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 = 0.1357

Drag z so that the shaded area under the standard normal curve, to the right of z, is 0.1357. Find z to two decimals.

A = 0.1357
z =
2.Normal distributionHard
μ = 32x = ?90%

The delivery times are normally distributed with mean 32 min and standard deviation 6 min. Find the value x below which 90% of them lie (the 90th percentile).

μ = 32σ = 6P = 90%
x = min
3.Confidence intervalsMedium
μ₁ − μ₂0LU = ?x̄₁ − x̄₂E99%

The battery life of two phone models is compared. Model 1: 80 phones, mean 114 h, known standard deviation 16 h. Model 2: 60 phones, mean 103 h, known standard deviation 17 h. Find the upper limit of the 99% confidence interval for μ₁ − μ₂.

n₁ = 80x̄₁ = 114 hσ₁ = 16 hn₂ = 60x̄₂ = 103 hσ₂ = 17 h99%
upper limit = h
4.Confidence intervalsHard
n = 15Lσ = ?s = 9.495%

The diameters of 15 machined shafts have a sample standard deviation of 9.4 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 = 9.4 µmχ² = 5.629, 26.119
σ = µm
5.Hypothesis tests & confidence intervalsMedium
t −1.99801.998

A sample of 64 students has a mean score of 46.7 and a sample standard deviation of 12. Find the test statistic t for H₀: μ = 50.

μ₀ = 50x̄ = 46.7s = 12n = 64
t =
6.Hypothesis tests & confidence intervalsHard
z −1.9601.96

A sample of 81 measurements has mean 73.7. The population standard deviation is 4. Find the lower limit L of the 95% confidence interval for the mean.

x̄ = 73.7σ = 4n = 8195%
L =
7.Linear regression & correlationMedium
xy14301645184620522255xy1520304050

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 = 90Σy = 228Σx² = 1660Σy² = 10770Σxy = 4218
r =
8.Linear regression & correlationHard
xy155717581961217823772588xy15202560708090ŷ = 63

The table gives 6 paired observations of x = temperature in °C and y = ice creams sold. Use the least-squares line ŷ = a + b·x. For which value x does the line predict ŷ = 63?

n = 6ŷ = 63
x =
9.Chi-square testsMedium
busbikecarTotalstudents38411493staff21383998Total597953191

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

row total = 98column total = 79N = 191
E =
10.Chi-square testsHard

A chi-square test of independence is carried out on a contingency table with 2 rows and 6 columns (not counting the totals). How many degrees of freedom df does the test have?

2 × 6
df =

3Solutions

1.Normal distribution

Answer: z = 1.1

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 − 0.1357 = 0.8643

Read the z-table backwards

Find 0.8643 inside the table and read off z.

z = 1.10

Answer

z = 1.10
2.Normal distribution

Answer: x = 39.69 min

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) = 0.9 → z = 1.282

Back to x

x = 32 + 1.282·6 = 39.69 min

Answer

x = 39.69 min
3.Confidence intervals

Answer: upper limit = 18.29 h

Interval for a difference: estimate ± critical value × standard error. For two independent means with known σ: SE = √(σ₁²/n₁ + σ₂²/n₂) and z. With equal unknown variances: pooled s_p and t with n₁ + n₂ − 2 degrees of freedom. For paired data, work with the differences: d̄ ± t·s_d/√n.

Point estimate

x̄₁ − x̄₂ = 114 − 103 = 11

Standard error

SE = √(16²/80 + 17²/60) = 2.831

Margin of error

E = 2.576·2.831 = 7.294

Interval

11 ± 7.294 → [3.706, 18.29]

Answer

upper limit = 18.29 h

The interval does not contain 0, so the means differ significantly.

4.Confidence intervals

Answer: σ = 14.82 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 = 14·9.4² = 1237

Limits for σ²

1237/26.119 = 47.36
1237/5.629 = 219.8

Limit for σ

σ = √219.8 = 14.82 µm

Answer

σ = 14.82 µm

95% interval for σ: [6.882, 14.82] 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 = 12/√64 = 12/8 = 1.5

Test statistic

t = (46.7 − 50)/1.5 = −2.2

Answer

t = −2.2
t0.025; df = 63 = 1.998

|t| > 1.998, so H₀ is rejected at the 5% level (two-sided).

6.Hypothesis tests & confidence intervals

Answer: L = 72.83

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 = 73.7 − 0.8711 = 72.83

Answer

L = 72.83
7.Linear regression & correlation

Answer: r = 0.933

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 = 4218 − 90·228/5 = 114
Sxx = 1660 − 90²/5 = 40
Syy = 10770 − 228²/5 = 373.2

Correlation coefficient

r = 114/√(40·373.2) = 0.933

Answer

r = 0.933
8.Linear regression & correlation

Answer: x = 17.91

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 = 120/6 = 20
ȳ = 419/6 = 69.83

Sums of squares and products

Sxx = Σ(x − x̄)² = 70
Sxy = Σ(x − x̄)(y − ȳ) = 229

Regression line

b = 229/70 = 3.271
a = 69.83 − 3.271·20 = 4.405
ŷ = 4.405 + 3.271·x

Solve for x

x = (63 − 4.405)/3.271 = 17.91

Answer

x = 17.91
9.Chi-square tests

Answer: E = 40.53

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 = 98·79/191 = 40.53

Answer

E = 40.53
10.Chi-square tests

Answer: df = 5

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 = (2 − 1)·(6 − 1) = 5

Answer

df = 5

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