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
174.85188.85P = ?

The heights of adult men are normally distributed with mean 178 cm and standard deviation 7 cm. What fraction lies between 174.85 and 188.85 cm? Give P as a decimal.

μ = 178σ = 7x1 = 174.85x2 = 188.85
P =
2.Normal distributionHard
μ = 178x = ?5%

The heights of adult men are normally distributed with mean 178 cm and standard deviation 7 cm. Find the value x below which 5% of them lie (the 5th percentile).

μ = 178σ = 7P = 5%
x = cm
3.Confidence intervalsMedium
n = 21, s = 13.5LUx̄ = 15E = ?99%

A sample of 21 items has mean 15 and standard deviation s = 13.5. The population is roughly normal. Find the margin of error E of a 99% confidence interval for the mean.

n = 21x̄ = 15s = 13.599%t(20) = 2.845
E =
4.Confidence intervalsHard
n = 15Lσ = ?s = 895%

The diameters of 15 machined shafts have a sample standard deviation of 8 µm. Assuming the diameters are normally distributed, find the upper limit σ of the 95% confidence interval for the population standard deviation. With df = 14, 2.5% of the χ² distribution lies below 5.629 and 2.5% lies above 26.119.

n = 15s = 8 µmχ² = 5.629, 26.119
σ = µm
5.Hypothesis tests & confidence intervalsMedium
t −2.0302.03

A sample of 36 students has a mean score of 494.8 and a sample standard deviation of 10. Find the test statistic t for H₀: μ = 500.

μ₀ = 500x̄ = 494.8s = 10n = 36
t =
6.Hypothesis tests & confidence intervalsHard
z −1.64501.645

A sample of 36 measurements has mean 121.9. The population standard deviation is 8. Find the upper limit U of the 90% confidence interval for the mean.

x̄ = 121.9σ = 8n = 3690%
U =
7.Linear regression & correlationMedium
xy128226326423521619xy2462025

The table gives 6 paired observations of x = car age in years and y = price in k€. Find Pearson's correlation coefficient r.

n = 6
r =
8.Linear regression & correlationHard
xy356463571670776881xy468607080ŷ = 65

The table gives 6 paired observations of x = hours studied and y = exam score. Use the least-squares line ŷ = a + b·x. For which value x does the line predict ŷ = 65?

n = 6ŷ = 65
x =
9.Chi-square testsMedium
lowmediumhighTotalurban144660120rural16343080Total308090200

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 = 2002 × 3
χ² =
10.Chi-square testsHard
yesnoTotalage 18–305644100age 31–50303060age 51+461460Total13288220dfα = 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 = 2203 × 2α = 0.05
χ² =

3Solutions

1.Normal distribution

Answer: P = 0.6131

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 = (174.85 − 178)/7 = −0.45
z₂ = (188.85 − 178)/7 = 1.55

Area between

P = Φ(1.55) − Φ(−0.45) = 0.9394 − 0.3264 = 0.6131

Answer

P = 0.6131
2.Normal distribution

Answer: x = 166.5 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) = 0.05 → z = −1.645

Back to x

x = 178 + −1.645·7 = 166.49 cm

Answer

x = 166.49 cm
3.Confidence intervals

Answer: E = 8.381

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 = 20, t = 2.845

Margin of error

E = 2.845·13.5/√21 = 8.381

Answer

E = 8.381
4.Confidence intervals

Answer: σ = 12.62 µm

Interval for a variance: (n − 1)s²/σ² has a χ² distribution with n − 1 degrees of freedom. The 95% interval for σ² runs from (n − 1)s² divided by the upper χ² value (2.5% above it) to (n − 1)s² divided by the lower χ² value (2.5% below it); the interval for σ is the square root of both limits. It is not symmetric around s.

Sum of squares

(n − 1)s² = 14·8² = 896

Limits for σ²

896/26.119 = 34.3
896/5.629 = 159.2

Limit for σ

σ = √159.2 = 12.62 µm

Answer

σ = 12.62 µm

95% interval for σ: [5.857, 12.62] µm.

5.Hypothesis tests & confidence intervals

Answer: t = −3.12

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/√36 = 10/6 = 1.667

Test statistic

t = (494.8 − 500)/1.667 = −3.12

Answer

t = −3.12
t0.025; df = 35 = 2.03

|t| > 2.03, so H₀ is rejected at the 5% level (two-sided).

6.Hypothesis tests & confidence intervals

Answer: U = 124.1

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 = 1.645·8/√36 = 2.193

Interval limit

U = 121.9 + 2.193 = 124.1

Answer

U = 124.1
7.Linear regression & correlation

Answer: r = −0.9817

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 = 143/6 = 23.83

Sums of squares and products

Sxx = Σ(x − x̄)² = 17.5
Sxy = Σ(x − x̄)(y − ȳ) = −31.5
Syy = Σ(y − ȳ)² = 58.83

Correlation coefficient

r = −31.5/√(17.5·58.83) = −0.9817

Answer

r = −0.9817
8.Linear regression & correlation

Answer: x = 4.534

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 = 417/6 = 69.5

Sums of squares and products

Sxx = Σ(x − x̄)² = 17.5
Sxy = Σ(x − x̄)(y − ȳ) = 81.5

Regression line

b = 81.5/17.5 = 4.657
a = 69.5 − 4.657·5.5 = 43.89
ŷ = 43.89 + 4.657·x

Solve for x

x = (65 − 43.89)/4.657 = 4.534

Answer

x = 4.534
9.Chi-square tests

Answer: χ² = 4.097

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: 120·30/200 = 18, 120·80/200 = 48, 120·90/200 = 54
rural: 80·30/200 = 12, 80·80/200 = 32, 80·90/200 = 36

Chi-square statistic

χ² = (14 − 18)²/18 + (46 − 48)²/48 + (60 − 54)²/54 + (16 − 12)²/12 + (34 − 32)²/32 + (30 − 36)²/36
χ² = 4.097

Answer

χ² = 4.097

df = 2. At α = 0.05 the critical value is 5.991, so H₀ (independence) is not rejected.

10.Chi-square tests

Answer: χ² = 10.11

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: 60, 40
age 31–50: 36, 24
age 51+: 36, 24

Chi-square statistic

χ² = (56 − 60)²/60 + (44 − 40)²/40 + (30 − 36)²/36 + (30 − 24)²/24 + (46 − 36)²/36 + (14 − 24)²/24
χ² = 10.11

Decision

df = 2, χ²crit = 5.991
10.11 > 5.991

Answer

χ² = 10.11

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

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