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 study is preregistered. This means that:

  1. AThe results were checked and approved by a second, independent statistician before publication
  2. BParticipants register their consent online
  3. CThe study has already been published
  4. DThe hypotheses, methods and analysis plan are recorded publicly before data collection
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. BPublication bias means unpublished null studies may be missing
  3. CMeta-analyses cannot include randomised trials
  4. DMeta-analyses always give smaller effects than the single studies they include
3.Descriptive statisticsMedium
20253035404550

Find the sample standard deviation s of the data: 21, 19, 49, 52, 20, 52.

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

A quantity is normally distributed with mean 100 and standard deviation 12. What proportion P of values is below 107.2?

μ = 100σ = 12x = 107.2
P =
5.ProbabilityMedium
xp(x)10.220.3530.240.25μ = 2.5

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

x: 1, 2, 3, 4p(x): 0.2, 0.35, 0.2, 0.25
σ =
6.ProbabilityHard
€p(x)−100.8300.154900.05

A game costs €10 to play. You win €40 with probability 0.15 and €500 with probability 0.05; otherwise you win nothing. The stake is not returned. Find the expected net gain E(G) per game.

stake = €10€40: p = 0.15€500: p = 0.05
E(G) = €
7.Normal distributionMedium
−3−2−10123zshaded area to the right of z = 0.0336

Drag z so that the shaded area under the standard normal curve, to the right of z, is 0.0336. Find z to two decimals.

A = 0.0336
z =
8.Normal distributionHard
235240245250255260265gbottom 10%

The bag weights are normally distributed with mean 250 g and standard deviation 5 g. Drag the cut-off so the shaded bottom 10% is to the left of it. Find the cut-off x.

μ = 250 gσ = 5 g10%
x = g
9.Confidence intervalsMedium
n = 16, sd​ = 10LU = ?d̄ = 5.9E95%

The blood pressure of 16 patients is measured before and after a treatment. The differences d = before − after (mmHg) have mean 5.9 and standard deviation 1. Find the upper limit of the 95% confidence interval for the mean difference.

n = 16d̄ = 5.9sd = 195%t(15) = 2.131
upper limit = mmHg
10.Confidence intervalsHard
σ = 40, n = ?x̄ − Ex̄ + Ex̄E = 195%

The population standard deviation is 40. 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 1?

σ = 40E = 195%
n =
11.Hypothesis tests & confidence intervalsMedium
t −2.06402.064

A sample of 25 students has a mean score of 500.7 and a sample standard deviation of 5. Find the test statistic t for H₀: μ = 500.

μ₀ = 500x̄ = 500.7s = 5n = 25
t =
12.Hypothesis tests & confidence intervalsHard
z −2.57602.576

A sample of 49 measurements has mean 117.4. The population standard deviation is 12. Find the upper limit U of the 99% confidence interval for the mean.

x̄ = 117.4σ = 12n = 4999%
U =
13.Linear regression & correlationMedium
xy1283365467589751186xy510406080

The table gives 6 paired observations of x = advertising spend in k€ and y = sales in k€. Find Pearson's correlation coefficient r.

n = 6
r =
14.Linear regression & correlationHard
xy234661281410181222xy5101020x = 7

The table gives 6 paired observations of x = load in kN and y = extension in mm. Use the least-squares line ŷ = a + b·x. A new observation (7, 15) is made. Find its residual e = y − ŷ.

n = 6(7, 15)
e =
15.Chi-square testsMedium
lowmediumhighTotalurban17214280rural11212860Total284270140

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 = 1402 × 3
χ² =
16.Chi-square testsHard

A chi-square goodness-of-fit test compares observed counts in 8 categories with fully specified expected proportions (no parameters are estimated from the data). How many degrees of freedom df does the test have?

k = 8
df =

3Solutions

1.Research methods in psychology

Answer: D · The hypotheses, methods and analysis plan are recorded publicly before data collection

Why this is the answer

Preregistration separates confirmatory from exploratory analyses and reduces practices such as p-hacking and HARKing.

Answer

D · The hypotheses, methods and analysis plan are recorded publicly before data collection
2.Research methods in psychology

Answer: B · 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

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

Answer: s = 17.03

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 = 213/6 = 35.5

Squared deviations

Σ(x − x̄)² = (−14.5)² + (−16.5)² + 13.5² + 16.5² + (−15.5)² + 16.5² = 1450

Standard deviation

s = √(1450/5) = 17.03

Answer

s = 17.03
4.Descriptive statistics

Answer: P = 0.7257

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 = 0.6

Standard normal table

Φ(0.6) = 0.7257
P = 0.7257

Answer

P = 0.7257
5.Probability

Answer: σ = 1.072

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 = 1·0.2 + 2·0.35 + 3·0.2 + 4·0.25 = 2.5

Mean of the squares

E(X²) = 1·0.2 + 4·0.35 + 9·0.2 + 16·0.25 = 7.4

Variance

Var(X) = 7.4 − 2.5² = 1.15
σ = √1.15 = 1.072

Answer

σ = 1.072
6.Probability

Answer: E(G) = 21 €

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.

Net outcomes

G: 30 (p = 0.15), 490 (p = 0.05), −10 (p = 0.8)

Expected value

E(G) = 30·0.15 + 490·0.05 − 10·0.8 = 21 €

Answer

E(G) = 21 €

A positive value means you win money on average.

7.Normal distribution

Answer: z = 1.83

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 − 0.0336 = 0.9664

Read the z-table backwards

Find 0.9664 inside the table and read off z.

z = 1.83

Answer

z = 1.83
8.Normal distribution

Answer: x = 243.6 g

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) = 0.1

z from the table

z ≈ −1.28

Back to the original units

x = 250 − 1.28·5 = 243.6 g

Answer

x = 243.6 g
9.Confidence intervals

Answer: upper limit = 6.433 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 = 15, t = 2.131

Margin of error

E = 2.131·1/√16 = 0.5327

Interval

5.9 ± 0.5327 → [5.367, 6.433]

Answer

upper limit = 6.433 mmHg
10.Confidence intervals

Answer: n = 6147

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 = (1.96·40/1)² = 6147

Round up

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

n = 6147

Answer

n = 6147
11.Hypothesis tests & confidence intervals

Answer: t = 0.7

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 = 5/√25 = 5/5 = 1

Test statistic

t = (500.7 − 500)/1 = 0.7

Answer

t = 0.7
t0.025; df = 24 = 2.064

|t| < 2.064, so H₀ is not rejected at the 5% level (two-sided).

12.Hypothesis tests & confidence intervals

Answer: U = 121.8

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).

Margin of error

E = 2.576·12/√49 = 4.416

Interval limit

U = 117.4 + 4.416 = 121.8

Answer

U = 121.8
13.Linear regression & correlation

Answer: r = 0.9935

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 = 36/6 = 6
ȳ = 329/6 = 54.83

Sums of squares and products

Sxx = Σ(x − x̄)² = 70
Sxy = Σ(x − x̄)(y − ȳ) = 419
Syy = Σ(y − ȳ)² = 2541

Correlation coefficient

r = 419/√(70·2541) = 0.9935

Answer

r = 0.9935
14.Linear regression & correlation

Answer: e = 2.5

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 = 42/6 = 7
ȳ = 75/6 = 12.5

Sums of squares and products

Sxx = Σ(x − x̄)² = 70
Sxy = Σ(x − x̄)(y − ȳ) = 133

Regression line

b = 133/70 = 1.9
a = 12.5 − 1.9·7 = −0.8
ŷ = −0.8 + 1.9·x

Predicted value

ŷ = −0.8 + 1.9·7 = 12.5

Residual

e = 15 − 12.5 = 2.5

Answer

e = 2.5

The point lies above the line.

15.Chi-square tests

Answer: χ² = 1.254

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

urban: 80·28/140 = 16, 80·42/140 = 24, 80·70/140 = 40
rural: 60·28/140 = 12, 60·42/140 = 18, 60·70/140 = 30

Chi-square statistic

χ² = (17 − 16)²/16 + (21 − 24)²/24 + (42 − 40)²/40 + (11 − 12)²/12 + (21 − 18)²/18 + (28 − 30)²/30
χ² = 1.254

Answer

χ² = 1.254

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

16.Chi-square tests

Answer: df = 7

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 = 8 − 1 = 7

Answer

df = 7

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