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 0.2177. Find z to two decimals.
A = 0.2177
z =
2.Normal distributionHard
The exam scores are normally distributed with mean 65 points and standard deviation 10 points. Find the value x below which 95% of them lie (the 95th percentile).
μ = 65σ = 10P = 95%
x = points
3.Confidence intervalsMedium
A 99% confidence interval for a mean, based on 36 observations and a known σ, has a margin of error of 3. What is the margin of error E₂ of a 90% interval based on 144 observations?
99%n₁ = 36E₁ = 390%n₂ = 144
E₂ =
4.Confidence intervalsHard
The population standard deviation is 24. How large a sample n do you need so that a 95% confidence interval for the mean has a margin of error of at most 1.5?
σ = 24E = 1.595%
n =
5.Hypothesis tests & confidence intervalsMedium
A sample of 36 students has a mean score of 73.83 and a sample standard deviation of 5. Find the test statistic t for H₀: μ = 75.
μ₀ = 75x̄ = 73.83s = 5n = 36
t =
6.Hypothesis tests & confidence intervalsHard
In a trial, 42 of 120 patients on treatment A recover, and 18 of 80 patients on treatment B. Using the pooled proportion, find the test statistic z for H₀: p₁ = p₂.
x₁ = 42n₁ = 120x₂ = 18n₂ = 80
z =
7.Linear regression & correlationMedium
The table gives 6 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 5 paired observations of x = hours studied and y = exam score. Use the least-squares line ŷ = a + b·x. A new observation (5.5, 75) is made. Find its residual e = y − ŷ.
n = 5(5.5, 75)
e =
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, no defect”.
row total = 94column total = 74N = 185
E =
10.Chi-square testsHard
The table shows observed counts. Test at α = 0.01 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 = 1803 × 2α = 0.01
χ² =
3Solutions
1.Normal distribution
Answer: z = 0.78
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 − 0.2177 = 0.7823
Read the z-table backwards
Find 0.7823 inside the table and read off z.
z = 0.78
Answer
z = 0.78
2.Normal distribution
Answer: x = 81.45 points
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 for the percentile
Φ(z) = 0.95 → z = 1.645
Back to x
x = 65 + 1.645·10 = 81.45 points
Answer
x = 81.45 points
3.Confidence intervals
Answer: E₂ = 0.9579
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.
Ratio of margins
E₂/E₁ = (1.645/2.576)·√(36/144) = 0.3193
New margin
E₂ = 3·0.3193 = 0.9579
Answer
E₂ = 0.9579
4.Confidence intervals
Answer: n = 984
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 = (1.96·24/1.5)² = 983.4
Round up
A sample size is a whole number, and rounding down would make the margin too large.
n = 984
Answer
n = 984
5.Hypothesis tests & confidence intervals
Answer: t = −1.404
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 = 5/√36 = 5/6 = 0.8333
Test statistic
t = (73.83 − 75)/0.8333 = −1.404
Answer
t = −1.404
t0.025; df = 35 = 2.03
|t| < 2.03, so H₀ is not rejected at the 5% level (two-sided).
6.Hypothesis tests & confidence intervals
Answer: z = 1.89
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 = 42/120 = 0.35
p̂₂ = 18/80 = 0.225
p̂ = 60/200 = 0.3
Standard error under H₀
SE = √(0.3·0.7·(1/120 + 1/80)) = 0.06614
Test statistic
z = (0.35 − 0.225)/0.06614 = 1.89
Answer
z = 1.89
|z| < 1.96, so H₀ is not rejected at the 5% level (two-sided).
7.Linear regression & correlation
Answer: r = 0.9987
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 = 3036 − 42·373/6 = 425
Sxx = 364 − 42²/6 = 70
Syy = 25775 − 373²/6 = 2587
Correlation coefficient
r = 425/√(70·2587) = 0.9987
Answer
r = 0.9987
8.Linear regression & correlation
Answer: e = 1.7
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 = 25/5 = 5
ȳ = 351/5 = 70.2
Sums of squares and products
Sxx = Σ(x − x̄)² = 10
Sxy = Σ(x − x̄)(y − ȳ) = 62
Regression line
b = 62/10 = 6.2
a = 70.2 − 6.2·5 = 39.2
ŷ = 39.2 + 6.2·x
Predicted value
ŷ = 39.2 + 6.2·5.5 = 73.3
Residual
e = 75 − 73.3 = 1.7
Answer
e = 1.7
The point lies above the line.
9.Chi-square tests
Answer: E = 37.6
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.
Expected count
E = 94·74/185 = 37.6
Answer
E = 37.6
10.Chi-square tests
Answer: χ² = 10.07
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.