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 76 and 103 points? Give P as a decimal.
μ = 100σ = 15x1 = 76x2 = 103
P =
2.Normal distributionHard
The bag weights are normally distributed with mean 250 g and standard deviation 5 g. Drag the cut-off so the shaded bottom 95% is to the left of it. Find the cut-off x.
μ = 250 gσ = 5 g95%
x = g
3.Confidence intervalsMedium
The blood pressure of 11 patients is measured before and after a treatment. The differences d = before − after (mmHg) have mean 1.7 and standard deviation 0.6. Find the margin of error E of the 90% confidence interval for the mean difference.
n = 11d̄ = 1.7sd = 0.690%t(10) = 1.812
E = mmHg
4.Confidence intervalsHard
The population standard deviation is 16. 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 1.5?
σ = 16E = 1.599%
n =
5.Hypothesis tests & confidence intervalsMedium
A sample of 49 students has a mean score of 53.43 and a sample standard deviation of 10. Find the test statistic t for H₀: μ = 50.
μ₀ = 50x̄ = 53.43s = 10n = 49
t =
6.Hypothesis tests & confidence intervalsHard
A sample of 36 measurements has mean 501.5. The population standard deviation is 4. Find the lower limit L of the 90% confidence interval for the mean.
x̄ = 501.5σ = 4n = 3690%
L =
7.Linear regression & correlationMedium
The table gives 6 paired observations of x = hours studied and y = exam score. Their sums are given. Find Pearson's correlation coefficient r from these sums.
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 = 21.5.
n = 5x = 21.5
ŷ =
9.Chi-square testsMedium
The contingency table shows observed counts. For a chi-square test of independence, find the expected count E for the cell “staff, bike”.
row total = 45column total = 52N = 147
E =
10.Chi-square testsHard
The table shows observed counts. Test at α = 0.05 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 = 2203 × 2α = 0.05
χ² =
3Solutions
1.Normal distribution
Answer: P = 0.5245
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 = (76 − 100)/15 = −1.6
z₂ = (103 − 100)/15 = 0.2
Area between
P = Φ(0.2) − Φ(−1.6) = 0.5793 − 0.0548 = 0.5245
Answer
P = 0.5245
2.Normal distribution
Answer: x = 258.2 g
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
Φ(z) = 0.95
z from the table
z ≈ 1.64
Back to the original units
x = 250 + 1.64·5 = 258.2 g
Answer
x = 258.2 g
3.Confidence intervals
Answer: E = 0.3278 mmHg
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.
Critical value
df = 10, t = 1.812
Margin of error
E = 1.812·0.6/√11 = 0.3278
Answer
E = 0.3278 mmHg
4.Confidence intervals
Answer: n = 756
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 = (2.576·16/1.5)² = 755
Round up
A sample size is a whole number, and rounding down would make the margin too large.
n = 756
Answer
n = 756
5.Hypothesis tests & confidence intervals
Answer: t = 2.401
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/√49 = 10/7 = 1.429
Test statistic
t = (53.43 − 50)/1.429 = 2.401
Answer
t = 2.401
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 = 500.4
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 = 1.645·4/√36 = 1.097
Interval limit
L = 501.5 − 1.097 = 500.4
Answer
L = 500.4
7.Linear regression & correlation
Answer: r = 0.994
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 = 5531 − 48·633/6 = 467
Sxx = 454 − 48²/6 = 70
Syy = 69935 − 633²/6 = 3154
Correlation coefficient
r = 467/√(70·3154) = 0.994
Answer
r = 0.994
8.Linear regression & correlation
Answer: ŷ = 68.4
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 = 115/5 = 23
ȳ = 371/5 = 74.2
Sums of squares and products
Sxx = Σ(x − x̄)² = 90
Sxy = Σ(x − x̄)(y − ȳ) = 348
Regression line
b = 348/90 = 3.867
a = 74.2 − 3.867·23 = −14.73
ŷ = −14.73 + 3.867·x
Prediction
ŷ = −14.73 + 3.867·21.5 = 68.4
Answer
ŷ = 68.4
9.Chi-square tests
Answer: E = 15.92
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 count
E = 45·52/147 = 15.92
Answer
E = 15.92
10.Chi-square tests
Answer: χ² = 4.958
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.