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 bag weights are normally distributed with mean 250 g and standard deviation 5 g. What fraction lies between 244.25 and 257.5 g? Give P as a decimal.
μ = 250σ = 5x1 = 244.25x2 = 257.5
P =
2.Normal distributionHard
The battery lifetimes are normally distributed with mean 500 h and standard deviation 40 h. Find the value x below which 75% of them lie (the 75th percentile).
μ = 500σ = 40P = 75%
x = h
3.Confidence intervalsMedium
A 95% confidence interval for a mean, based on 64 observations and a known σ, has a margin of error of 8.3. What is the margin of error E₂ of a 99% interval from the same data?
95%n₁ = 64E₁ = 8.399%
E₂ =
4.Confidence intervalsHard
The population standard deviation is 37. 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 2?
σ = 37E = 299%
n =
5.Hypothesis tests & confidence intervalsMedium
A sample of 25 students has a mean score of 494.4 and a sample standard deviation of 20. Find the test statistic t for H₀: μ = 500.
μ₀ = 500x̄ = 494.4s = 20n = 25
t =
6.Hypothesis tests & confidence intervalsHard
A sample of 16 measurements has mean 96.3. The population standard deviation is 10. Find the upper limit U of the 99% confidence interval for the mean.
x̄ = 96.3σ = 10n = 1699%
U =
7.Linear regression & correlationMedium
The table gives 6 paired observations of x = hours studied and y = exam score. Find Pearson's correlation coefficient r.
n = 6
r =
8.Linear regression & correlationHard
The table gives 5 paired observations of x = temperature in °C and y = ice creams sold. Use the least-squares line ŷ = a + b·x. Predict ŷ for x = 17.
n = 5x = 17
ŷ =
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 = 1402 × 3
χ² =
10.Chi-square testsHard
The table shows observed counts. Test at α = 0.01 whether the two variables are independent (no continuity correction). Compute the test statistic χ², then compare it with the critical value from the table.
N = 1403 × 2α = 0.01
χ² =
3Solutions
1.Normal distribution
Answer: P = 0.8081
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 = (244.25 − 250)/5 = −1.15
z₂ = (257.5 − 250)/5 = 1.5
Area between
P = Φ(1.5) − Φ(−1.15) = 0.9332 − 0.1251 = 0.8081
Answer
P = 0.8081
2.Normal distribution
Answer: x = 527 h
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) = 0.75 → z = 0.6745
Back to x
x = 500 + 0.6745·40 = 526.98 h
Answer
x = 526.98 h
3.Confidence intervals
Answer: E₂ = 10.91
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 = (2.576/1.96)·√(64/64) = 1.314
New margin
E₂ = 8.3·1.314 = 10.91
Answer
E₂ = 10.91
4.Confidence intervals
Answer: n = 2272
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 = (2.576·37/2)² = 2271
Round up
A sample size is a whole number, and rounding down would make the margin too large.
n = 2272
Answer
n = 2272
5.Hypothesis tests & confidence intervals
Answer: t = −1.4
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).
Standard error
SE = 20/√25 = 20/5 = 4
Test statistic
t = (494.4 − 500)/4 = −1.4
Answer
t = −1.4
t0.025; df = 24 = 2.064
|t| < 2.064, so H₀ is not rejected at the 5% level (two-sided).
6.Hypothesis tests & confidence intervals
Answer: U = 102.7
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 = 2.576·10/√16 = 6.44
Interval limit
U = 96.3 + 6.44 = 102.7
Answer
U = 102.7
7.Linear regression & correlation
Answer: r = 0.9609
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 = 27/6 = 4.5
ȳ = 389/6 = 64.83
Sums of squares and products
Sxx = Σ(x − x̄)² = 17.5
Sxy = Σ(x − x̄)(y − ȳ) = 90.5
Syy = Σ(y − ȳ)² = 506.8
Correlation coefficient
r = 90.5/√(17.5·506.8) = 0.9609
Answer
r = 0.9609
8.Linear regression & correlation
Answer: ŷ = 60.4
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 = 100/5 = 20
ȳ = 368/5 = 73.6
Sums of squares and products
Sxx = Σ(x − x̄)² = 40
Sxy = Σ(x − x̄)(y − ȳ) = 176
Regression line
b = 176/40 = 4.4
a = 73.6 − 4.4·20 = −14.4
ŷ = −14.4 + 4.4·x
Prediction
ŷ = −14.4 + 4.4·17 = 60.4
Answer
ŷ = 60.4
9.Chi-square tests
Answer: χ² = 4.69
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: χ² = 7.143
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.