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 0.7813. Find z to two decimals.
A = 0.7813
z =
2.Normal distributionHard
The heights of adult men are normally distributed with mean 178 cm and standard deviation 7 cm. Drag the cut-off so the shaded top 85% is to the right of it. Find the cut-off x.
μ = 178 cmσ = 7 cm85%
x = cm
3.Confidence intervalsMedium
A sample of 10 items has mean 31 and standard deviation s = 5.5. The population is roughly normal. Find the margin of error E of a 90% confidence interval for the mean.
n = 10x̄ = 31s = 5.590%t(9) = 1.833
E =
4.Confidence intervalsHard
The population standard deviation is 36. 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 5?
σ = 36E = 599%
n =
5.Hypothesis tests & confidence intervalsMedium
A sample of 25 students has a mean score of 62.6 and a sample standard deviation of 20. Find the test statistic t for H₀: μ = 75.
μ₀ = 75x̄ = 62.6s = 20n = 25
t =
6.Hypothesis tests & confidence intervalsHard
In a trial, 53 of 150 patients on treatment A recover, and 42 of 200 patients on treatment B. Using the pooled proportion, find the test statistic z for H₀: p₁ = p₂.
x₁ = 53n₁ = 150x₂ = 42n₂ = 200
z =
7.Linear regression & correlationMedium
The table gives 5 paired observations of x = advertising spend in k€ and y = sales in k€. 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 = 29.5.
n = 6x = 29.5
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 = 2002 × 3
χ² =
10.Chi-square testsHard
A chi-square goodness-of-fit test checks whether counts in 8 categories follow a Poisson distribution whose mean is estimated from the same data. How many degrees of freedom df does the test have?
k = 8
df =
3Solutions
1.Normal distribution
Answer: z = 1.23
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 + 0.7813)/2 = 0.8907
Read the z-table backwards
Find 0.8907 inside the table and read off z.
z = 1.23
Answer
z = 1.23
2.Normal distribution
Answer: x = 170.7 cm
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) = 0.15
z from the table
z ≈ −1.04
Back to the original units
x = 178 − 1.04·7 = 170.7 cm
Answer
x = 170.7 cm
3.Confidence intervals
Answer: E = 3.188
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 = 9, t = 1.833
Margin of error
E = 1.833·5.5/√10 = 3.188
Answer
E = 3.188
4.Confidence intervals
Answer: n = 344
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 = (2.576·36/5)² = 344
Round up
A sample size is a whole number, and rounding down would make the margin too large.
n = 344
Answer
n = 344
5.Hypothesis tests & confidence intervals
Answer: t = −3.1
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 = (62.6 − 75)/4 = −3.1
Answer
t = −3.1
t0.025; df = 24 = 2.064
|t| > 2.064, so H₀ is rejected at the 5% level (two-sided).
6.Hypothesis tests & confidence intervals
Answer: z = 2.984
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.
Sample proportions
p̂₁ = 53/150 = 0.3533
p̂₂ = 42/200 = 0.21
p̂ = 95/350 = 0.2714
Standard error under H₀
SE = √(0.2714·0.7286·(1/150 + 1/200)) = 0.04803
Test statistic
z = (0.3533 − 0.21)/0.04803 = 2.984
Answer
z = 2.984
|z| > 1.96, so H₀ is rejected at the 5% level (two-sided).
7.Linear regression & correlation
Answer: r = 0.8941
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 = 1314 − 25·257/5 = 29
Sxx = 135 − 25²/5 = 10
Syy = 13315 − 257²/5 = 105.2
Correlation coefficient
r = 29/√(10·105.2) = 0.8941
Answer
r = 0.8941
8.Linear regression & correlation
Answer: ŷ = 124.8
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 = 586/6 = 97.67
Sums of squares and products
Sxx = Σ(x − x̄)² = 157.5
Sxy = Σ(x − x̄)(y − ȳ) = 711
Regression line
b = 711/157.5 = 4.514
a = 97.67 − 4.514·23.5 = −8.419
ŷ = −8.419 + 4.514·x
Prediction
ŷ = −8.419 + 4.514·29.5 = 124.8
Answer
ŷ = 124.8
9.Chi-square tests
Answer: χ² = 4.516
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: df = 6
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.