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 to the right of z = 0.9884

Drag z so that the shaded area under the standard normal curve, to the right of z, is 0.9884. Find z to two decimals.

A = 0.9884
z =
2.Normal distributionHard
235240245250255260265gtop 10%

The bag weights are normally distributed with mean 250 g and standard deviation 5 g. Drag the cut-off so the shaded top 10% is to the right of it. Find the cut-off x.

μ = 250 gσ = 5 g10%
x = g
3.Confidence intervalsMedium
μ₁ − μ₂0LU = ?x̄₁ − x̄₂E99%

The battery life of two phone models is compared. Model 1: 30 phones, mean 124.5 h, known standard deviation 3 h. Model 2: 100 phones, mean 114.5 h, known standard deviation 15 h. Find the upper limit of the 99% confidence interval for μ₁ − μ₂.

n₁ = 30x̄₁ = 124.5 hσ₁ = 3 hn₂ = 100x̄₂ = 114.5 hσ₂ = 15 h99%
upper limit = h
4.Confidence intervalsHard
σ = 16, n = ?x̄ − Ex̄ + Ex̄E = 1.590%

The population standard deviation is 16. How large a sample n do you need so that a 90% confidence interval for the mean has a margin of error of at most 1.5?

σ = 16E = 1.590%
n =
5.Hypothesis tests & confidence intervalsMedium
t −2.0302.03

A sample of 36 students has a mean score of 101.1 and a sample standard deviation of 4. Find the test statistic t for H₀: μ = 100.

μ₀ = 100x̄ = 101.1s = 4n = 36
t =
6.Hypothesis tests & confidence intervalsHard
μ₀x̄c

A one-sided z test of H₀: μ = 200 against H₁: μ > 200 is done at α = 0.10 with a sample of 25; the standard deviation is known to be 7.5. In reality the mean is 203.75. Find the probability of a Type II error β in percent.

μ₀ = 200μ = 203.75σ = 7.5n = 25α = 0.10
β = %
7.Linear regression & correlationMedium
xy2941462182710341241xy51010203040

The table gives 6 paired observations of x = load in kN and y = extension in mm. Find Pearson's correlation coefficient r.

n = 6
r =
8.Linear regression & correlationHard
xy341442553660759866xy468405060x = 3.5

The table gives 6 paired observations of x = advertising spend in k€ and y = sales in k€. Use the least-squares line ŷ = a + b·x. Predict ŷ for x = 3.5.

n = 6x = 3.5
ŷ =
9.Chi-square testsMedium
no defectminormajorTotalday shift2719955night shift27381681Total545725136

The contingency table shows observed counts. For a chi-square test of independence, find the expected count E for the cell “day shift, minor”.

row total = 55column total = 57N = 136
E =
10.Chi-square testsHard
yesnoTotalplant A354580plant B285280plant C334780Total96144240dfα = 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 = 2403 × 2α = 0.01
χ² =

3Solutions

1.Normal distribution

Answer: z = −2.27

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 − 0.9884 = 0.0116

Read the z-table backwards

Find 0.0116 inside the table and read off z.

z = −2.27

Answer

z = −2.27
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: upper limit = 14.11 h

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.

Point estimate

x̄₁ − x̄₂ = 124.5 − 114.5 = 10

Standard error

SE = √(3²/30 + 15²/100) = 1.597

Margin of error

E = 2.576·1.597 = 4.114

Interval

10 ± 4.114 → [5.886, 14.11]

Answer

upper limit = 14.11 h

The interval does not contain 0, so the means differ significantly.

4.Confidence intervals

Answer: n = 308

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 = (1.645·16/1.5)² = 307.9

Round up

A sample size is a whole number, and rounding down would make the margin too large.

n = 308

Answer

n = 308
5.Hypothesis tests & confidence intervals

Answer: t = 1.65

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 = 4/√36 = 4/6 = 0.6667

Test statistic

t = (101.1 − 100)/0.6667 = 1.65

Answer

t = 1.65
t0.025; df = 35 = 2.03

|t| < 2.03, so H₀ is not rejected at the 5% level (two-sided).

6.Hypothesis tests & confidence intervals

Answer: β = 11.16 %

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 = 7.5/√25 = 1.5
x̄c = 200 + 1.282·1.5 = 201.9

z under the true mean

z = (201.9 − 203.75)/1.5 = −1.218

Type II error

power = P(Z > −1.218) = 88.84%
β = 100% − 88.84% = 11.16%

Answer

β = 11.16%
7.Linear regression & correlation

Answer: r = 0.9989

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 = 146/6 = 24.33

Sums of squares and products

Sxx = Σ(x − x̄)² = 70
Sxy = Σ(x − x̄)(y − ȳ) = 226
Syy = Σ(y − ȳ)² = 731.3

Correlation coefficient

r = 226/√(70·731.3) = 0.9989

Answer

r = 0.9989
8.Linear regression & correlation

Answer: ŷ = 43.04

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 = 321/6 = 53.5

Sums of squares and products

Sxx = Σ(x − x̄)² = 17.5
Sxy = Σ(x − x̄)(y − ȳ) = 91.5

Regression line

b = 91.5/17.5 = 5.229
a = 53.5 − 5.229·5.5 = 24.74
ŷ = 24.74 + 5.229·x

Prediction

ŷ = 24.74 + 5.229·3.5 = 43.04

Answer

ŷ = 43.04
9.Chi-square tests

Answer: E = 23.05

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 = 55·57/136 = 23.05

Answer

E = 23.05
10.Chi-square tests

Answer: χ² = 1.354

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: 32, 48
plant B: 32, 48
plant C: 32, 48

Chi-square statistic

χ² = (35 − 32)²/32 + (45 − 48)²/48 + (28 − 32)²/32 + (52 − 48)²/48 + (33 − 32)²/32 + (47 − 48)²/48
χ² = 1.354

Decision

df = 2, χ²crit = 9.210
1.354 < 9.210

Answer

χ² = 1.354

χ² is below the critical value, so H₀ is not rejected: there is no significant evidence of a relationship.

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