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 = 0.8132

Drag z so that the shaded area between −z and z under the standard normal curve is 0.8132. Find z to two decimals.

A = 0.8132
z =
2.Normal distributionHard
157164171178185192199cmbottom 40%

The heights of adult men are normally distributed with mean 178 cm and standard deviation 7 cm. Drag the cut-off so the shaded bottom 40% is to the left of it. Find the cut-off x.

μ = 178 cmσ = 7 cm40%
x = cm
3.Confidence intervalsMedium
n = 13, sd​ = 2.30L = ?Ud̄ = 5.8E95%

The blood pressure of 13 patients is measured before and after a treatment. The differences d = before − after (mmHg) have mean 5.8 and standard deviation 2.3. Find the lower limit of the 95% confidence interval for the mean difference.

n = 13d̄ = 5.8sd = 2.395%t(12) = 2.179
lower limit = mmHg
4.Confidence intervalsHard
σ = 18, n = ?x̄ − Ex̄ + Ex̄E = 199%

The population standard deviation is 18. How large a sample n do you need so that a 99% confidence interval for the mean has a margin of error of at most 1?

σ = 18E = 199%
n =
5.Hypothesis tests & confidence intervalsMedium
t −1.9901.99

A sample of 81 students has a mean score of 46.56 and a sample standard deviation of 10. Find the test statistic t for H₀: μ = 50.

μ₀ = 50x̄ = 46.56s = 10n = 81
t =
6.Hypothesis tests & confidence intervalsHard
μ₀x̄c

A one-sided z test of H₀: μ = 50 against H₁: μ < 50 is done at α = 0.10 with a sample of 64; the standard deviation is known to be 20. In reality the mean is 46.25. Find the power 1 − β of the test in percent.

μ₀ = 50μ = 46.25σ = 20n = 64α = 0.10
1 − β = %
7.Linear regression & correlationMedium
xy1771198521942311025107xy202580100

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 = 105Σy = 467Σx² = 2245Σy² = 44651Σxy = 10001
r =
8.Linear regression & correlationHard
xy1475168418912010422119xy152080100120ŷ = 78

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. For which value x does the line predict ŷ = 78?

n = 5ŷ = 78
x =
9.Chi-square testsMedium
busbikecarTotalstudents9262560staff6142040Total154045100

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 = 1002 × 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 = 0.01.

α = 0.01
χ²crit =

3Solutions

1.Normal distribution

Answer: z = 1.32

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 + 0.8132)/2 = 0.9066

Read the z-table backwards

Find 0.9066 inside the table and read off z.

z = 1.32

Answer

z = 1.32
2.Normal distribution

Answer: x = 176.3 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) = 0.4

z from the table

z ≈ −0.25

Back to the original units

x = 178 − 0.25·7 = 176.3 cm

Answer

x = 176.3 cm
3.Confidence intervals

Answer: lower limit = 4.41 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 = 12, t = 2.179

Margin of error

E = 2.179·2.3/√13 = 1.39

Interval

5.8 ± 1.39 → [4.41, 7.19]

Answer

lower limit = 4.41 mmHg
4.Confidence intervals

Answer: n = 2150

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 = (2.576·18/1)² = 2150

Round up

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

n = 2150

Answer

n = 2150
5.Hypothesis tests & confidence intervals

Answer: t = −3.096

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 = 10/√81 = 10/9 = 1.111

Test statistic

t = (46.56 − 50)/1.111 = −3.096

Answer

t = −3.096
t0.025; df = 80 = 1.99

|t| > 1.99, so H₀ is rejected at the 5% level (two-sided).

6.Hypothesis tests & confidence intervals

Answer: 1 − β = 58.63 %

Power of a test: power = 1 − β is the probability of rejecting H₀ when it is false. Find the cutoff x̄c from μ₀ and z_α, then compute the probability of landing beyond x̄c using the true mean: z = (x̄c − μ)/(σ/√n).

Critical sample mean

SE = 20/√64 = 2.5
x̄c = 50 − 1.282·2.5 = 46.8

z under the true mean

z = (46.8 − 46.25)/2.5 = 0.218

Power

power = P(Z < 0.218) = 58.63%
1 − β = 58.63%

Answer

1 − β = 58.63%
7.Linear regression & correlation

Answer: r = 0.9543

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 = 10001 − 105·467/5 = 194
Sxx = 2245 − 105²/5 = 40
Syy = 44651 − 467²/5 = 1033

Correlation coefficient

r = 194/√(40·1033) = 0.9543

Answer

r = 0.9543
8.Linear regression & correlation

Answer: x = 14.93

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 = 90/5 = 18
ȳ = 473/5 = 94.6

Sums of squares and products

Sxx = Σ(x − x̄)² = 40
Sxy = Σ(x − x̄)(y − ȳ) = 216

Regression line

b = 216/40 = 5.4
a = 94.6 − 5.4·18 = −2.6
ŷ = −2.6 + 5.4·x

Solve for x

x = (78 − (−2.6))/5.4 = 14.93

Answer

x = 14.93
9.Chi-square tests

Answer: χ² = 0.787

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: 60·15/100 = 9, 60·40/100 = 24, 60·45/100 = 27
staff: 40·15/100 = 6, 40·40/100 = 16, 40·45/100 = 18

Chi-square statistic

χ² = (9 − 9)²/9 + (26 − 24)²/24 + (25 − 27)²/27 + (6 − 6)²/6 + (14 − 16)²/16 + (20 − 18)²/18
χ² = 0.787

Answer

χ² = 0.787

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

10.Chi-square tests

Answer: χ²crit = 13.28

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 = 0.01) = 13.277

Answer

χ²crit = 13.277

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

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