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 0.8969. Find z to two decimals.
A = 0.8969
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 95% is to the right of it. Find the cut-off x.
μ = 178 cmσ = 7 cm95%
x = cm
3.Confidence intervalsMedium
A sample of 7 items has mean 81.5 and standard deviation s = 12. The population is roughly normal. Find the upper limit of the 90% confidence interval for the mean.
n = 7x̄ = 81.5s = 1290%t(6) = 1.943
upper limit =
4.Confidence intervalsHard
The diameters of 15 machined shafts have a sample standard deviation of 10.8 µm. Assuming the diameters are normally distributed, find the upper limit σ of the 95% confidence interval for the population standard deviation. With df = 14, 2.5% of the χ² distribution lies below 5.629 and 2.5% lies above 26.119.
n = 15s = 10.8 µmχ² = 5.629, 26.119
σ = µm
5.Hypothesis tests & confidence intervalsMedium
A sample of 25 students has a mean score of 111.2 and a sample standard deviation of 20. Find the test statistic t for H₀: μ = 120.
μ₀ = 120x̄ = 111.2s = 20n = 25
t =
6.Hypothesis tests & confidence intervalsHard
A sample of 36 measurements has mean 112.5. The population standard deviation is 15. Find the upper limit U of the 99% confidence interval for the mean.
x̄ = 112.5σ = 15n = 3699%
U =
7.Linear regression & correlationMedium
The table gives 5 paired observations of x = hours studied and y = exam score. Their sums are given. Find Pearson's correlation coefficient r from these sums.
The table gives 6 paired observations of x = load in kN and y = extension in mm. Use the least-squares line ŷ = a + b·x. Predict ŷ for x = 7.
n = 6x = 7
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 × 2
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. Using the table, find the critical value χ²crit at significance level α = 0.01.
α = 0.01
χ²crit =
3Solutions
1.Normal distribution
Answer: z = 1.63
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 + 0.8969)/2 = 0.9484
Read the z-table backwards
Find 0.9484 inside the table and read off z.
z = 1.63
Answer
z = 1.63
2.Normal distribution
Answer: x = 166.5 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) = 0.05
z from the table
z ≈ −1.64
Back to the original units
x = 178 − 1.64·7 = 166.5 cm
Answer
x = 166.5 cm
3.Confidence intervals
Answer: upper limit = 90.31
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 = 6, t = 1.943
Margin of error
E = 1.943·12/√7 = 8.813
Interval
81.5 ± 8.813 → [72.69, 90.31]
Answer
upper limit = 90.31
4.Confidence intervals
Answer: σ = 17.03 µm
Interval for a variance: (n − 1)s²/σ² has a χ² distribution with n − 1 degrees of freedom. The 95% interval for σ² runs from (n − 1)s² divided by the upper χ² value (2.5% above it) to (n − 1)s² divided by the lower χ² value (2.5% below it); the interval for σ is the square root of both limits. It is not symmetric around s.
Sum of squares
(n − 1)s² = 14·10.8² = 1633
Limits for σ²
1633/26.119 = 62.52
1633/5.629 = 290.1
Limit for σ
σ = √290.1 = 17.03 µm
Answer
σ = 17.03 µm
95% interval for σ: [7.907, 17.03] µm.
5.Hypothesis tests & confidence intervals
Answer: t = −2.2
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 = (111.2 − 120)/4 = −2.2
Answer
t = −2.2
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: U = 118.9
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).
Margin of error
E = 2.576·15/√36 = 6.44
Interval limit
U = 112.5 + 6.44 = 118.9
Answer
U = 118.9
7.Linear regression & correlation
Answer: r = 0.9993
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 = 1939 − 25·337/5 = 254
Sxx = 165 − 25²/5 = 40
Syy = 24329 − 337²/5 = 1615
Correlation coefficient
r = 254/√(40·1615) = 0.9993
Answer
r = 0.9993
8.Linear regression & correlation
Answer: ŷ = 11.83
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 = 42/6 = 7
ȳ = 71/6 = 11.83
Sums of squares and products
Sxx = Σ(x − x̄)² = 70
Sxy = Σ(x − x̄)(y − ȳ) = 111
Regression line
b = 111/70 = 1.586
a = 11.83 − 1.586·7 = 0.7333
ŷ = 0.7333 + 1.586·x
Prediction
ŷ = 0.7333 + 1.586·7 = 11.83
Answer
ŷ = 11.83
9.Chi-square tests
Answer: χ² = 0.4689
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 = 1. At α = 0.05 the critical value is 3.841, 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 = 8 − 1 − 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.