Research Methods and Statistics for Psychology: cram pack

Formula sheets, the key method steps for every topic and 16 practice problems, with full solutions at the back.

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1Formulas and methods

Descriptive statistics

Mean
x̄ = Σx/n
Sample standard deviation
s = √(Σ(x − x̄)²/(n − 1))
Median
middle value of the sorted data
z-score
z = (x − μ)/σ
Normal distribution
68% within ±1σ, 95% within ±2σ

Probability

Complement
P(not A) = 1 − P(A)
Addition rule
P(A ∪ B) = P(A) + P(B) − P(A ∩ B)
Independent events
P(A ∩ B) = P(A)·P(B)
Conditional probability
P(A|B) = P(A ∩ B)/P(B)
Binomial
P(X = k) = C(n,k)·pk(1 − p)n−k

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.Research methods in psychologyMedium

In a repeated measures design, order effects can be controlled by:

  1. ACounterbalancing
  2. BUsing a larger sample only
  3. CSingle-blind procedures
  4. DRandom sampling
2.Research methods in psychologyHard

Increasing the sample size in a study most directly:

  1. AGuarantees a statistically significant result if any effect exists
  2. BIncreases statistical power to detect a real effect
  3. CRemoves all bias from the sample
  4. DIncreases the effect size
3.Descriptive statisticsMedium
152025303540455055

Find the sample standard deviation s of the data: 30, 29, 59, 54, 14.

n = 5
s =
4.Descriptive statisticsHard
z −2−1012

Exam scores are normally distributed with mean 170 and standard deviation 12. Find the z-score z of a score of 155.6.

μ = 170σ = 12x = 155.6
z =
5.ProbabilityMedium
xp(x)00.310.3520.230.15μ = 1.2

For the distribution shown, find the standard deviation σ of X.

x: 0, 1, 2, 3p(x): 0.3, 0.35, 0.2, 0.15
σ =
6.ProbabilityHard
n = 11, p = 0.4: P(X = 3) = ?k01234567891011

A test has 11 multiple-choice questions. A student who guesses gets each one right with probability 0.4, independently. What is the probability P of exactly 3 correct answers?

n = 11p = 0.4k = 3
P =
7.Normal distributionMedium
125 kg126.1 kgT = X₁ + … + X₅?

The masses of bags of cement are independent and normal with mean 25 kg and standard deviation 0.5 kg. Find the probability P that the total mass of 5 of them exceeds 126.1 kg. Round z to two decimals.

μ = 25 kgσ = 0.5 kgn = 5> 126.1 kg
P =
8.Normal distributionHard
μ = 65x = ?10%

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

μ = 65σ = 10P = 10%
x = points
9.Confidence intervalsMedium
x̄E₁ = 4.790%, n = 64x̄E₂ = ?90%, n = 576

A 90% confidence interval for a mean, based on 64 observations and a known σ, has a margin of error of 4.7. What is the margin of error E₂ if the same study uses 576 observations instead?

90%n₁ = 64E₁ = 4.7n₂ = 576
E₂ =
10.Confidence intervalsHard
σ = 24, n = ?x̄ − Ex̄ + Ex̄E = 1.590%

The population standard deviation is 24. How large a sample n do you need so that a 90% confidence interval for the mean has a margin of error of at most 1.5?

σ = 24E = 1.590%
n =
11.Hypothesis tests & confidence intervalsMedium
t −1.99801.998

A sample of 64 students has a mean score of 50.85 and a sample standard deviation of 4. Find the test statistic t for H₀: μ = 50.

μ₀ = 50x̄ = 50.85s = 4n = 64
t =
12.Hypothesis tests & confidence intervalsHard
H₀H₁μ₀x̄cμ₀ + 5α = 0.01βσ = 26, power 80%, n = ?

A one-sided z test at α = 0.01 must detect a shift of the mean of 5 units with a power of 80%. The standard deviation is 26. Find the minimum sample size n.

α = 0.01power = 80%δ = 5σ = 26zα = 2.326zβ = 0.842
n =
13.Linear regression & correlationMedium
xy18952111624128271393016033186xy202530100150

The table gives 6 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 = 6Σx = 153Σy = 824Σx² = 4059Σy² = 118382Σxy = 21909
r =
14.Linear regression & correlationHard
xy124222318417513610xy24610152025x = 3.5

The table gives 6 paired observations of x = car age in years and y = price in k€. Use the least-squares line ŷ = a + b·x. Predict ŷ for x = 3.5.

n = 6x = 3.5
ŷ =
15.Chi-square testsMedium
improvednot improvedTotalnew drug322860placebo164460Total4872120

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 = 1202 × 2
χ² =
16.Chi-square testsHard
yesnoTotalplant A301040plant B372360plant C172340Total8456140dfα = 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 = 0.01 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 = 1403 × 2α = 0.01
χ² =

3Solutions

1.Research methods in psychology

Answer: A · Counterbalancing

Why this is the answer

Counterbalancing distributes practice and fatigue effects evenly across conditions.

Answer

A · Counterbalancing
2.Research methods in psychology

Answer: B · Increases statistical power to detect a real effect

Why this is the answer

Larger samples reduce standard error, increasing power. They do not change the true effect size or remove systematic bias.

Answer

B · Increases statistical power to detect a real effect
3.Descriptive statistics

Answer: s = 18.81

Descriptive statistics: the mean is the sum divided by n; the median is the middle value of the sorted data (the average of the two middle values when n is even). The sample standard deviation is s = √(Σ(x − x̄)²/(n − 1)). A z-score z = (x − μ)/σ says how many standard deviations a value lies from the mean.

Mean

x̄ = 186/5 = 37.2

Squared deviations

Σ(x − x̄)² = (−7.2)² + (−8.2)² + 21.8² + 16.8² + (−23.2)² = 1415

Standard deviation

s = √(1415/4) = 18.81

Answer

s = 18.81
4.Descriptive statistics

Answer: z = −1.2

Descriptive statistics: the mean is the sum divided by n; the median is the middle value of the sorted data (the average of the two middle values when n is even). The sample standard deviation is s = √(Σ(x − x̄)²/(n − 1)). A z-score z = (x − μ)/σ says how many standard deviations a value lies from the mean.

Standardise

z = (155.6 − 170)/12 = −1.2

Answer

z = −1.2
5.Probability

Answer: σ = 1.03

Discrete random variables: a probability distribution lists every value x a variable can take with its probability p(x); the probabilities are between 0 and 1 and add up to 1. The expected value (mean) is E(X) = Σx·p(x), the long-run average. The variance is Var(X) = Σ(x − μ)²·p(x) = E(X²) − μ², and the standard deviation σ = √Var(X). For a linear change Y = aX + b: E(Y) = a·E(X) + b, Var(Y) = a²·Var(X) and σY = |a|·σX; adding a constant does not change the spread.

Mean

μ = 0·0.3 + 1·0.35 + 2·0.2 + 3·0.15 = 1.2

Mean of the squares

E(X²) = 0·0.3 + 1·0.35 + 4·0.2 + 9·0.15 = 2.5

Variance

Var(X) = 2.5 − 1.2² = 1.06
σ = √1.06 = 1.03

Answer

σ = 1.03
6.Probability

Answer: P = 0.1774

Probability: with equally likely outcomes, P(A) = favourable/total. For independent events multiply: P(A and B) = P(A)·P(B). The complement often helps: P(at least one) = 1 − P(none). Binomial: in n independent trials with success probability p, P(X = k) = C(n, k)·pᵏ·(1 − p)ⁿ⁻ᵏ.

Binomial formula

P(X = 3) = 165·0.43·0.68 = 165·0.064·0.0168 = 0.1774

Answer

P = 0.1774
7.Normal distribution

Answer: P = 0.1635

Sums and differences of normal variables: if X and Y are independent and normal, X + Y and X − Y are normal too. Means add or subtract: E(X ± Y) = μX ± μY. Variances always add, also for a difference: Var(X ± Y) = σX² + σY², so σ = √(σX² + σY²). Standard deviations never add directly. The total of n independent items with the same distribution has mean n·μ and standard deviation σ·√n. Then standardise as usual: z = (x − mean)/σ and read Φ(z).

Distribution of the total

μT = 125 kg, σT = 0.5·√5 = 1.118 kg

Standardise

z = (126.1 − 125)/1.118 = 0.98

Probability

P = 1 − Φ(0.98) = 0.1635

Answer

P = 0.1635
8.Normal distribution

Answer: x = 52.18 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) = 0.1 → z = −1.282

Back to x

x = 65 + −1.282·10 = 52.18 points

Answer

x = 52.18 points
9.Confidence intervals

Answer: E₂ = 1.567

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.

Ratio of margins

E₂/E₁ = (1.645/1.645)·√(64/576) = 0.3333

New margin

E₂ = 4.7·0.3333 = 1.567

Answer

E₂ = 1.567
10.Confidence intervals

Answer: n = 693

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 = (1.645·24/1.5)² = 692.7

Round up

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

n = 693

Answer

n = 693
11.Hypothesis tests & confidence intervals

Answer: t = 1.7

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 = 4/√64 = 4/8 = 0.5

Test statistic

t = (50.85 − 50)/0.5 = 1.7

Answer

t = 1.7
t0.025; df = 63 = 1.998

|t| < 1.998, so H₀ is not rejected at the 5% level (two-sided).

12.Hypothesis tests & confidence intervals

Answer: n = 272

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)·26/5)² = 271.4

Round up

n = 272

Answer

n = 272
13.Linear regression & correlation

Answer: r = 0.9893

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 = 21909 − 153·824/6 = 897
Sxx = 4059 − 153²/6 = 157.5
Syy = 118382 − 824²/6 = 5219

Correlation coefficient

r = 897/√(157.5·5219) = 0.9893

Answer

r = 0.9893
14.Linear regression & correlation

Answer: ŷ = 17.33

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 = 21/6 = 3.5
ȳ = 104/6 = 17.33

Sums of squares and products

Sxx = Σ(x − x̄)² = 17.5
Sxy = Σ(x − x̄)(y − ȳ) = −49

Regression line

b = −49/17.5 = −2.8
a = 17.33 − (−2.8)·3.5 = 27.13
ŷ = 27.13 − 2.8·x

Prediction

ŷ = 27.13 + (−2.8)·3.5 = 17.33

Answer

ŷ = 17.33
15.Chi-square tests

Answer: χ² = 8.889

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

new drug: 60·48/120 = 24, 60·72/120 = 36
placebo: 60·48/120 = 24, 60·72/120 = 36

Chi-square statistic

χ² = (32 − 24)²/24 + (28 − 36)²/36 + (16 − 24)²/24 + (44 − 36)²/36
χ² = 8.889

Answer

χ² = 8.889

df = 1. At α = 0.05 the critical value is 3.841, so H₀ (independence) is rejected.

16.Chi-square tests

Answer: χ² = 8.924

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

plant A: 24, 16
plant B: 36, 24
plant C: 24, 16

Chi-square statistic

χ² = (30 − 24)²/24 + (10 − 16)²/16 + (37 − 36)²/36 + (23 − 24)²/24 + (17 − 24)²/24 + (23 − 16)²/16
χ² = 8.924

Decision

df = 2, χ²crit = 9.210
8.924 < 9.210

Answer

χ² = 8.924

χ² is below the critical value, so H₀ is not rejected: there is no significant evidence of a relationship.

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