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
−3−2−10123zshaded area between −z and z = 0.9786

Drag z so that the shaded area between −z and z under the standard normal curve is 0.9786. Find z to two decimals.

A = 0.9786
z =
2.Normal distributionHard
235240245250255260265gbottom 90%

The bag weights are normally distributed with mean 250 g and standard deviation 5 g. Drag the cut-off so the shaded bottom 90% is to the left of it. Find the cut-off x.

μ = 250 gσ = 5 g90%
x = g
3.Confidence intervalsMedium
n = 15, s = 3.5LUx̄ = 15E = ?99%

A sample of 15 items has mean 15 and standard deviation s = 3.5. The population is roughly normal. Find the margin of error E of a 99% confidence interval for the mean.

n = 15x̄ = 15s = 3.599%t(14) = 2.977
E =
4.Confidence intervalsHard
p ≈ 0.1, n = ?p̂ − Ep̂ + Ep̂E = 0.0590%

A pilot study suggests that about 0.1 of products are defective. How many products n must be inspected so that a 90% confidence interval for the proportion has a margin of error of at most 0.05?

p ≈ 0.1E = 0.0590%
n =
5.Hypothesis tests & confidence intervalsMedium
2.04p = ?

The mean of a population is claimed to be 250; the population standard deviation is known to be 10. A random sample of 100 observations has a mean of 252.04. Test H₀: μ = 250 against H₁: μ > 250. Find the p-value p.

μ₀ = 250σ = 10n = 100x̄ = 252.04H₁: μ > μ₀
p =
6.Hypothesis tests & confidence intervalsHard
z −1.64501.645

A sample of 81 measurements has mean 101. The population standard deviation is 15. Find the upper limit U of the 90% confidence interval for the mean.

x̄ = 101σ = 15n = 8190%
U =
7.Linear regression & correlationMedium
xy357451561667768867xy468506070

The table gives 6 paired observations of x = hours studied and y = exam score. Find Pearson's correlation coefficient r.

n = 6
r =
8.Linear regression & correlationHard
xy126223324420522620xy24620222426x = 2.5

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. Predict ŷ for x = 2.5.

n = 6x = 2.5
ŷ =
9.Chi-square testsMedium
lowmediumhighTotalurban214039100rural2121740Total424256140

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 = 1402 × 3
χ² =
10.Chi-square testsHard
passfailTotalplant A243660plant B433780plant C134760Total80120200dfα = 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 = 0.05 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 = 2003 × 2α = 0.05
χ² =

3Solutions

1.Normal distribution

Answer: z = 2.3

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 + 0.9786)/2 = 0.9893

Read the z-table backwards

Find 0.9893 inside the table and read off z.

z = 2.30

Answer

z = 2.30
2.Normal distribution

Answer: x = 256.4 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) = 0.9

z from the table

z ≈ 1.28

Back to the original units

x = 250 + 1.28·5 = 256.4 g

Answer

x = 256.4 g
3.Confidence intervals

Answer: E = 2.69

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 = 14, t = 2.977

Margin of error

E = 2.977·3.5/√15 = 2.69

Answer

E = 2.69
4.Confidence intervals

Answer: n = 98

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 = 1.645²·0.1·0.9/0.05² = 97.42

Round up

n = 98

Answer

n = 98
5.Hypothesis tests & confidence intervals

Answer: p = 0.02068

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.

Test statistic

SE = σ/√n = 10/√100 = 1
z = (252.04 − 250)/1 = 2.04

Tail area

Φ(2.04) = 0.9793
1 − Φ(2.04) = 0.0207

p-value

p = 0.02068

Answer

p = 0.02068

p ≤ 0.05, so H₀ is rejected at the 5% level.

6.Hypothesis tests & confidence intervals

Answer: U = 103.7

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 = 1.645·15/√81 = 2.742

Interval limit

U = 101 + 2.742 = 103.7

Answer

U = 103.7
7.Linear regression & correlation

Answer: r = 0.8381

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 = 371/6 = 61.83

Sums of squares and products

Sxx = Σ(x − x̄)² = 17.5
Sxy = Σ(x − x̄)(y − ȳ) = 53.5
Syy = Σ(y − ȳ)² = 232.8

Correlation coefficient

r = 53.5/√(17.5·232.8) = 0.8381

Answer

r = 0.8381
8.Linear regression & correlation

Answer: ŷ = 23.56

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 = 135/6 = 22.5

Sums of squares and products

Sxx = Σ(x − x̄)² = 17.5
Sxy = Σ(x − x̄)(y − ȳ) = −18.5

Regression line

b = −18.5/17.5 = −1.057
a = 22.5 − (−1.057)·3.5 = 26.2
ŷ = 26.2 − 1.057·x

Prediction

ŷ = 26.2 + (−1.057)·2.5 = 23.56

Answer

ŷ = 23.56
9.Chi-square tests

Answer: χ² = 21.2

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

urban: 100·42/140 = 30, 100·42/140 = 30, 100·56/140 = 40
rural: 40·42/140 = 12, 40·42/140 = 12, 40·56/140 = 16

Chi-square statistic

χ² = (21 − 30)²/30 + (40 − 30)²/30 + (39 − 40)²/40 + (21 − 12)²/12 + (2 − 12)²/12 + (17 − 16)²/16
χ² = 21.2

Answer

χ² = 21.2

df = 2. At α = 0.05 the critical value is 5.991, so H₀ (independence) is rejected.

10.Chi-square tests

Answer: χ² = 14.7

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

plant A: 24, 36
plant B: 32, 48
plant C: 24, 36

Chi-square statistic

χ² = (24 − 24)²/24 + (36 − 36)²/36 + (43 − 32)²/32 + (37 − 48)²/48 + (13 − 24)²/24 + (47 − 36)²/36
χ² = 14.7

Decision

df = 2, χ²crit = 5.991
14.7 > 5.991

Answer

χ² = 14.7

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

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