Statistics 2: Statistical Inference: cram pack

Formula sheets, the key method steps for every topic and 10 practice problems, with full solutions at the back.

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1Formulas and methods

Normal distribution

z-score
z = (x − μ)/σ
Below x
P(X < x) = Φ(z)
Above x
P(X > x) = 1 − Φ(z)
Between
Φ(z2) − Φ(z1)
Percentile
x = μ + z·σ

Method

  1. Convert the value to a z-score: z = (x − μ)/σ.
  2. Look up Φ(z), the area to the left of z.
  3. For the area to the right use 1 − Φ(z); for an interval subtract the two left areas.
  4. For a percentile, work backwards: find z for the area, then x = μ + z·σ.

Confidence intervals

σ known
x̄ ± z·σ/√n
σ unknown
x̄ ± t·s/√n, df = n − 1
Critical z
90%: 1.645, 95%: 1.96, 99%: 2.576
Sample size
n = (z·σ/E)², rounded up

Method

  1. Decide between z (population σ known) and t (only the sample s, with n − 1 degrees of freedom).
  2. Find the critical value for the confidence level.
  3. Margin of error: E = critical value × standard deviation/√n.
  4. Interval: x̄ ± E.

Hypothesis tests & confidence intervals

z-test for a mean
z = (x̄ − μ0)/(σ/√n)
t-test for a mean
t = (x̄ − μ0)/(s/√n), df = n − 1
Two-sided at 5%
reject H0 if |z| > 1.96
Confidence interval
x̄ ± z·σ/√n

Method

  1. State H₀ and H₁ and decide whether the test is one- or two-sided.
  2. Calculate the test statistic: (x̄ − μ₀)/(σ/√n), or with s and t.
  3. Compare with the critical value, or find the p-value.
  4. Reject H₀ if the statistic is beyond the critical value (p below α), and state the conclusion in words.

Linear regression & correlation

Sums of squares
Sxx = Σ(x − x̄)² = Σx² − (Σx)²/n, Sxy = Σxy − ΣxΣy/n
Slope
b = Sxy/Sxx
Intercept
a = ȳ − b·x̄
Correlation
r = Sxy/√(SxxSyy)
Residual
e = y − ŷ

Chi-square tests

Test statistic
χ² = Σ(O − E)²/E
Goodness of fit
E = n·p, df = k − 1 − (estimated parameters)
Independence
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
244.25254.25P = ?

The bag weights are normally distributed with mean 250 g and standard deviation 5 g. What fraction lies between 244.25 and 254.25 g? Give P as a decimal.

μ = 250σ = 5x1 = 244.25x2 = 254.25
P =
2.Normal distributionHard
μ = 250x = ?25%

The bag weights are normally distributed with mean 250 g and standard deviation 5 g. Find the value x below which 25% of them lie (the 25th percentile).

μ = 250σ = 5P = 25%
x = g
3.Confidence intervalsMedium
n = 10, sd​ = 6.20LUd̄ = 7.9E = ?95%

The blood pressure of 10 patients is measured before and after a treatment. The differences d = before − after (mmHg) have mean 7.9 and standard deviation 6.2. Find the margin of error E of the 95% confidence interval for the mean difference.

n = 10d̄ = 7.9sd = 6.295%t(9) = 2.262
E = mmHg
4.Confidence intervalsHard
p = 0.5, n = ?p̂ − Ep̂ + Ep̂E = 0.0199%

Nothing is known in advance about the proportion of defective products. How many products n must be inspected so that a 99% confidence interval for the proportion has a margin of error of at most 0.01?

p = 0.5E = 0.0199%
n =
5.Hypothesis tests & confidence intervalsMedium
−2.2282.228t = ?df = 10

11 employees take a typing test before and after a training course. For each person the difference d = after − before (words per minute) is recorded. The mean difference is 5.89 and the standard deviation of the differences is 9.3. Find the paired test statistic t for H₀: μd = 0.

n = 11d̄ = 5.89sd = 9.3
t =
6.Hypothesis tests & confidence intervalsHard
−1.961.96z = ?

In a trial, 125 of 250 patients on treatment A recover, and 92 of 150 patients on treatment B. Using the pooled proportion, find the test statistic z for H₀: p₁ = p₂.

x₁ = 125n₁ = 250x₂ = 92n₂ = 150
z =
7.Linear regression & correlationMedium
xy340443552656767870xy46840506070

The table gives 6 paired observations of x = advertising spend in k€ and y = sales in k€. Find Pearson's correlation coefficient r.

n = 6
r =
8.Linear regression & correlationHard
xy12322231941751169xy246101520ŷ = 20

The table gives 6 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 = 20?

n = 6ŷ = 20
x =
9.Chi-square testsMedium
no defectminormajorTotalday shift2625960night shift31143681Total573945141

The contingency table shows observed counts. For a chi-square test of independence, find the expected count E for the cell “night shift, no defect”.

row total = 81column total = 57N = 141
E =
10.Chi-square testsHard
yesnoTotalage 18–30413980age 31–504159100age 51+225880Total104156260dfα = 0.05α = 0.0113.8416.63525.9919.21037.81511.34549.48813.277511.07015.086612.59216.812714.06718.475815.50720.090916.91921.6661018.30723.209

The table shows observed counts. Test at α = 0.01 whether the two variables are independent (no continuity correction). Compute the test statistic χ², then compare it with the critical value from the table.

N = 2603 × 2α = 0.01
χ² =

3Solutions

1.Normal distribution

Answer: P = 0.6773

Normal distribution: convert a value x to a z-score, z = (x − μ)/σ, which counts standard deviations from the mean. A z-table (or Φ(z)) gives the area to the left of z, which is P(X < x). The area to the right is 1 − Φ(z), and the area between two values is the difference of their two areas.

z-scores

z₁ = (244.25 − 250)/5 = −1.15
z₂ = (254.25 − 250)/5 = 0.85

Area between

P = Φ(0.85) − Φ(−1.15) = 0.8023 − 0.1251 = 0.6773

Answer

P = 0.6773
2.Normal distribution

Answer: x = 246.6 g

Normal distribution: convert a value x to a z-score, z = (x − μ)/σ, which counts standard deviations from the mean. A z-table (or Φ(z)) gives the area to the left of z, which is P(X < x). The area to the right is 1 − Φ(z), and the area between two values is the difference of their two areas.

z for the percentile

Φ(z) = 0.25 → z = −0.6745

Back to x

x = 250 + −0.6745·5 = 246.63 g

Answer

x = 246.63 g
3.Confidence intervals

Answer: E = 4.435 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 = 9, t = 2.262

Margin of error

E = 2.262·6.2/√10 = 4.435

Answer

E = 4.435 mmHg
4.Confidence intervals

Answer: n = 16590

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 = 2.576²·0.5·0.5/0.01² = 16589

Round up

n = 16590

Answer

n = 16590
5.Hypothesis tests & confidence intervals

Answer: t = 2.101

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 of the mean difference

SE = 9.3/√11 = 2.804

Test statistic

t = 5.89/2.804 = 2.101

Answer

t = 2.101

With df = 10, the two-sided 5% critical value is 2.228; |t| is smaller, so H₀ is not rejected.

6.Hypothesis tests & confidence intervals

Answer: z = −2.203

Hypothesis test: the test statistic = (estimate − value under H₀)/(standard error under H₀). The p-value is the probability, assuming H₀ is true, of a statistic at least as extreme as the one observed: the area in one tail for a one-sided H₁, twice the tail area for a two-sided H₁. Reject H₀ when p ≤ α.

Sample proportions

p̂₁ = 125/250 = 0.5
p̂₂ = 92/150 = 0.6133
p̂ = 217/400 = 0.5425

Standard error under H₀

SE = √(0.5425·0.4575·(1/250 + 1/150)) = 0.05145

Test statistic

z = (0.5 − 0.6133)/0.05145 = −2.203

Answer

z = −2.203

|z| > 1.96, so H₀ is rejected at the 5% level (two-sided).

7.Linear regression & correlation

Answer: r = 0.9881

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 = 33/6 = 5.5
ȳ = 328/6 = 54.67

Sums of squares and products

Sxx = Σ(x − x̄)² = 17.5
Sxy = Σ(x − x̄)(y − ȳ) = 113
Syy = Σ(y − ȳ)² = 747.3

Correlation coefficient

r = 113/√(17.5·747.3) = 0.9881

Answer

r = 0.9881
8.Linear regression & correlation

Answer: x = 2.444

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 = 21/6 = 3.5
ȳ = 101/6 = 16.83

Sums of squares and products

Sxx = Σ(x − x̄)² = 17.5
Sxy = Σ(x − x̄)(y − ȳ) = −52.5

Regression line

b = −52.5/17.5 = −3
a = 16.83 − (−3)·3.5 = 27.33
ŷ = 27.33 − 3·x

Solve for x

x = (20 − 27.33)/(−3) = 2.444

Answer

x = 2.444
9.Chi-square tests

Answer: E = 32.74

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 = 81·57/141 = 32.74

Answer

E = 32.74
10.Chi-square tests

Answer: χ² = 9.469

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 counts

age 18–30: 32, 48
age 31–50: 40, 60
age 51+: 32, 48

Chi-square statistic

χ² = (41 − 32)²/32 + (39 − 48)²/48 + (41 − 40)²/40 + (59 − 60)²/60 + (22 − 32)²/32 + (58 − 48)²/48
χ² = 9.469

Decision

df = 2, χ²crit = 9.210
9.469 > 9.210

Answer

χ² = 9.469

χ² is above the critical value, so H₀ is rejected: the variables appear to be related.

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