 64E₁ = 5.290%n₂ = 16
E₂ =
4.Confidence intervalsHard
The population standard deviation is 21. 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 = 21E = 190%
n =
5.Hypothesis tests & confidence intervalsMedium
A sample of 49 students has a mean score of 115.5 and a sample standard deviation of 12. Find the test statistic t for H₀: μ = 120.
μ₀ = 120x̄ = 115.5s = 12n = 49
t =
6.Hypothesis tests & confidence intervalsHard
A sample of 25 measurements has mean 113.7. The population standard deviation is 15. Find the lower limit L of the 95% confidence interval for the mean.
x̄ = 113.7σ = 15n = 2595%
L =
7.Linear regression & correlationMedium
The table gives 5 paired observations of x = temperature in °C and y = ice creams sold. Find the coefficient of determination r² of the least-squares line.
n = 5
r² =
8.Linear regression & correlationHard
The table gives 6 paired observations of x = car age in years and y = price in k€. Use the least-squares line ŷ = a + b·x. For which value x does the line predict ŷ = 26?
n = 6ŷ = 26
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 = 1602 × 2
χ² =
10.Chi-square testsHard
A chi-square goodness-of-fit test checks whether counts in 4 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 = 0.01.
α = 0.01
χ²crit =
3Solutions
1.Normal distribution
Answer: P = 0.7796
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.
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) = 0.99 → z = 2.326
Back to x
x = 100 + 2.326·15 = 134.89 points
Answer
x = 134.89 points
3.Confidence intervals
Answer: E₂ = 6.641
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 = (1.645/2.576)·√(64/16) = 1.277
New margin
E₂ = 5.2·1.277 = 6.641
Answer
E₂ = 6.641
4.Confidence intervals
Answer: n = 1194
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 = (1.645·21/1)² = 1193
Round up
A sample size is a whole number, and rounding down would make the margin too large.
n = 1194
Answer
n = 1194
5.Hypothesis tests & confidence intervals
Answer: t = −2.625
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/√49 = 12/7 = 1.714
Test statistic
t = (115.5 − 120)/1.714 = −2.625
Answer
t = −2.625
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: L = 107.8
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 = 1.96·15/√25 = 5.88
Interval limit
L = 113.7 − 5.88 = 107.8
Answer
L = 107.8
7.Linear regression & correlation
Answer: r² = 0.9644
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 = 110/5 = 22
ȳ = 361/5 = 72.2
Sums of squares and products
Sxx = Σ(x − x̄)² = 40
Sxy = Σ(x − x̄)(y − ȳ) = 154
Syy = Σ(y − ȳ)² = 614.8
Coefficient of determination
r² = 154²/(40·614.8) = 0.9644
Answer
r² = 0.9644
About 96.44% of the variation in y is explained by the line.
8.Linear regression & correlation
Answer: x = 1.344
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 = 122/6 = 20.33
Sums of squares and products
Sxx = Σ(x − x̄)² = 17.5
Sxy = Σ(x − x̄)(y − ȳ) = −46
Regression line
b = −46/17.5 = −2.629
a = 20.33 − (−2.629)·3.5 = 29.53
ŷ = 29.53 − 2.629·x
Solve for x
x = (26 − 29.53)/(−2.629) = 1.344
Answer
x = 1.344
9.Chi-square tests
Answer: χ² = 6.533
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 = 1. At α = 0.05 the critical value is 3.841, so H₀ (independence) is rejected.
10.Chi-square tests
Answer: χ²crit = 9.21
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 = 4 − 1 − 1 = 2
Table
χ²crit(df = 2, α = 0.01) = 9.210
Answer
χ²crit = 9.210
Reject H₀ if the computed χ² is larger than this value.