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
Drag z so that the shaded area between −z and z under the standard normal curve is 0.4907. Find z to two decimals.
A = 0.4907
z =
2.Normal distributionHard
The IQ scores are normally distributed with mean 100 points and standard deviation 15 points. Drag the cut-off so the shaded top 25% is to the right of it. Find the cut-off x.
μ = 100 pointsσ = 15 points25%
x = points
3.Confidence intervalsMedium
The battery life of two phone models is compared. Model 1: 60 phones, mean 42.5 h, known standard deviation 13 h. Model 2: 60 phones, mean 36 h, known standard deviation 11 h. Find the lower limit of the 95% confidence interval for μ₁ − μ₂.
The population standard deviation is 40. 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?
σ = 40E = 199%
n =
5.Hypothesis tests & confidence intervalsMedium
16 employees take a typing test before and after a training course. For each person the difference d = after − before (words per minute) is recorded. The mean difference is −3.25 and the standard deviation of the differences is 6.5. Find the paired test statistic t for H₀: μd = 0.
n = 16d̄ = −3.25sd = 6.5
t =
6.Hypothesis tests & confidence intervalsHard
A sample of 36 measurements has mean 72.3. The population standard deviation is 8. Find the lower limit L of the 99% confidence interval for the mean.
x̄ = 72.3σ = 8n = 3699%
L =
7.Linear regression & correlationMedium
The table gives 5 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 6 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 = 20.5.
n = 6x = 20.5
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 = 1802 × 3
χ² =
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). Using the table, find the critical value χ²crit at significance level α = 0.01.
α = 0.01
χ²crit =
3Solutions
1.Normal distribution
Answer: z = 0.66
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 of z
Φ(z) = (1 + 0.4907)/2 = 0.7454
Read the z-table backwards
Find 0.7454 inside the table and read off z.
z = 0.66
Answer
z = 0.66
2.Normal distribution
Answer: x = 110.1 points
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 = 100 + 0.67·15 = 110.1 points
Answer
x = 110.1 points
3.Confidence intervals
Answer: lower limit = 2.191 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 = 42.5 − 36 = 6.5
Standard error
SE = √(13²/60 + 11²/60) = 2.198
Margin of error
E = 1.96·2.198 = 4.309
Interval
6.5 ± 4.309 → [2.191, 10.81]
Answer
lower limit = 2.191 h
The interval does not contain 0, so the means differ significantly.
4.Confidence intervals
Answer: n = 10618
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 = (2.576·40/1)² = 10617
Round up
A sample size is a whole number, and rounding down would make the margin too large.
n = 10618
Answer
n = 10618
5.Hypothesis tests & confidence intervals
Answer: t = −2
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 of the mean difference
SE = 6.5/√16 = 1.625
Test statistic
t = −3.25/1.625 = −2
Answer
t = −2
With df = 15, the two-sided 5% critical value is 2.131; |t| is smaller, so H₀ is not rejected.
6.Hypothesis tests & confidence intervals
Answer: L = 68.87
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 = 2.576·8/√36 = 3.435
Interval limit
L = 72.3 − 3.435 = 68.87
Answer
L = 68.87
7.Linear regression & correlation
Answer: r = 0.9898
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 = 2374 − 30·362/5 = 202
Sxx = 220 − 30²/5 = 40
Syy = 27250 − 362²/5 = 1041
Correlation coefficient
r = 202/√(40·1041) = 0.9898
Answer
r = 0.9898
8.Linear regression & correlation
Answer: ŷ = 69.01
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 = 141/6 = 23.5
ȳ = 489/6 = 81.5
Sums of squares and products
Sxx = Σ(x − x̄)² = 157.5
Sxy = Σ(x − x̄)(y − ȳ) = 655.5
Regression line
b = 655.5/157.5 = 4.162
a = 81.5 − 4.162·23.5 = −16.3
ŷ = −16.3 + 4.162·x
Prediction
ŷ = −16.3 + 4.162·20.5 = 69.01
Answer
ŷ = 69.01
9.Chi-square tests
Answer: χ² = 15.87
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 rejected.
10.Chi-square tests
Answer: χ²crit = 21.67
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)·(4 − 1) = 9
Table
χ²crit(df = 9, α = 0.01) = 21.666
Answer
χ²crit = 21.666
Reject H₀ if the computed χ² is larger than this value.