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 = 0.8485

Drag z so that the shaded area under the standard normal curve, to the right of z, is 0.8485. Find z to two decimals.

A = 0.8485
z =
2.Normal distributionHard
μ = 250x = ?10%

The bag weights are normally distributed with mean 250 g and standard deviation 5 g. Find the value x below which 10% of them lie (the 10th percentile).

μ = 250σ = 5P = 10%
x = g
3.Confidence intervalsMedium
n = 11, s = 13.5L = ?Ux̄ = 69.5E90%

A sample of 11 items has mean 69.5 and standard deviation s = 13.5. The population is roughly normal. Find the lower limit of the 90% confidence interval for the mean.

n = 11x̄ = 69.5s = 13.590%t(10) = 1.812
lower limit =
4.Confidence intervalsHard
n = 16Lσ = ?s = 6.995%

The diameters of 16 machined shafts have a sample standard deviation of 6.9 µm. Assuming the diameters are normally distributed, find the upper 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 = 6.9 µmχ² = 6.262, 27.488
σ = µm
5.Hypothesis tests & confidence intervalsMedium
t −1.99801.998

A sample of 64 students has a mean score of 51.06 and a sample standard deviation of 5. Find the test statistic t for H₀: μ = 50.

μ₀ = 50x̄ = 51.06s = 5n = 64
t =
6.Hypothesis tests & confidence intervalsHard
σ = 12, n = ?x̄ − Ex̄ + Ex̄E = 1.599%

How large must a sample be so that the 99% confidence interval for a mean has a margin of error of at most 1.5? The standard deviation is 12. Find the smallest sample size n.

σ = 12E = 1.599%
n =
7.Linear regression & correlationMedium
xy1580178419922110323110xy15208090100110

The table gives 5 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 = 5Σx = 95Σy = 469Σx² = 1845Σy² = 44629Σxy = 9069
r =
8.Linear regression & correlationHard
xy121221317413513612xy2461520x = 5.5

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. Predict ŷ for x = 5.5.

n = 6x = 5.5
ŷ =
9.Chi-square testsMedium
yesnoTotalmen56460women283260Total8436120

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
yesnoTotalplant A6238100plant B261440plant C3268100Total120120240dfα = 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 = −1.03

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 − 0.8485 = 0.1515

Read the z-table backwards

Find 0.1515 inside the table and read off z.

z = −1.03

Answer

z = −1.03
2.Normal distribution

Answer: x = 243.6 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) = 0.1 → z = −1.282

Back to x

x = 250 + −1.282·5 = 243.59 g

Answer

x = 243.59 g
3.Confidence intervals

Answer: lower limit = 62.12

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 = 10, t = 1.812

Margin of error

E = 1.812·13.5/√11 = 7.376

Interval

69.5 ± 7.376 → [62.12, 76.88]

Answer

lower limit = 62.12
4.Confidence intervals

Answer: σ = 10.68 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 = 15·6.9² = 714.2

Limits for σ²

714.2/27.488 = 25.98
714.2/6.262 = 114

Limit for σ

σ = √114 = 10.68 µm

Answer

σ = 10.68 µm

95% interval for σ: [5.097, 10.68] µm.

5.Hypothesis tests & confidence intervals

Answer: t = 1.696

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 = 5/√64 = 5/8 = 0.625

Test statistic

t = (51.06 − 50)/0.625 = 1.696

Answer

t = 1.696
t0.025; df = 63 = 1.998

|t| < 1.998, so H₀ is not rejected at the 5% level (two-sided).

6.Hypothesis tests & confidence intervals

Answer: n = 425

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 = (2.576·12/1.5)² = 424.7

Round up

n = 425

Answer

n = 425
7.Linear regression & correlation

Answer: r = 0.99

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 = 9069 − 95·469/5 = 158
Sxx = 1845 − 95²/5 = 40
Syy = 44629 − 469²/5 = 636.8

Correlation coefficient

r = 158/√(40·636.8) = 0.99

Answer

r = 0.99
8.Linear regression & correlation

Answer: ŷ = 12

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 = 97/6 = 16.17

Sums of squares and products

Sxx = Σ(x − x̄)² = 17.5
Sxy = Σ(x − x̄)(y − ȳ) = −36.5

Regression line

b = −36.5/17.5 = −2.086
a = 16.17 − (−2.086)·3.5 = 23.47
ŷ = 23.47 − 2.086·x

Prediction

ŷ = 23.47 + (−2.086)·5.5 = 12

Answer

ŷ = 12
9.Chi-square tests

Answer: χ² = 31.11

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

men: 60·84/120 = 42, 60·36/120 = 18
women: 60·84/120 = 42, 60·36/120 = 18

Chi-square statistic

χ² = (56 − 42)²/42 + (4 − 18)²/18 + (28 − 42)²/42 + (32 − 18)²/18
χ² = 31.11

Answer

χ² = 31.11

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

10.Chi-square tests

Answer: χ² = 22.32

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: 50, 50
plant B: 20, 20
plant C: 50, 50

Chi-square statistic

χ² = (62 − 50)²/50 + (38 − 50)²/50 + (26 − 20)²/20 + (14 − 20)²/20 + (32 − 50)²/50 + (68 − 50)²/50
χ² = 22.32

Decision

df = 2, χ²crit = 5.991
22.32 > 5.991

Answer

χ² = 22.32

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

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