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.

New problemsBack to the course

In the print window, choose “Save as PDF” to keep a copy on your device.

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

A researcher studies the effect of gender on memory scores. Why is this a quasi-experiment rather than a true experiment?

  1. AThe DV is not measured
  2. BThe IV is a pre-existing characteristic, so participants cannot be randomly allocated to it
  3. CThere are only two conditions, and a true experiment needs at least three levels of the IV
  4. DThe study takes place in a laboratory
2.Research methods in psychologyHard

The "replication crisis" in psychology refers to:

  1. AJournals being legally required to reject all replication studies submitted
  2. BCopying of questionnaires without permission
  3. CMany published findings failing to reproduce in new studies
  4. DA shortage of participants for experiments
3.Descriptive statisticsMedium
25303540455055

Find the sample standard deviation s of the data: 42, 47, 55, 30, 22, 23.

n = 6
s =
4.Descriptive statisticsHard
z −2−1012

A quantity is normally distributed with mean 50 and standard deviation 5. What proportion P of values is above 60?

μ = 50σ = 5x = 60
P =
5.ProbabilityMedium
xp(x)00.2510.1520.2530.35μ = 1.7

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

x: 0, 1, 2, 3p(x): 0.25, 0.15, 0.25, 0.35
σ =
6.ProbabilityHard
n = 8, p = 0.4: P(X ≤ 1) = ?k012345678

8 components are tested, and each one fails with probability 0.4, independently. What is the probability P that at most one fails?

n = 8p = 0.4
P =
7.Normal distributionMedium
502572P = ?

The battery lifetimes are normally distributed with mean 500 h and standard deviation 40 h. What fraction lies between 502 and 572 h? Give P as a decimal.

μ = 500σ = 40x1 = 502x2 = 572
P =
8.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
9.Confidence intervalsMedium
x̄E₁ = 3.295%, n = 64x̄E₂ = ?90%, n = 64

A 95% confidence interval for a mean, based on 64 observations and a known σ, has a margin of error of 3.2. What is the margin of error E₂ of a 90% interval from the same data?

95%n₁ = 64E₁ = 3.290%
E₂ =
10.Confidence intervalsHard
σ = 30, n = ?x̄ − Ex̄ + Ex̄E = 499%

The population standard deviation is 30. 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 4?

σ = 30E = 499%
n =
11.Hypothesis tests & confidence intervalsMedium
t −1.9901.99

A sample of 81 students has a mean score of 200.4 and a sample standard deviation of 5. Find the test statistic t for H₀: μ = 200.

μ₀ = 200x̄ = 200.4s = 5n = 81
t =
12.Hypothesis tests & confidence intervalsHard
H₀H₁μ₀x̄cμ₀ + 3α = 0.01βσ = 23, power 90%, n = ?

A one-sided z test at α = 0.01 must detect a shift of the mean of 3 units with a power of 90%. The standard deviation is 23. Find the minimum sample size n.

α = 0.01power = 90%δ = 3σ = 23zα = 2.326zβ = 1.282
n =
13.Linear regression & correlationMedium
xy125226320417516613xy246152025

The table gives 6 paired observations of x = car age in years and y = price in k€. Find Pearson's correlation coefficient r.

n = 6
r =
14.Linear regression & correlationHard
xy123223321420515xy241520ŷ = 23

The table gives 5 paired observations of x = car age in years and y = price in k€. Use the least-squares line ŷ = a + b·x. For which value x does the line predict ŷ = 23?

n = 5ŷ = 23
x =
15.Chi-square testsMedium
improvednot improvedTotalnew drug174360placebo312960Total4872120

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 A76480plant B271340plant C411960Total14436180dfα = 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 = 1803 × 2α = 0.01
χ² =

3Solutions

1.Research methods in psychology

Answer: B · The IV is a pre-existing characteristic, so participants cannot be randomly allocated to it

Why this is the answer

In a quasi-experiment the IV is a naturally occurring difference between people. Without random allocation, other differences between groups may confound the results.

Answer

B · The IV is a pre-existing characteristic, so participants cannot be randomly allocated to it
2.Research methods in psychology

Answer: C · Many published findings failing to reproduce in new studies

Why this is the answer

Large replication projects found many effects were smaller or absent, linked to publication bias, small samples and flexible analysis. Pre-registration is one response.

Answer

C · Many published findings failing to reproduce in new studies
3.Descriptive statistics

Answer: s = 13.55

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 = 219/6 = 36.5

Squared deviations

Σ(x − x̄)² = 5.5² + 10.5² + 18.5² + (−6.5)² + (−14.5)² + (−13.5)² = 917.5

Standard deviation

s = √(917.5/5) = 13.55

Answer

s = 13.55
4.Descriptive statistics

Answer: P = 0.02275

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 = 2

Standard normal table

Φ(2) = 0.9772
P = 1 − 0.9772 = 0.02275

Answer

P = 0.02275
5.Probability

Answer: σ = 1.187

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 = 0·0.25 + 1·0.15 + 2·0.25 + 3·0.35 = 1.7

Mean of the squares

E(X²) = 0·0.25 + 1·0.15 + 4·0.25 + 9·0.35 = 4.3

Variance

Var(X) = 4.3 − 1.7² = 1.41
σ = √1.41 = 1.187

Answer

σ = 1.187
6.Probability

Answer: P = 0.1064

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.

Zero and one failure

P(X = 0) = 0.68 = 0.0168
P(X = 1) = 8·0.4·0.67 = 0.08958

Add them

P = 0.0168 + 0.08958 = 0.1064

Answer

P = 0.1064
7.Normal distribution

Answer: P = 0.4441

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-scores

z₁ = (502 − 500)/40 = 0.05
z₂ = (572 − 500)/40 = 1.8

Area between

P = Φ(1.8) − Φ(0.05) = 0.9641 − 0.5199 = 0.4441

Answer

P = 0.4441
8.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
9.Confidence intervals

Answer: E₂ = 2.686

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 = (1.645/1.96)·√(64/64) = 0.8393

New margin

E₂ = 3.2·0.8393 = 2.686

Answer

E₂ = 2.686
10.Confidence intervals

Answer: n = 374

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 = (2.576·30/4)² = 373.3

Round up

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

n = 374

Answer

n = 374
11.Hypothesis tests & confidence intervals

Answer: t = 0.72

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 = 5/√81 = 5/9 = 0.5556

Test statistic

t = (200.4 − 200)/0.5556 = 0.72

Answer

t = 0.72
t0.025; df = 80 = 1.99

|t| < 1.99, so H₀ is not rejected at the 5% level (two-sided).

12.Hypothesis tests & confidence intervals

Answer: n = 766

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 + 1.282)·23/3)² = 765.1

Round up

n = 766

Answer

n = 766
13.Linear regression & correlation

Answer: r = −0.962

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 = 117/6 = 19.5

Sums of squares and products

Sxx = Σ(x − x̄)² = 17.5
Sxy = Σ(x − x̄)(y − ȳ) = −46.5
Syy = Σ(y − ȳ)² = 133.5

Correlation coefficient

r = −46.5/√(17.5·133.5) = −0.962

Answer

r = −0.962
14.Linear regression & correlation

Answer: x = 1.632

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 = 15/5 = 3
ȳ = 102/5 = 20.4

Sums of squares and products

Sxx = Σ(x − x̄)² = 10
Sxy = Σ(x − x̄)(y − ȳ) = −19

Regression line

b = −19/10 = −1.9
a = 20.4 − (−1.9)·3 = 26.1
ŷ = 26.1 − 1.9·x

Solve for x

x = (23 − 26.1)/(−1.9) = 1.632

Answer

x = 1.632
15.Chi-square tests

Answer: χ² = 6.806

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 = (17 − 24)²/24 + (43 − 36)²/36 + (31 − 24)²/24 + (29 − 36)²/36
χ² = 6.806

Answer

χ² = 6.806

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

16.Chi-square tests

Answer: χ² = 20.26

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: 64, 16
plant B: 32, 8
plant C: 48, 12

Chi-square statistic

χ² = (76 − 64)²/64 + (4 − 16)²/16 + (27 − 32)²/32 + (13 − 8)²/8 + (41 − 48)²/48 + (19 − 12)²/12
χ² = 20.26

Decision

df = 2, χ²crit = 9.210
20.26 > 9.210

Answer

χ² = 20.26

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

ExamProblems · examproblems.com/courses/research-methods-psychology