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 0.9656. Find z to two decimals.
A = 0.9656
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 97.5% of them lie (the 97.5th percentile).
μ = 65σ = 10P = 97.5%
x = points
3.Confidence intervalsMedium
The blood pressure of 12 patients is measured before and after a treatment. The differences d = before − after (mmHg) have mean −6.7 and standard deviation 3.1. Find the lower limit of the 99% confidence interval for the mean difference.
n = 12d̄ = −6.7sd = 3.199%t(11) = 3.106
lower limit = mmHg
4.Confidence intervalsHard
The population standard deviation is 37. 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 4?
σ = 37E = 495%
n =
5.Hypothesis tests & confidence intervalsMedium
10 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 1.57 and the standard deviation of the differences is 3.1. Find the paired test statistic t for H₀: μd = 0.
n = 10d̄ = 1.57sd = 3.1
t =
6.Hypothesis tests & confidence intervalsHard
In a trial, 53 of 150 patients on treatment A recover, and 26 of 120 patients on treatment B. Using the pooled proportion, find the test statistic z for H₀: p₁ = p₂.
x₁ = 53n₁ = 150x₂ = 26n₂ = 120
z =
7.Linear regression & correlationMedium
The table gives 5 paired observations of x = temperature in °C and y = ice creams sold. Their sums are given. Find Pearson's correlation coefficient r from these sums.
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. Predict ŷ for x = 23.
n = 5x = 23
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 = 1602 × 3
χ² =
10.Chi-square testsHard
A chi-square goodness-of-fit test checks whether counts in 6 categories follow a Poisson distribution whose mean is estimated from the same data. Using the table, find the critical value χ²crit at significance level α = 0.05.
α = 0.05
χ²crit =
3Solutions
1.Normal distribution
Answer: z = −1.82
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 − 0.9656 = 0.0344
Read the z-table backwards
Find 0.0344 inside the table and read off z.
z = −1.82
Answer
z = −1.82
2.Normal distribution
Answer: x = 84.6 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) = 0.975 → z = 1.96
Back to x
x = 65 + 1.96·10 = 84.6 points
Answer
x = 84.6 points
3.Confidence intervals
Answer: lower limit = −9.48 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 = 11, t = 3.106
Margin of error
E = 3.106·3.1/√12 = 2.78
Interval
−6.7 ± 2.78 → [−9.48, −3.92]
Answer
lower limit = −9.48 mmHg
4.Confidence intervals
Answer: n = 329
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 = (1.96·37/4)² = 328.7
Round up
A sample size is a whole number, and rounding down would make the margin too large.
n = 329
Answer
n = 329
5.Hypothesis tests & confidence intervals
Answer: t = 1.602
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 = 3.1/√10 = 0.9803
Test statistic
t = 1.57/0.9803 = 1.602
Answer
t = 1.602
With df = 9, the two-sided 5% critical value is 2.262; |t| is smaller, so H₀ is not rejected.
6.Hypothesis tests & confidence intervals
Answer: z = 2.453
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 = 26/120 = 0.2167
p̂ = 79/270 = 0.2926
Standard error under H₀
SE = √(0.2926·0.7074·(1/150 + 1/120)) = 0.05572
Test statistic
z = (0.3533 − 0.2167)/0.05572 = 2.453
Answer
z = 2.453
|z| > 1.96, so H₀ is rejected at the 5% level (two-sided).
7.Linear regression & correlation
Answer: r = 0.9837
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 = 12285 − 120·498/5 = 333
Sxx = 2970 − 120²/5 = 90
Syy = 50874 − 498²/5 = 1273
Correlation coefficient
r = 333/√(90·1273) = 0.9837
Answer
r = 0.9837
8.Linear regression & correlation
Answer: ŷ = 74.75
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 = 100/5 = 20
ȳ = 316/5 = 63.2
Sums of squares and products
Sxx = Σ(x − x̄)² = 40
Sxy = Σ(x − x̄)(y − ȳ) = 154
Regression line
b = 154/40 = 3.85
a = 63.2 − 3.85·20 = −13.8
ŷ = −13.8 + 3.85·x
Prediction
ŷ = −13.8 + 3.85·23 = 74.75
Answer
ŷ = 74.75
9.Chi-square tests
Answer: χ² = 0.1778
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: χ²crit = 9.488
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 = 6 − 1 − 1 = 4
Table
χ²crit(df = 4, α = 0.05) = 9.488
Answer
χ²crit = 9.488
Reject H₀ if the computed χ² is larger than this value.