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 = 0.2358

Drag z so that the shaded area between −z and z under the standard normal curve is 0.2358. Find z to two decimals.

A = 0.2358
z =
2.Normal distributionHard
235240245250255260265gtop 10%

The bag weights are normally distributed with mean 250 g and standard deviation 5 g. Drag the cut-off so the shaded top 10% is to the right of it. Find the cut-off x.

μ = 250 gσ = 5 g10%
x = g
3.Confidence intervalsMedium
n = 30, s = 2.5LU = ?x̄ = 90E95%

A sample of 30 items has mean 90 and standard deviation s = 2.5. The population is roughly normal. Find the upper limit of the 95% confidence interval for the mean.

n = 30x̄ = 90s = 2.595%t(29) = 2.045
upper limit =
4.Confidence intervalsHard
n = 11Lσ = ?s = 5.695%

The diameters of 11 machined shafts have a sample standard deviation of 5.6 µm. Assuming the diameters are normally distributed, find the upper limit σ of the 95% confidence interval for the population standard deviation. With df = 10, 2.5% of the χ² distribution lies below 3.247 and 2.5% lies above 20.483.

n = 11s = 5.6 µmχ² = 3.247, 20.483
σ = µm
5.Hypothesis tests & confidence intervalsMedium
t −2.06402.064

A sample of 25 students has a mean score of 102.1 and a sample standard deviation of 15. Find the test statistic t for H₀: μ = 100.

μ₀ = 100x̄ = 102.1s = 15n = 25
t =
6.Hypothesis tests & confidence intervalsHard
H₀H₁μ₀x̄cμ₀ + 4α = 0.01βσ = 21, power 80%, n = ?

A one-sided z test at α = 0.01 must detect a shift of the mean of 4 units with a power of 80%. The standard deviation is 21. Find the minimum sample size n.

α = 0.01power = 80%δ = 4σ = 21zα = 2.326zβ = 0.842
n =
7.Linear regression & correlationMedium
xy166819792293251182813131142xy15202530100150

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

n = 6
r =
8.Linear regression & correlationHard
xy1393465567599691170xy51040506070x = 6

The table gives 6 paired observations of x = advertising spend in k€ and y = sales in k€. Use the least-squares line ŷ = a + b·x. A new observation (6, 55) is made. Find its residual e = y − ŷ.

n = 6(6, 55)
e =
9.Chi-square testsMedium
busbikecarTotalstudents10111940staff223761120Total324880160

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 × 3
χ² =
10.Chi-square testsHard
yesnoTotalage 18–30431760age 31–508713100age 51+503080Total18060240dfα = 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 = 0.3

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 + 0.2358)/2 = 0.6179

Read the z-table backwards

Find 0.6179 inside the table and read off z.

z = 0.30

Answer

z = 0.30
2.Normal distribution

Answer: x = 256.4 g

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) = 0.9

z from the table

z ≈ 1.28

Back to the original units

x = 250 + 1.28·5 = 256.4 g

Answer

x = 256.4 g
3.Confidence intervals

Answer: upper limit = 90.93

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 = 29, t = 2.045

Margin of error

E = 2.045·2.5/√30 = 0.9334

Interval

90 ± 0.9334 → [89.07, 90.93]

Answer

upper limit = 90.93
4.Confidence intervals

Answer: σ = 9.828 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 = 10·5.6² = 313.6

Limits for σ²

313.6/20.483 = 15.31
313.6/3.247 = 96.58

Limit for σ

σ = √96.58 = 9.828 µm

Answer

σ = 9.828 µm

95% interval for σ: [3.913, 9.828] µm.

5.Hypothesis tests & confidence intervals

Answer: t = 0.7

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 = (102.1 − 100)/3 = 0.7

Answer

t = 0.7
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: n = 277

Sample size for a test: to detect a true shift δ with power 1 − β in a one-sided z test at level α, the standard error must satisfy δ = (z_α + z_β)·σ/√n, so n = ((z_α + z_β)·σ/δ)², rounded up.

Solve for n

n = ((2.326 + 0.842)·21/4)² = 276.6

Round up

n = 277

Answer

n = 277
7.Linear regression & correlation

Answer: r = 0.9925

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 = 141/6 = 23.5
ȳ = 631/6 = 105.2

Sums of squares and products

Sxx = Σ(x − x̄)² = 157.5
Sxy = Σ(x − x̄)(y − ȳ) = 826.5
Syy = Σ(y − ȳ)² = 4403

Correlation coefficient

r = 826.5/√(157.5·4403) = 0.9925

Answer

r = 0.9925
8.Linear regression & correlation

Answer: e = −1.5

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 = 36/6 = 6
ȳ = 339/6 = 56.5

Sums of squares and products

Sxx = Σ(x − x̄)² = 70
Sxy = Σ(x − x̄)(y − ȳ) = 227

Regression line

b = 227/70 = 3.243
a = 56.5 − 3.243·6 = 37.04
ŷ = 37.04 + 3.243·x

Predicted value

ŷ = 37.04 + 3.243·6 = 56.5

Residual

e = 55 − 56.5 = −1.5

Answer

e = −1.5

The point lies below the line.

9.Chi-square tests

Answer: χ² = 0.8444

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

students: 40·32/160 = 8, 40·48/160 = 12, 40·80/160 = 20
staff: 120·32/160 = 24, 120·48/160 = 36, 120·80/160 = 60

Chi-square statistic

χ² = (10 − 8)²/8 + (11 − 12)²/12 + (19 − 20)²/20 + (22 − 24)²/24 + (37 − 36)²/36 + (61 − 60)²/60
χ² = 0.8444

Answer

χ² = 0.8444

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

10.Chi-square tests

Answer: χ² = 14.7

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

age 18–30: 45, 15
age 31–50: 75, 25
age 51+: 60, 20

Chi-square statistic

χ² = (43 − 45)²/45 + (17 − 15)²/15 + (87 − 75)²/75 + (13 − 25)²/25 + (50 − 60)²/60 + (30 − 20)²/20
χ² = 14.7

Decision

df = 2, χ²crit = 5.991
14.7 > 5.991

Answer

χ² = 14.7

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

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