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 = 0.9656

Drag z so that the shaded area under the standard normal curve, to the right of z, is 0.9656. Find z to two decimals.

A = 0.9656
z =
2.Normal distributionHard
μ = 65x = ?97.5%

The exam scores are normally distributed with mean 65 points and standard deviation 10 points. Find the value x below which 97.5% of them lie (the 97.5th percentile).

μ = 65σ = 10P = 97.5%
x = points
3.Confidence intervalsMedium
n = 12, sd​ = 3.10L = ?Ud̄ = −6.7E99%

The blood pressure of 12 patients is measured before and after a treatment. The differences d = before − after (mmHg) have mean −6.7 and standard deviation 3.1. Find the lower limit of the 99% confidence interval for the mean difference.

n = 12d̄ = −6.7sd = 3.199%t(11) = 3.106
lower limit = mmHg
4.Confidence intervalsHard
σ = 37, n = ?x̄ − Ex̄ + Ex̄E = 495%

The population standard deviation is 37. How large a sample n do you need so that a 95% confidence interval for the mean has a margin of error of at most 4?

σ = 37E = 495%
n =
5.Hypothesis tests & confidence intervalsMedium
−2.2622.262t = ?df = 9

10 employees take a typing test before and after a training course. For each person the difference d = after − before (words per minute) is recorded. The mean difference is 1.57 and the standard deviation of the differences is 3.1. Find the paired test statistic t for H₀: μd = 0.

n = 10d̄ = 1.57sd = 3.1
t =
6.Hypothesis tests & confidence intervalsHard
−1.961.96z = ?

In a trial, 53 of 150 patients on treatment A recover, and 26 of 120 patients on treatment B. Using the pooled proportion, find the test statistic z for H₀: p₁ = p₂.

x₁ = 53n₁ = 150x₂ = 26n₂ = 120
z =
7.Linear regression & correlationMedium
xy18752190241002711530118xy20253080100120

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 = 120Σy = 498Σx² = 2970Σy² = 50874Σxy = 12285
r =
8.Linear regression & correlationHard
xy16421862206422732475xy204060x = 23

The table gives 5 paired observations of x = temperature in °C and y = ice creams sold. Use the least-squares line ŷ = a + b·x. Predict ŷ for x = 23.

n = 5x = 23
ŷ =
9.Chi-square testsMedium
busbikecarTotalstudents8112140staff243759120Total324880160

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
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 6 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: z = −1.82

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 − 0.9656 = 0.0344

Read the z-table backwards

Find 0.0344 inside the table and read off z.

z = −1.82

Answer

z = −1.82
2.Normal distribution

Answer: x = 84.6 points

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 = 65 + 1.96·10 = 84.6 points

Answer

x = 84.6 points
3.Confidence intervals

Answer: lower limit = −9.48 mmHg

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.

Critical value

df = 11, t = 3.106

Margin of error

E = 3.106·3.1/√12 = 2.78

Interval

−6.7 ± 2.78 → [−9.48, −3.92]

Answer

lower limit = −9.48 mmHg
4.Confidence intervals

Answer: n = 329

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.

Solve for n

n = (1.96·37/4)² = 328.7

Round up

A sample size is a whole number, and rounding down would make the margin too large.

n = 329

Answer

n = 329
5.Hypothesis tests & confidence intervals

Answer: t = 1.602

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 of the mean difference

SE = 3.1/√10 = 0.9803

Test statistic

t = 1.57/0.9803 = 1.602

Answer

t = 1.602

With df = 9, the two-sided 5% critical value is 2.262; |t| is smaller, so H₀ is not rejected.

6.Hypothesis tests & confidence intervals

Answer: z = 2.453

Hypothesis test: the test statistic = (estimate − value under H₀)/(standard error under H₀). The p-value is the probability, assuming H₀ is true, of a statistic at least as extreme as the one observed: the area in one tail for a one-sided H₁, twice the tail area for a two-sided H₁. Reject H₀ when p ≤ α.

Sample proportions

p̂₁ = 53/150 = 0.3533
p̂₂ = 26/120 = 0.2167
p̂ = 79/270 = 0.2926

Standard error under H₀

SE = √(0.2926·0.7074·(1/150 + 1/120)) = 0.05572

Test statistic

z = (0.3533 − 0.2167)/0.05572 = 2.453

Answer

z = 2.453

|z| > 1.96, so H₀ is rejected at the 5% level (two-sided).

7.Linear regression & correlation

Answer: r = 0.9837

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 = 12285 − 120·498/5 = 333
Sxx = 2970 − 120²/5 = 90
Syy = 50874 − 498²/5 = 1273

Correlation coefficient

r = 333/√(90·1273) = 0.9837

Answer

r = 0.9837
8.Linear regression & correlation

Answer: ŷ = 74.75

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 = 100/5 = 20
ȳ = 316/5 = 63.2

Sums of squares and products

Sxx = Σ(x − x̄)² = 40
Sxy = Σ(x − x̄)(y − ȳ) = 154

Regression line

b = 154/40 = 3.85
a = 63.2 − 3.85·20 = −13.8
ŷ = −13.8 + 3.85·x

Prediction

ŷ = −13.8 + 3.85·23 = 74.75

Answer

ŷ = 74.75
9.Chi-square tests

Answer: χ² = 0.1778

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 = (8 − 8)²/8 + (11 − 12)²/12 + (21 − 20)²/20 + (24 − 24)²/24 + (37 − 36)²/36 + (59 − 60)²/60
χ² = 0.1778

Answer

χ² = 0.1778

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

10.Chi-square tests

Answer: χ²crit = 9.488

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 = 6 − 1 − 1 = 4

Table

χ²crit(df = 4, α = 0.05) = 9.488

Answer

χ²crit = 9.488

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

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