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
24.836.8P = ?

The delivery times are normally distributed with mean 32 min and standard deviation 6 min. What fraction lies between 24.8 and 36.8 min? Give P as a decimal.

μ = 32σ = 6x1 = 24.8x2 = 36.8
P =
2.Normal distributionHard
μ = 250x = ?97.5%

The bag weights are normally distributed with mean 250 g and standard deviation 5 g. Find the value x below which 97.5% of them lie (the 97.5th percentile).

μ = 250σ = 5P = 97.5%
x = g
3.Confidence intervalsMedium
n = 12, s = 15LU = ?x̄ = 47E95%

A sample of 12 items has mean 47 and standard deviation s = 15. The population is roughly normal. Find the upper limit of the 95% confidence interval for the mean.

n = 12x̄ = 47s = 1595%t(11) = 2.201
upper limit =
4.Confidence intervalsHard
n = 16σ = ?Us = 1.595%

The diameters of 16 machined shafts have a sample standard deviation of 1.5 µm. Assuming the diameters are normally distributed, find the lower limit σ of the 95% confidence interval for the population standard deviation. With df = 15, 2.5% of the χ² distribution lies below 6.262 and 2.5% lies above 27.488.

n = 16s = 1.5 µmχ² = 6.262, 27.488
σ = µm
5.Hypothesis tests & confidence intervalsMedium
t −2.06402.064

A sample of 25 students has a mean score of 55.1 and a sample standard deviation of 15. Find the test statistic t for H₀: μ = 50.

μ₀ = 50x̄ = 55.1s = 15n = 25
t =
6.Hypothesis tests & confidence intervalsHard
z −1.9601.96

A sample of 81 measurements has mean 199.6. The population standard deviation is 4. Find the lower limit L of the 95% confidence interval for the mean.

x̄ = 199.6σ = 4n = 8195%
L =
7.Linear regression & correlationMedium
xy1671198722100251132813131151xy15202530100150

The table gives 6 paired observations of x = temperature in °C and y = ice creams sold. Their sums are given. Find Pearson's correlation coefficient r from these sums.

n = 6Σx = 141Σy = 653Σx² = 3471Σy² = 75341Σxy = 16163
r =
8.Linear regression & correlationHard
xy124225318416516610xy24610152025x = 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. A new observation (2.5, 25) is made. Find its residual e = y − ŷ.

n = 6(2.5, 25)
e =
9.Chi-square testsMedium
no defectminormajorTotalday shift20321668night shift44341290Total646628158

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

row total = 90column total = 28N = 158
E =
10.Chi-square testsHard
dfα = 0.05α = 0.0113.8416.63525.9919.21037.81511.34549.48813.277511.07015.086612.59216.812714.06718.475815.50720.090916.91921.6661018.30723.209

A chi-square goodness-of-fit test checks whether counts in 9 categories follow a Poisson distribution whose mean is estimated from the same data. Using the table, find the critical value χ²crit at significance level α = 0.05.

α = 0.05
χ²crit =

3Solutions

1.Normal distribution

Answer: P = 0.6731

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 = (24.8 − 32)/6 = −1.2
z₂ = (36.8 − 32)/6 = 0.8

Area between

P = Φ(0.8) − Φ(−1.2) = 0.7881 − 0.1151 = 0.6731

Answer

P = 0.6731
2.Normal distribution

Answer: x = 259.8 g

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) = 0.975 → z = 1.96

Back to x

x = 250 + 1.96·5 = 259.8 g

Answer

x = 259.8 g
3.Confidence intervals

Answer: upper limit = 56.53

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 = 11, t = 2.201

Margin of error

E = 2.201·15/√12 = 9.531

Interval

47 ± 9.531 → [37.47, 56.53]

Answer

upper limit = 56.53
4.Confidence intervals

Answer: σ = 1.108 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 = 15·1.5² = 33.75

Limits for σ²

33.75/27.488 = 1.228
33.75/6.262 = 5.39

Limit for σ

σ = √1.228 = 1.108 µm

Answer

σ = 1.108 µm

95% interval for σ: [1.108, 2.322] µm.

5.Hypothesis tests & confidence intervals

Answer: t = 1.7

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).

Standard error

SE = 15/√25 = 15/5 = 3

Test statistic

t = (55.1 − 50)/3 = 1.7

Answer

t = 1.7
t0.025; df = 24 = 2.064

|t| < 2.064, so H₀ is not rejected at the 5% level (two-sided).

6.Hypothesis tests & confidence intervals

Answer: L = 198.7

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 = 1.96·4/√81 = 0.8711

Interval limit

L = 199.6 − 0.8711 = 198.7

Answer

L = 198.7
7.Linear regression & correlation

Answer: r = 0.9965

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.

Centred sums

Sxy = 16163 − 141·653/6 = 817.5
Sxx = 3471 − 141²/6 = 157.5
Syy = 75341 − 653²/6 = 4273

Correlation coefficient

r = 817.5/√(157.5·4273) = 0.9965

Answer

r = 0.9965
8.Linear regression & correlation

Answer: e = 4.005

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 = 109/6 = 18.17

Sums of squares and products

Sxx = Σ(x − x̄)² = 17.5
Sxy = Σ(x − x̄)(y − ȳ) = −49.5

Regression line

b = −49.5/17.5 = −2.829
a = 18.17 − (−2.829)·3.5 = 28.07
ŷ = 28.07 − 2.829·x

Predicted value

ŷ = 28.07 + (−2.829)·2.5 = 21

Residual

e = 25 − 21 = 4.005

Answer

e = 4.005

The point lies above the line.

9.Chi-square tests

Answer: E = 15.95

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 = 90·28/158 = 15.95

Answer

E = 15.95
10.Chi-square tests

Answer: χ²crit = 14.07

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 = 9 − 1 − 1 = 7

Table

χ²crit(df = 7, α = 0.05) = 14.067

Answer

χ²crit = 14.067

Reject H₀ if the computed χ² is larger than this value.

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