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 0.8132. Find z to two decimals.
A = 0.8132
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 bottom 40% is to the left of it. Find the cut-off x.
μ = 178 cmσ = 7 cm40%
x = cm
3.Confidence intervalsMedium
The blood pressure of 13 patients is measured before and after a treatment. The differences d = before − after (mmHg) have mean 5.8 and standard deviation 2.3. Find the lower limit of the 95% confidence interval for the mean difference.
n = 13d̄ = 5.8sd = 2.395%t(12) = 2.179
lower limit = mmHg
4.Confidence intervalsHard
The population standard deviation is 18. 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 1?
σ = 18E = 199%
n =
5.Hypothesis tests & confidence intervalsMedium
A sample of 81 students has a mean score of 46.56 and a sample standard deviation of 10. Find the test statistic t for H₀: μ = 50.
μ₀ = 50x̄ = 46.56s = 10n = 81
t =
6.Hypothesis tests & confidence intervalsHard
A one-sided z test of H₀: μ = 50 against H₁: μ < 50 is done at α = 0.10 with a sample of 64; the standard deviation is known to be 20. In reality the mean is 46.25. Find the power 1 − β of the test in percent.
μ₀ = 50μ = 46.25σ = 20n = 64α = 0.10
1 − β = %
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. For which value x does the line predict ŷ = 78?
n = 5ŷ = 78
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 = 1002 × 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 = 0.01.
α = 0.01
χ²crit =
3Solutions
1.Normal distribution
Answer: z = 1.32
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 + 0.8132)/2 = 0.9066
Read the z-table backwards
Find 0.9066 inside the table and read off z.
z = 1.32
Answer
z = 1.32
2.Normal distribution
Answer: x = 176.3 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) = 0.4
z from the table
z ≈ −0.25
Back to the original units
x = 178 − 0.25·7 = 176.3 cm
Answer
x = 176.3 cm
3.Confidence intervals
Answer: lower limit = 4.41 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 = 12, t = 2.179
Margin of error
E = 2.179·2.3/√13 = 1.39
Interval
5.8 ± 1.39 → [4.41, 7.19]
Answer
lower limit = 4.41 mmHg
4.Confidence intervals
Answer: n = 2150
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 = (2.576·18/1)² = 2150
Round up
A sample size is a whole number, and rounding down would make the margin too large.
n = 2150
Answer
n = 2150
5.Hypothesis tests & confidence intervals
Answer: t = −3.096
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 = 10/√81 = 10/9 = 1.111
Test statistic
t = (46.56 − 50)/1.111 = −3.096
Answer
t = −3.096
t0.025; df = 80 = 1.99
|t| > 1.99, so H₀ is rejected at the 5% level (two-sided).
6.Hypothesis tests & confidence intervals
Answer: 1 − β = 58.63 %
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 = 20/√64 = 2.5
x̄c = 50 − 1.282·2.5 = 46.8
z under the true mean
z = (46.8 − 46.25)/2.5 = 0.218
Power
power = P(Z < 0.218) = 58.63%
1 − β = 58.63%
Answer
1 − β = 58.63%
7.Linear regression & correlation
Answer: r = 0.9543
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 = 10001 − 105·467/5 = 194
Sxx = 2245 − 105²/5 = 40
Syy = 44651 − 467²/5 = 1033
Correlation coefficient
r = 194/√(40·1033) = 0.9543
Answer
r = 0.9543
8.Linear regression & correlation
Answer: x = 14.93
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 = 90/5 = 18
ȳ = 473/5 = 94.6
Sums of squares and products
Sxx = Σ(x − x̄)² = 40
Sxy = Σ(x − x̄)(y − ȳ) = 216
Regression line
b = 216/40 = 5.4
a = 94.6 − 5.4·18 = −2.6
ŷ = −2.6 + 5.4·x
Solve for x
x = (78 − (−2.6))/5.4 = 14.93
Answer
x = 14.93
9.Chi-square tests
Answer: χ² = 0.787
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 = 13.28
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 = 0.01) = 13.277
Answer
χ²crit = 13.277
Reject H₀ if the computed χ² is larger than this value.