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 = 0.8969

Drag z so that the shaded area between −z and z under the standard normal curve is 0.8969. Find z to two decimals.

A = 0.8969
z =
2.Normal distributionHard
157164171178185192199cmtop 95%

The heights of adult men are normally distributed with mean 178 cm and standard deviation 7 cm. Drag the cut-off so the shaded top 95% is to the right of it. Find the cut-off x.

μ = 178 cmσ = 7 cm95%
x = cm
3.Confidence intervalsMedium
n = 7, s = 12LU = ?x̄ = 81.5E90%

A sample of 7 items has mean 81.5 and standard deviation s = 12. The population is roughly normal. Find the upper limit of the 90% confidence interval for the mean.

n = 7x̄ = 81.5s = 1290%t(6) = 1.943
upper limit =
4.Confidence intervalsHard
n = 15Lσ = ?s = 10.895%

The diameters of 15 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 = 14, 2.5% of the χ² distribution lies below 5.629 and 2.5% lies above 26.119.

n = 15s = 10.8 µmχ² = 5.629, 26.119
σ = µm
5.Hypothesis tests & confidence intervalsMedium
t −2.06402.064

A sample of 25 students has a mean score of 111.2 and a sample standard deviation of 20. Find the test statistic t for H₀: μ = 120.

μ₀ = 120x̄ = 111.2s = 20n = 25
t =
6.Hypothesis tests & confidence intervalsHard
z −2.57602.576

A sample of 36 measurements has mean 112.5. The population standard deviation is 15. Find the upper limit U of the 99% confidence interval for the mean.

x̄ = 112.5σ = 15n = 3699%
U =
7.Linear regression & correlationMedium
xy142354568781992xy5406080

The table gives 5 paired observations of x = hours studied and y = exam score. Their sums are given. Find Pearson's correlation coefficient r from these sums.

n = 5Σx = 25Σy = 337Σx² = 165Σy² = 24329Σxy = 1939
r =
8.Linear regression & correlationHard
xy244661181410171219xy5105101520x = 7

The table gives 6 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 = 7.

n = 6x = 7
ŷ =
9.Chi-square testsMedium
yesnoTotalmen3763100women194160Total56104160

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 = 1602 × 2
χ² =
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 8 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 = 0.01.

α = 0.01
χ²crit =

3Solutions

1.Normal distribution

Answer: z = 1.63

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 + 0.8969)/2 = 0.9484

Read the z-table backwards

Find 0.9484 inside the table and read off z.

z = 1.63

Answer

z = 1.63
2.Normal distribution

Answer: x = 166.5 cm

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

Φ(z) = 0.05

z from the table

z ≈ −1.64

Back to the original units

x = 178 − 1.64·7 = 166.5 cm

Answer

x = 166.5 cm
3.Confidence intervals

Answer: upper limit = 90.31

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 = 6, t = 1.943

Margin of error

E = 1.943·12/√7 = 8.813

Interval

81.5 ± 8.813 → [72.69, 90.31]

Answer

upper limit = 90.31
4.Confidence intervals

Answer: σ = 17.03 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 = 14·10.8² = 1633

Limits for σ²

1633/26.119 = 62.52
1633/5.629 = 290.1

Limit for σ

σ = √290.1 = 17.03 µm

Answer

σ = 17.03 µm

95% interval for σ: [7.907, 17.03] µm.

5.Hypothesis tests & confidence intervals

Answer: t = −2.2

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 = 20/√25 = 20/5 = 4

Test statistic

t = (111.2 − 120)/4 = −2.2

Answer

t = −2.2
t0.025; df = 24 = 2.064

|t| > 2.064, so H₀ is rejected at the 5% level (two-sided).

6.Hypothesis tests & confidence intervals

Answer: U = 118.9

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/√36 = 6.44

Interval limit

U = 112.5 + 6.44 = 118.9

Answer

U = 118.9
7.Linear regression & correlation

Answer: r = 0.9993

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 = 1939 − 25·337/5 = 254
Sxx = 165 − 25²/5 = 40
Syy = 24329 − 337²/5 = 1615

Correlation coefficient

r = 254/√(40·1615) = 0.9993

Answer

r = 0.9993
8.Linear regression & correlation

Answer: ŷ = 11.83

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 = 42/6 = 7
ȳ = 71/6 = 11.83

Sums of squares and products

Sxx = Σ(x − x̄)² = 70
Sxy = Σ(x − x̄)(y − ȳ) = 111

Regression line

b = 111/70 = 1.586
a = 11.83 − 1.586·7 = 0.7333
ŷ = 0.7333 + 1.586·x

Prediction

ŷ = 0.7333 + 1.586·7 = 11.83

Answer

ŷ = 11.83
9.Chi-square tests

Answer: χ² = 0.4689

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: 100·56/160 = 35, 100·104/160 = 65
women: 60·56/160 = 21, 60·104/160 = 39

Chi-square statistic

χ² = (37 − 35)²/35 + (63 − 65)²/65 + (19 − 21)²/21 + (41 − 39)²/39
χ² = 0.4689

Answer

χ² = 0.4689

df = 1. At α = 0.05 the critical value is 3.841, so H₀ (independence) is not rejected.

10.Chi-square tests

Answer: χ²crit = 16.81

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 = 8 − 1 − 1 = 6

Table

χ²crit(df = 6, α = 0.01) = 16.812

Answer

χ²crit = 16.812

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

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