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 0.8849. Find z to two decimals.
A = 0.8849
z =
2.Normal distributionHard
The bag weights are normally distributed with mean 250 g and standard deviation 5 g. Drag the cut-off so the shaded bottom 75% is to the left of it. Find the cut-off x.
μ = 250 gσ = 5 g75%
x = g
3.Confidence intervalsMedium
A 90% confidence interval for a mean, based on 36 observations and a known σ, has a margin of error of 7.2. What is the margin of error E₂ of a 95% interval based on 72 observations?
90%n₁ = 36E₁ = 7.295%n₂ = 72
E₂ =
4.Confidence intervalsHard
The diameters of 21 machined shafts have a sample standard deviation of 11.8 µm. Assuming the diameters are normally distributed, find the lower limit σ of the 95% confidence interval for the population standard deviation. With df = 20, 2.5% of the χ² distribution lies below 9.591 and 2.5% lies above 34.17.
n = 21s = 11.8 µmχ² = 9.591, 34.17
σ = µm
5.Hypothesis tests & confidence intervalsMedium
A sample of 25 students has a mean score of 44.72 and a sample standard deviation of 12. Find the test statistic t for H₀: μ = 50.
μ₀ = 50x̄ = 44.72s = 12n = 25
t =
6.Hypothesis tests & confidence intervalsHard
A sample of 81 measurements has mean 100.3. The population standard deviation is 5. Find the lower limit L of the 90% confidence interval for the mean.
x̄ = 100.3σ = 5n = 8190%
L =
7.Linear regression & correlationMedium
The table gives 6 paired observations of x = temperature in °C and y = ice creams sold. Find Pearson's correlation coefficient r.
n = 6
r =
8.Linear regression & correlationHard
The table gives 5 paired observations of x = car age in years and y = price in k€. Use the least-squares line ŷ = a + b·x. For which value x does the line predict ŷ = 24?
n = 5ŷ = 24
x =
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 “urban, medium”.
row total = 105column total = 51N = 172
E =
10.Chi-square testsHard
A chi-square goodness-of-fit test checks whether counts in 5 categories follow a Poisson distribution whose mean is estimated from the same data. How many degrees of freedom df does the test have?
k = 5
df =
3Solutions
1.Normal distribution
Answer: z = −1.2
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 − 0.8849 = 0.1151
Read the z-table backwards
Find 0.1151 inside the table and read off z.
z = −1.20
Answer
z = −1.20
2.Normal distribution
Answer: x = 253.4 g
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) = 0.75
z from the table
z ≈ 0.67
Back to the original units
x = 250 + 0.67·5 = 253.4 g
Answer
x = 253.4 g
3.Confidence intervals
Answer: E₂ = 6.066
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 = (1.96/1.645)·√(36/72) = 0.8425
New margin
E₂ = 7.2·0.8425 = 6.066
Answer
E₂ = 6.066
4.Confidence intervals
Answer: σ = 9.028 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 = 20·11.8² = 2785
Limits for σ²
2785/34.17 = 81.5
2785/9.591 = 290.4
Limit for σ
σ = √81.5 = 9.028 µm
Answer
σ = 9.028 µm
95% interval for σ: [9.028, 17.04] 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 = 12/√25 = 12/5 = 2.4
Test statistic
t = (44.72 − 50)/2.4 = −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: L = 99.39
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 = 1.645·5/√81 = 0.9139
Interval limit
L = 100.3 − 0.9139 = 99.39
Answer
L = 99.39
7.Linear regression & correlation
Answer: r = 0.9871
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 = 126/6 = 21
ȳ = 559/6 = 93.17
Sums of squares and products
Sxx = Σ(x − x̄)² = 70
Sxy = Σ(x − x̄)(y − ȳ) = 315
Syy = Σ(y − ȳ)² = 1455
Correlation coefficient
r = 315/√(70·1455) = 0.9871
Answer
r = 0.9871
8.Linear regression & correlation
Answer: x = 1.615
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 = 102/5 = 20.4
Sums of squares and products
Sxx = Σ(x − x̄)² = 10
Sxy = Σ(x − x̄)(y − ȳ) = −26
Regression line
b = −26/10 = −2.6
a = 20.4 − (−2.6)·3 = 28.2
ŷ = 28.2 − 2.6·x
Solve for x
x = (24 − 28.2)/(−2.6) = 1.615
Answer
x = 1.615
9.Chi-square tests
Answer: E = 31.13
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 = 105·51/172 = 31.13
Answer
E = 31.13
10.Chi-square tests
Answer: df = 3
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.