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 0.2912. Find z to two decimals.
A = 0.2912
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 40% is to the right of it. Find the cut-off x.
μ = 178 cmσ = 7 cm40%
x = cm
3.Confidence intervalsMedium
The battery life of two phone models is compared. Model 1: 50 phones, mean 62.5 h, known standard deviation 14 h. Model 2: 80 phones, mean 68 h, known standard deviation 6 h. Find the upper limit of the 95% confidence interval for μ₁ − μ₂.
The population standard deviation is 27. 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 = 27E = 599%
n =
5.Hypothesis tests & confidence intervalsMedium
A sample of 16 students has a mean score of 193 and a sample standard deviation of 20. Find the test statistic t for H₀: μ = 200.
μ₀ = 200x̄ = 193s = 20n = 16
t =
6.Hypothesis tests & confidence intervalsHard
A one-sided z test at α = 0.01 must detect a shift of the mean of 5 units with a power of 90%. The standard deviation is 13. Find the minimum sample size n.
The table gives 6 paired observations of x = hours studied and y = exam score. Find Pearson's correlation coefficient r.
n = 6
r =
8.Linear regression & correlationHard
The table gives 5 paired observations of x = hours studied and y = exam score. Use the least-squares line ŷ = a + b·x. For which value x does the line predict ŷ = 72?
n = 5ŷ = 72
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 = 2002 × 3
χ² =
10.Chi-square testsHard
A chi-square test of independence is carried out on a contingency table with 3 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 = 0.55
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 − 0.2912 = 0.7088
Read the z-table backwards
Find 0.7088 inside the table and read off z.
z = 0.55
Answer
z = 0.55
2.Normal distribution
Answer: x = 179.8 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) = 0.6
z from the table
z ≈ 0.25
Back to the original units
x = 178 + 0.25·7 = 179.8 cm
Answer
x = 179.8 cm
3.Confidence intervals
Answer: upper limit = −1.403 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 = 62.5 − 68 = −5.5
Standard error
SE = √(14²/50 + 6²/80) = 2.09
Margin of error
E = 1.96·2.09 = 4.097
Interval
−5.5 ± 4.097 → [−9.597, −1.403]
Answer
upper limit = −1.403 h
The interval does not contain 0, so the means differ significantly.
4.Confidence intervals
Answer: n = 194
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 = (2.576·27/5)² = 193.5
Round up
A sample size is a whole number, and rounding down would make the margin too large.
n = 194
Answer
n = 194
5.Hypothesis tests & confidence intervals
Answer: t = −1.4
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/√16 = 20/4 = 5
Test statistic
t = (193 − 200)/5 = −1.4
Answer
t = −1.4
t0.025; df = 15 = 2.131
|t| < 2.131, so H₀ is not rejected at the 5% level (two-sided).
6.Hypothesis tests & confidence intervals
Answer: n = 88
Sample size for a test: to detect a true shift δ with power 1 − β in a one-sided z test at level α, the standard error must satisfy δ = (z_α + z_β)·σ/√n, so n = ((z_α + z_β)·σ/δ)², rounded up.
Solve for n
n = ((2.326 + 1.282)·13/5)² = 88
Round up
n = 88
Answer
n = 88
7.Linear regression & correlation
Answer: r = 0.9942
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 = 48/6 = 8
ȳ = 617/6 = 102.8
Sums of squares and products
Sxx = Σ(x − x̄)² = 70
Sxy = Σ(x − x̄)(y − ȳ) = 509
Syy = Σ(y − ȳ)² = 3745
Correlation coefficient
r = 509/√(70·3745) = 0.9942
Answer
r = 0.9942
8.Linear regression & correlation
Answer: x = 3.545
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 = 15/5 = 3
ȳ = 342/5 = 68.4
Sums of squares and products
Sxx = Σ(x − x̄)² = 10
Sxy = Σ(x − x̄)(y − ȳ) = 66
Regression line
b = 66/10 = 6.6
a = 68.4 − 6.6·3 = 48.6
ŷ = 48.6 + 6.6·x
Solve for x
x = (72 − 48.6)/6.6 = 3.545
Answer
x = 3.545
9.Chi-square tests
Answer: χ² = 4.194
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 = 16.81
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 = (3 − 1)·(4 − 1) = 6
Table
χ²crit(df = 6, α = 0.01) = 16.812
Answer
χ²crit = 16.812
Reject H₀ if the computed χ² is larger than this value.