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 = 0.8385

Drag z so that the shaded area between −z and z under the standard normal curve is 0.8385. Find z to two decimals.

A = 0.8385
z =
2.Normal distributionHard
μ = 178x = ?20%

The heights of adult men are normally distributed with mean 178 cm and standard deviation 7 cm. Find the value x below which 20% of them lie (the 20th percentile).

μ = 178σ = 7P = 20%
x = cm
3.Confidence intervalsMedium
μ₁ − μ₂0LUx̄₁ − x̄₂E = ?99%

The battery life of two phone models is compared. Model 1: 50 phones, mean 133 h, known standard deviation 16 h. Model 2: 80 phones, mean 128 h, known standard deviation 12 h. Find the margin of error E of the 99% confidence interval for μ₁ − μ₂.

n₁ = 50x̄₁ = 133 hσ₁ = 16 hn₂ = 80x̄₂ = 128 hσ₂ = 12 h99%
E = h
4.Confidence intervalsHard
p ≈ 0.8, n = ?p̂ − Ep̂ + Ep̂E = 0.0495%

A pilot study suggests that about 0.8 of products are defective. How many products n must be inspected so that a 95% confidence interval for the proportion has a margin of error of at most 0.04?

p ≈ 0.8E = 0.0495%
n =
5.Hypothesis tests & confidence intervalsMedium
t −1.99801.998

A sample of 64 students has a mean score of 503.9 and a sample standard deviation of 15. Find the test statistic t for H₀: μ = 500.

μ₀ = 500x̄ = 503.9s = 15n = 64
t =
6.Hypothesis tests & confidence intervalsHard
σ = 12, n = ?x̄ − Ex̄ + Ex̄E = 390%

How large must a sample be so that the 90% confidence interval for a mean has a margin of error of at most 3? The standard deviation is 12. Find the smallest sample size n.

σ = 12E = 390%
n =
7.Linear regression & correlationMedium
xy254146178241030xy510102030

The table gives 5 paired observations of x = load in kN and y = extension in mm. Find the coefficient of determination r² of the least-squares line.

n = 5
r² =
8.Linear regression & correlationHard
xy185120532261245926622878xy202550607080x = 21

The table gives 6 paired observations of x = temperature in °C and y = ice creams sold. Use the least-squares line ŷ = a + b·x. Predict ŷ for x = 21.

n = 6x = 21
ŷ =
9.Chi-square testsMedium
yesnoTotalmen215980women352560Total5684140

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 × 2
χ² =
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 = 1.4

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 + 0.8385)/2 = 0.9192

Read the z-table backwards

Find 0.9192 inside the table and read off z.

z = 1.40

Answer

z = 1.40
2.Normal distribution

Answer: x = 172.1 cm

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) = 0.2 → z = −0.8416

Back to x

x = 178 + −0.8416·7 = 172.11 cm

Answer

x = 172.11 cm
3.Confidence intervals

Answer: E = 6.776 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 = 133 − 128 = 5

Standard error

SE = √(16²/50 + 12²/80) = 2.631

Margin of error

E = 2.576·2.631 = 6.776

Answer

E = 6.776 h

The interval contains 0, so the data do not show a significant difference.

4.Confidence intervals

Answer: n = 385

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 = 1.96²·0.8·0.2/0.04² = 384.2

Round up

n = 385

Answer

n = 385
5.Hypothesis tests & confidence intervals

Answer: t = 2.08

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 = 15/√64 = 15/8 = 1.875

Test statistic

t = (503.9 − 500)/1.875 = 2.08

Answer

t = 2.08
t0.025; df = 63 = 1.998

|t| > 1.998, so H₀ is rejected at the 5% level (two-sided).

6.Hypothesis tests & confidence intervals

Answer: n = 44

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).

Solve for n

n = (1.645·12/3)² = 43.3

Round up

n = 44

Answer

n = 44
7.Linear regression & correlation

Answer: r² = 0.9836

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 = 30/5 = 6
ȳ = 90/5 = 18

Sums of squares and products

Sxx = Σ(x − x̄)² = 40
Sxy = Σ(x − x̄)(y − ȳ) = 120
Syy = Σ(y − ȳ)² = 366

Coefficient of determination

r² = 120²/(40·366) = 0.9836

Answer

r² = 0.9836

About 98.36% of the variation in y is explained by the line.

8.Linear regression & correlation

Answer: ŷ = 56.1

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 = 138/6 = 23
ȳ = 364/6 = 60.67

Sums of squares and products

Sxx = Σ(x − x̄)² = 70
Sxy = Σ(x − x̄)(y − ȳ) = 160

Regression line

b = 160/70 = 2.286
a = 60.67 − 2.286·23 = 8.095
ŷ = 8.095 + 2.286·x

Prediction

ŷ = 8.095 + 2.286·21 = 56.1

Answer

ŷ = 56.1
9.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

men: 80·56/140 = 32, 80·84/140 = 48
women: 60·56/140 = 24, 60·84/140 = 36

Chi-square statistic

χ² = (21 − 32)²/32 + (59 − 48)²/48 + (35 − 24)²/24 + (25 − 36)²/36
χ² = 14.7

Answer

χ² = 14.7

df = 1. At α = 0.05 the critical value is 3.841, so H₀ (independence) is rejected.

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.

Degrees of freedom

df = 5 − 1 − 1 = 3

Answer

df = 3

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