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 under the standard normal curve, to the right of z, is 0.2843. Find z to two decimals.
A = 0.2843
z =
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 top 40% is to the right of it. Find the cut-off x.
μ = 250 gσ = 5 g40%
x = g
3.Confidence intervalsMedium
A sample of 13 items has mean 24.5 and standard deviation s = 11. The population is roughly normal. Find the margin of error E of a 99% confidence interval for the mean.
n = 13x̄ = 24.5s = 1199%t(12) = 3.055
E =
4.Confidence intervalsHard
Two fertilisers are tested. Plot group 1: 6 plots, mean yield 70.5 kg, standard deviation 6 kg. Plot group 2: 8 plots, mean 72.5 kg, standard deviation 8.5 kg. Assume equal variances (pooled t, df = n₁ + n₂ − 2). Find the lower limit of the 90% confidence interval for μ₁ − μ₂.
Two teaching methods are compared. Group 1: 13 students, mean score 120, standard deviation 11.5. Group 2: 8 students, mean score 115.3, standard deviation 12. Assume equal population variances and use the pooled two-sample t test. Find the test statistic t for H₀: μ₁ = μ₂, taking the difference as x̄₁ − x̄₂ (group 1 minus group 2).
n₁ = 13x̄₁ = 120s₁ = 11.5n₂ = 8x̄₂ = 115.3s₂ = 12
t =
6.Hypothesis tests & confidence intervalsHard
A sample of 36 measurements has mean 197.6. The population standard deviation is 5. Find the lower limit L of the 90% confidence interval for the mean.
x̄ = 197.6σ = 5n = 3690%
L =
7.Linear regression & correlationMedium
The table gives 5 paired observations of x = hours studied and y = exam score. Find Pearson's correlation coefficient r.
n = 5
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. For which value x does the line predict ŷ = 89?
n = 5ŷ = 89
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 = 1002 × 2
χ² =
10.Chi-square testsHard
A chi-square test of independence is carried out on a contingency table with 2 rows and 11 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 = 0.57
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 − 0.2843 = 0.7157
Read the z-table backwards
Find 0.7157 inside the table and read off z.
z = 0.57
Answer
z = 0.57
2.Normal distribution
Answer: x = 251.3 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) = 0.6
z from the table
z ≈ 0.25
Back to the original units
x = 250 + 0.25·5 = 251.3 g
Answer
x = 251.3 g
3.Confidence intervals
Answer: E = 9.32
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.
Critical value
df = n − 1 = 12, t = 3.055
Margin of error
E = 3.055·11/√13 = 9.32
Answer
E = 9.32
4.Confidence intervals
Answer: lower limit = −9.275 kg
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.
Pooled standard deviation
sp² = (5·6² + 7·8.5²)/12 = 57.15
sp = 7.559
Margin of error
E = 1.782·7.559·√(1/6 + 1/8) = 7.275
Interval
−2 ± 7.275 → [−9.275, 5.275]
Answer
lower limit = −9.275 kg
5.Hypothesis tests & confidence intervals
Answer: t = 0.895
Hypothesis test: the test statistic = (estimate − value under H₀)/(standard error under H₀). The p-value is the probability, assuming H₀ is true, of a statistic at least as extreme as the one observed: the area in one tail for a one-sided H₁, twice the tail area for a two-sided H₁. Reject H₀ when p ≤ α.
Pooled variance
sp² = (12·11.5² + 7·12²)/19 = 136.6
sp = 11.69
Standard error
SE = 11.69·√(1/13 + 1/8) = 5.252
Test statistic
t = (120 − 115.3)/5.252 = 0.895
Answer
t = 0.895
With df = 19, the two-sided 5% critical value is 2.093; |t| is smaller, so H₀ is not rejected.
6.Hypothesis tests & confidence intervals
Answer: L = 196.2
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).
Margin of error
E = 1.645·5/√36 = 1.371
Interval limit
L = 197.6 − 1.371 = 196.2
Answer
L = 196.2
7.Linear regression & correlation
Answer: r = 0.983
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 = 35/5 = 7
ȳ = 374/5 = 74.8
Sums of squares and products
Sxx = Σ(x − x̄)² = 40
Sxy = Σ(x − x̄)(y − ȳ) = 200
Syy = Σ(y − ȳ)² = 1035
Correlation coefficient
r = 200/√(40·1035) = 0.983
Answer
r = 0.983
8.Linear regression & correlation
Answer: x = 24.01
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 = 105/5 = 21
ȳ = 390/5 = 78
Sums of squares and products
Sxx = Σ(x − x̄)² = 40
Sxy = Σ(x − x̄)(y − ȳ) = 146
Regression line
b = 146/40 = 3.65
a = 78 − 3.65·21 = 1.35
ŷ = 1.35 + 3.65·x
Solve for x
x = (89 − 1.35)/3.65 = 24.01
Answer
x = 24.01
9.Chi-square tests
Answer: χ² = 4.34
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 = 23.21
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 = (2 − 1)·(11 − 1) = 10
Table
χ²crit(df = 10, α = 0.01) = 23.209
Answer
χ²crit = 23.209
Reject H₀ if the computed χ² is larger than this value.