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 245 and 257.5 g? Give P as a decimal.
μ = 250σ = 5x1 = 245x2 = 257.5
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 75% is to the left of it. Find the cut-off x.
μ = 250 gσ = 5 g75%
x = g
3.Confidence intervalsMedium
The blood pressure of 20 patients is measured before and after a treatment. The differences d = before − after (mmHg) have mean −2.1 and standard deviation 3.1. Find the upper limit of the 99% confidence interval for the mean difference.
n = 20d̄ = −2.1sd = 3.199%t(19) = 2.861
upper limit = mmHg
4.Confidence intervalsHard
The population standard deviation is 32. 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 4?
σ = 32E = 490%
n =
5.Hypothesis tests & confidence intervalsMedium
A sample of 81 students has a mean score of 123.2 and a sample standard deviation of 10. Find the test statistic t for H₀: μ = 120.
μ₀ = 120x̄ = 123.2s = 10n = 81
t =
6.Hypothesis tests & confidence intervalsHard
A sample of 49 measurements has mean 50.2. The population standard deviation is 4. Find the lower limit L of the 99% confidence interval for the mean.
x̄ = 50.2σ = 4n = 4999%
L =
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 = load in kN and y = extension in mm. Use the least-squares line ŷ = a + b·x. Predict ŷ for x = 9.
n = 5x = 9
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 “night shift, major”.
row total = 76column total = 51N = 144
E =
10.Chi-square testsHard
A chi-square test of independence is carried out on a contingency table with 4 rows and 4 columns (not counting the totals). How many degrees of freedom df does the test have?
4 × 4
df =
3Solutions
1.Normal distribution
Answer: P = 0.7745
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 = (245 − 250)/5 = −1
z₂ = (257.5 − 250)/5 = 1.5
Area between
P = Φ(1.5) − Φ(−1) = 0.9332 − 0.1587 = 0.7745
Answer
P = 0.7745
2.Normal distribution
Answer: x = 253.4 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) = 0.75
z from the table
z ≈ 0.67
Back to the original units
x = 250 + 0.67·5 = 253.4 g
Answer
x = 253.4 g
3.Confidence intervals
Answer: upper limit = −0.1168 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 = 19, t = 2.861
Margin of error
E = 2.861·3.1/√20 = 1.983
Interval
−2.1 ± 1.983 → [−4.083, −0.1168]
Answer
upper limit = −0.1168 mmHg
4.Confidence intervals
Answer: n = 174
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 = (1.645·32/4)² = 173.2
Round up
A sample size is a whole number, and rounding down would make the margin too large.
n = 174
Answer
n = 174
5.Hypothesis tests & confidence intervals
Answer: t = 2.88
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/√81 = 10/9 = 1.111
Test statistic
t = (123.2 − 120)/1.111 = 2.88
Answer
t = 2.88
t0.025; df = 80 = 1.99
|t| > 1.99, so H₀ is rejected at the 5% level (two-sided).
6.Hypothesis tests & confidence intervals
Answer: L = 48.73
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 = 2.576·4/√49 = 1.472
Interval limit
L = 50.2 − 1.472 = 48.73
Answer
L = 48.73
7.Linear regression & correlation
Answer: r = 0.9702
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 = 48/6 = 8
ȳ = 368/6 = 61.33
Sums of squares and products
Sxx = Σ(x − x̄)² = 70
Sxy = Σ(x − x̄)(y − ȳ) = 236
Syy = Σ(y − ȳ)² = 845.3
Correlation coefficient
r = 236/√(70·845.3) = 0.9702
Answer
r = 0.9702
8.Linear regression & correlation
Answer: ŷ = 15.4
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 = 30/5 = 6
ȳ = 59/5 = 11.8
Sums of squares and products
Sxx = Σ(x − x̄)² = 40
Sxy = Σ(x − x̄)(y − ȳ) = 48
Regression line
b = 48/40 = 1.2
a = 11.8 − 1.2·6 = 4.6
ŷ = 4.6 + 1.2·x
Prediction
ŷ = 4.6 + 1.2·9 = 15.4
Answer
ŷ = 15.4
9.Chi-square tests
Answer: E = 26.92
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 = 76·51/144 = 26.92
Answer
E = 26.92
10.Chi-square tests
Answer: df = 9
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.