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 battery lifetimes are normally distributed with mean 500 h and standard deviation 40 h. Drag the cut-off so the shaded top 85% is to the right of it. Find the cut-off x.
μ = 500 hσ = 40 h85%
x = h
3.Confidence intervalsMedium
A sample of 6 items has mean 85 and standard deviation s = 1. The population is roughly normal. Find the lower limit of the 95% confidence interval for the mean.
n = 6x̄ = 85s = 195%t(5) = 2.571
lower limit =
4.Confidence intervalsHard
Nothing is known in advance about the proportion of defective products. How many products n must be inspected so that a 90% confidence interval for the proportion has a margin of error of at most 0.01?
p = 0.5E = 0.0190%
n =
5.Hypothesis tests & confidence intervalsMedium
A sample of 36 students has a mean score of 115.5 and a sample standard deviation of 15. Find the test statistic t for H₀: μ = 120.
μ₀ = 120x̄ = 115.5s = 15n = 36
t =
6.Hypothesis tests & confidence intervalsHard
A one-sided z test of H₀: μ = 200 against H₁: μ > 200 is done at α = 0.01 with a sample of 100; the standard deviation is known to be 15. In reality the mean is 205.25. Find the power 1 − β of the test in percent.
μ₀ = 200μ = 205.25σ = 15n = 100α = 0.01
1 − β = %
7.Linear regression & correlationMedium
The table gives 5 paired observations of x = load in kN and y = extension in mm. Find Pearson's correlation coefficient r.
n = 5
r =
8.Linear regression & correlationHard
The table gives 5 paired observations of x = advertising spend in k€ and y = sales in k€. Use the least-squares line ŷ = a + b·x. Predict ŷ for x = 5.
n = 5x = 5
ŷ =
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, major”.
row total = 61column total = 59N = 135
E =
10.Chi-square testsHard
A chi-square goodness-of-fit test compares observed counts in 11 categories with fully specified expected proportions (no parameters are estimated from the data). How many degrees of freedom df does the test have?
k = 11
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 = 458 h
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 = 500 − 1.04·40 = 458 h
Answer
x = 458 h
3.Confidence intervals
Answer: lower limit = 83.95
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 = 5, t = 2.571
Margin of error
E = 2.571·1/√6 = 1.05
Interval
85 ± 1.05 → [83.95, 86.05]
Answer
lower limit = 83.95
4.Confidence intervals
Answer: n = 6766
Confidence interval for a proportion: p̂ ± z·√(p̂(1 − p̂)/n), where p̂ = x/n. To reach a margin E you need n ≥ z²·p(1 − p)/E², rounded up; without a prior estimate use p = 0.5, which gives the largest n.
Solve for n
n = 1.645²·0.5·0.5/0.01² = 6765
Round up
n = 6766
Answer
n = 6766
5.Hypothesis tests & confidence intervals
Answer: t = −1.8
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 = 15/√36 = 15/6 = 2.5
Test statistic
t = (115.5 − 120)/2.5 = −1.8
Answer
t = −1.8
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: 1 − β = 87.98 %
Power of a test: power = 1 − β is the probability of rejecting H₀ when it is false. Find the cutoff x̄c from μ₀ and z_α, then compute the probability of landing beyond x̄c using the true mean: z = (x̄c − μ)/(σ/√n).
Critical sample mean
SE = 15/√100 = 1.5
x̄c = 200 + 2.326·1.5 = 203.5
z under the true mean
z = (203.5 − 205.25)/1.5 = −1.174
Power
power = P(Z > −1.174) = 87.98%
1 − β = 87.98%
Answer
1 − β = 87.98%
7.Linear regression & correlation
Answer: r = 0.9912
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 = 30/5 = 6
ȳ = 88/5 = 17.6
Sums of squares and products
Sxx = Σ(x − x̄)² = 40
Sxy = Σ(x − x̄)(y − ȳ) = 112
Syy = Σ(y − ȳ)² = 319.2
Correlation coefficient
r = 112/√(40·319.2) = 0.9912
Answer
r = 0.9912
8.Linear regression & correlation
Answer: ŷ = 52.55
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 = 30/5 = 6
ȳ = 287/5 = 57.4
Sums of squares and products
Sxx = Σ(x − x̄)² = 40
Sxy = Σ(x − x̄)(y − ȳ) = 194
Regression line
b = 194/40 = 4.85
a = 57.4 − 4.85·6 = 28.3
ŷ = 28.3 + 4.85·x
Prediction
ŷ = 28.3 + 4.85·5 = 52.55
Answer
ŷ = 52.55
9.Chi-square tests
Answer: E = 26.66
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 = 61·59/135 = 26.66
Answer
E = 26.66
10.Chi-square tests
Answer: df = 10
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.