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

Which test is suitable for comparing the means of two independent groups with interval data that are roughly normal?

  1. AWilcoxon signed-rank test
  2. BIndependent samples t-test
  3. CPearson correlation
  4. DChi-square test
2.Research methods in psychologyHard

A meta-analysis combines 40 studies of a therapy. Why might its overall effect size still be too large?

  1. ACombining studies always removes all bias
  2. BMeta-analyses always give smaller effects than the single studies they include
  3. CPublication bias means unpublished null studies may be missing
  4. DMeta-analyses cannot include randomised trials
3.Descriptive statisticsMedium
303540455055

Find the sample standard deviation s of the data: 26, 46, 33, 58, 53, 41.

n = 6
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 185.

μ = 170σ = 12x = 185
z =
5.ProbabilityMedium
red: 6blue: 8

A bag holds 6 red and 8 blue balls. You draw two balls without putting the first one back. What is the probability P that both are red?

6 / 8
P =
6.ProbabilityHard
xp(x)00.2520.240.2560.3μ

X has the distribution shown, and Y = 10X − 3. Find the standard deviation σY of Y.

x: 0, 2, 4, 6p(x): 0.25, 0.2, 0.25, 0.3Y = 10X − 3
σY =
7.Normal distributionMedium
−3−2−10123zshaded area between −z and z = 0.4647

Drag z so that the shaded area between −z and z under the standard normal curve is 0.4647. Find z to two decimals.

A = 0.4647
z =
8.Normal distributionHard
380420460500540580620htop 30%

The battery lifetimes are normally distributed with mean 500 h and standard deviation 40 h. Drag the cut-off so the shaded top 30% is to the right of it. Find the cut-off x.

μ = 500 hσ = 40 h30%
x = h
9.Confidence intervalsMedium
n = 25, s = 8LUx̄ = 28.5E = ?95%

A sample of 25 items has mean 28.5 and standard deviation s = 8. The population is roughly normal. Find the margin of error E of a 95% confidence interval for the mean.

n = 25x̄ = 28.5s = 895%t(24) = 2.064
E =
10.Confidence intervalsHard
μ₁ − μ₂, t(24) = 2.0640LU = ?x̄₁ − x̄₂E95%

Two fertilisers are tested. Plot group 1: 8 plots, mean yield 36 kg, standard deviation 4 kg. Plot group 2: 18 plots, mean 37.5 kg, standard deviation 7.5 kg. Assume equal variances (pooled t, df = n₁ + n₂ − 2). Find the upper limit of the 95% confidence interval for μ₁ − μ₂.

n₁ = 8x̄₁ = 36s₁ = 4n₂ = 18x̄₂ = 37.5s₂ = 7.595%t(24) = 2.064
upper limit = kg
11.Hypothesis tests & confidence intervalsMedium
t −2.13102.131

A sample of 16 students has a mean score of 55.1 and a sample standard deviation of 12. Find the test statistic t for H₀: μ = 50.

μ₀ = 50x̄ = 55.1s = 12n = 16
t =
12.Hypothesis tests & confidence intervalsHard
H₀H₁μ₀x̄cμ₀ + 2.5α = 0.01βσ = 17, power 80%, n = ?

A one-sided z test at α = 0.01 must detect a shift of the mean of 2.5 units with a power of 80%. The standard deviation is 17. Find the minimum sample size n.

α = 0.01power = 80%δ = 2.5σ = 17zα = 2.326zβ = 0.842
n =
13.Linear regression & correlationMedium
xy243342450555656768xy24640506070

The table gives 6 paired observations of x = advertising spend in k€ and y = sales in k€. Find the coefficient of determination r² of the least-squares line.

n = 6
r² =
14.Linear regression & correlationHard
xy126224323422519616xy2462025x = 1.5

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. A new observation (1.5, 28) is made. Find its residual e = y − ŷ.

n = 6(1.5, 28)
e =
15.Chi-square testsMedium
improvednot improvedTotalnew drug352560placebo372360Total7248120

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
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 test of independence is carried out on a contingency table with 4 rows and 4 columns (not counting the totals). Using the table, find the critical value χ²crit at significance level α = 0.01.

α = 0.01
χ²crit =

3Solutions

1.Research methods in psychology

Answer: B · Independent samples t-test

Why this is the answer

The independent t-test compares two group means. Chi-square is for frequencies; Wilcoxon is for related samples with non-parametric data.

Answer

B · Independent samples t-test
2.Research methods in psychology

Answer: C · Publication bias means unpublished null studies may be missing

Why this is the answer

If null results are less often published, the pooled effect is inflated. Funnel plots and searches for unpublished data help detect and reduce this.

Answer

C · Publication bias means unpublished null studies may be missing
3.Descriptive statistics

Answer: s = 12.06

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 = 257/6 = 42.83

Squared deviations

Σ(x − x̄)² = (−16.83)² + 3.167² + (−9.833)² + 15.17² + 10.17² + (−1.833)² = 726.8

Standard deviation

s = √(726.8/5) = 12.06

Answer

s = 12.06
4.Descriptive statistics

Answer: z = 1.25

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 = (185 − 170)/12 = 1.25

Answer

z = 1.25
5.Probability

Answer: P = 0.1648

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.

First and second draw

P = 6/14 · 5/13 = 30/182 = 0.1648

Answer

P = 0.1648
6.Probability

Answer: σY = 23.15

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.

Spread of X

μ = 3.2, Var(X) = 5.36, σX = 2.315

Linear change

σY = |10|·2.315 = 23.15

Answer

σY = 23.15

The mean does move: E(Y) = 10·3.2 − 3 = 29.

7.Normal distribution

Answer: z = 0.62

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 + 0.4647)/2 = 0.7324

Read the z-table backwards

Find 0.7324 inside the table and read off z.

z = 0.62

Answer

z = 0.62
8.Normal distribution

Answer: x = 521 h

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) = 0.7

z from the table

z ≈ 0.52

Back to the original units

x = 500 + 0.52·40 = 521 h

Answer

x = 521 h
9.Confidence intervals

Answer: E = 3.302

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.

Critical value

df = n − 1 = 24, t = 2.064

Margin of error

E = 2.064·8/√25 = 3.302

Answer

E = 3.302
10.Confidence intervals

Answer: upper limit = 4.351 kg

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.

Pooled standard deviation

sp² = (7·4² + 17·7.5²)/24 = 44.51
sp = 6.672

Margin of error

E = 2.064·6.672·√(1/8 + 1/18) = 5.851

Interval

−1.5 ± 5.851 → [−7.351, 4.351]

Answer

upper limit = 4.351 kg
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 = 12/√16 = 12/4 = 3

Test statistic

t = (55.1 − 50)/3 = 1.7

Answer

t = 1.7
t0.025; df = 15 = 2.131

|t| < 2.131, so H₀ is not rejected at the 5% level (two-sided).

12.Hypothesis tests & confidence intervals

Answer: n = 465

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)·17/2.5)² = 464.1

Round up

n = 465

Answer

n = 465
13.Linear regression & correlation

Answer: r² = 0.9082

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 = 27/6 = 4.5
ȳ = 314/6 = 52.33

Sums of squares and products

Sxx = Σ(x − x̄)² = 17.5
Sxy = Σ(x − x̄)(y − ȳ) = 86
Syy = Σ(y − ȳ)² = 465.3

Coefficient of determination

r² = 86²/(17.5·465.3) = 0.9082

Answer

r² = 0.9082

About 90.82% of the variation in y is explained by the line.

14.Linear regression & correlation

Answer: e = 2.562

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 = 130/6 = 21.67

Sums of squares and products

Sxx = Σ(x − x̄)² = 17.5
Sxy = Σ(x − x̄)(y − ȳ) = −33

Regression line

b = −33/17.5 = −1.886
a = 21.67 − (−1.886)·3.5 = 28.27
ŷ = 28.27 − 1.886·x

Predicted value

ŷ = 28.27 + (−1.886)·1.5 = 25.44

Residual

e = 28 − 25.44 = 2.562

Answer

e = 2.562

The point lies above the line.

15.Chi-square tests

Answer: χ² = 0.1389

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

Chi-square statistic

χ² = (35 − 36)²/36 + (25 − 24)²/24 + (37 − 36)²/36 + (23 − 24)²/24
χ² = 0.1389

Answer

χ² = 0.1389

df = 1. At α = 0.05 the critical value is 3.841, so H₀ (independence) is not rejected.

16.Chi-square tests

Answer: χ²crit = 21.67

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 = (4 − 1)·(4 − 1) = 9

Table

χ²crit(df = 9, α = 0.01) = 21.666

Answer

χ²crit = 21.666

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

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