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 = 0.9803

Drag z so that the shaded area under the standard normal curve, to the right of z, is 0.9803. Find z to two decimals.

A = 0.9803
z =
2.Normal distributionHard
μ = 100x = ?80%

The IQ scores are normally distributed with mean 100 points and standard deviation 15 points. Find the value x below which 80% of them lie (the 80th percentile).

μ = 100σ = 15P = 80%
x = points
3.Confidence intervalsMedium
x̄E₁ = 11.195%, n = 50x̄E₂ = ?99%, n = 50

A 95% confidence interval for a mean, based on 50 observations and a known σ, has a margin of error of 11.1. What is the margin of error E₂ of a 99% interval from the same data?

95%n₁ = 50E₁ = 11.199%
E₂ =
4.Confidence intervalsHard
n = 10Lσ = ?s = 10.895%

The diameters of 10 machined shafts have a sample standard deviation of 10.8 µm. Assuming the diameters are normally distributed, find the upper limit σ of the 95% confidence interval for the population standard deviation. With df = 9, 2.5% of the χ² distribution lies below 2.7 and 2.5% lies above 19.023.

n = 10s = 10.8 µmχ² = 2.7, 19.023
σ = µm
5.Hypothesis tests & confidence intervalsMedium
t −2.0302.03

A sample of 36 students has a mean score of 43.8 and a sample standard deviation of 12. Find the test statistic t for H₀: μ = 50.

μ₀ = 50x̄ = 43.8s = 12n = 36
t =
6.Hypothesis tests & confidence intervalsHard
z −2.57602.576

A sample of 16 measurements has mean 117. The population standard deviation is 15. Find the lower limit L of the 99% confidence interval for the mean.

x̄ = 117σ = 15n = 1699%
L =
7.Linear regression & correlationMedium
xy34355676096211691377xy5106080

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
xy2114196238341039xy51010203040x = 3

The table gives 5 paired observations of x = load in kN and y = extension in mm. Use the least-squares line ŷ = a + b·x. Predict ŷ for x = 3.

n = 5x = 3
ŷ =
9.Chi-square testsMedium
improvednot improvedTotalnew drug305080placebo36440Total6654120

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 = 1202 × 2
χ² =
10.Chi-square testsHard
yesnoTotalage 18–30144660age 31–503862100age 51+206080Total72168240dfα = 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 = 2403 × 2α = 0.05
χ² =

3Solutions

1.Normal distribution

Answer: z = −2.06

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 − 0.9803 = 0.0197

Read the z-table backwards

Find 0.0197 inside the table and read off z.

z = −2.06

Answer

z = −2.06
2.Normal distribution

Answer: x = 112.6 points

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) = 0.8 → z = 0.8416

Back to x

x = 100 + 0.8416·15 = 112.62 points

Answer

x = 112.62 points
3.Confidence intervals

Answer: E₂ = 14.59

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.

Ratio of margins

E₂/E₁ = (2.576/1.96)·√(50/50) = 1.314

New margin

E₂ = 11.1·1.314 = 14.59

Answer

E₂ = 14.59
4.Confidence intervals

Answer: σ = 19.72 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 = 9·10.8² = 1050

Limits for σ²

1050/19.023 = 55.18
1050/2.7 = 388.8

Limit for σ

σ = √388.8 = 19.72 µm

Answer

σ = 19.72 µm

95% interval for σ: [7.429, 19.72] µm.

5.Hypothesis tests & confidence intervals

Answer: t = −3.1

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).

Standard error

SE = 12/√36 = 12/6 = 2

Test statistic

t = (43.8 − 50)/2 = −3.1

Answer

t = −3.1
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: L = 107.3

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 = 2.576·15/√16 = 9.66

Interval limit

L = 117 − 9.66 = 107.3

Answer

L = 107.3
7.Linear regression & correlation

Answer: r = 0.9737

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 = 48/6 = 8
ȳ = 367/6 = 61.17

Sums of squares and products

Sxx = Σ(x − x̄)² = 70
Sxy = Σ(x − x̄)(y − ȳ) = 211
Syy = Σ(y − ȳ)² = 670.8

Correlation coefficient

r = 211/√(70·670.8) = 0.9737

Answer

r = 0.9737
8.Linear regression & correlation

Answer: ŷ = 14.55

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 = 126/5 = 25.2

Sums of squares and products

Sxx = Σ(x − x̄)² = 40
Sxy = Σ(x − x̄)(y − ȳ) = 142

Regression line

b = 142/40 = 3.55
a = 25.2 − 3.55·6 = 3.9
ŷ = 3.9 + 3.55·x

Prediction

ŷ = 3.9 + 3.55·3 = 14.55

Answer

ŷ = 14.55
9.Chi-square tests

Answer: χ² = 29.7

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

new drug: 80·66/120 = 44, 80·54/120 = 36
placebo: 40·66/120 = 22, 40·54/120 = 18

Chi-square statistic

χ² = (30 − 44)²/44 + (50 − 36)²/36 + (36 − 22)²/22 + (4 − 18)²/18
χ² = 29.7

Answer

χ² = 29.7

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

10.Chi-square tests

Answer: χ² = 5.27

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: 18, 42
age 31–50: 30, 70
age 51+: 24, 56

Chi-square statistic

χ² = (14 − 18)²/18 + (46 − 42)²/42 + (38 − 30)²/30 + (62 − 70)²/70 + (20 − 24)²/24 + (60 − 56)²/56
χ² = 5.27

Decision

df = 2, χ²crit = 5.991
5.27 < 5.991

Answer

χ² = 5.27

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

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