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

Experts check whether the items of a maths exam cover all topics in the syllabus. This assesses:

  1. AContent validity
  2. BEcological validity
  3. CTest-retest reliability
  4. DPredictive validity
2.Research methods in psychologyHard

Which non-parametric test compares two related samples, such as before-and-after scores on an ordinal scale?

  1. AChi-square test of independence
  2. BMann–Whitney U test
  3. CIndependent t-test
  4. DWilcoxon signed-rank test
3.Descriptive statisticsMedium
2025303540455055

Find the sample standard deviation s of the data: 43, 52, 57, 44, 17, 28.

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

Exam scores are normally distributed with mean 100 and standard deviation 20. Find the z-score z of a score of 112.

μ = 100σ = 20x = 112
z =
5.ProbabilityMedium
xp(x)100.2200.35300.3400.15μ = 24

For the distribution shown, find the variance Var(X) of X.

x: 10, 20, 30, 40p(x): 0.2, 0.35, 0.3, 0.15
Var(X) =
6.ProbabilityHard
€p(x)−30.7590.21470.05

A game costs €3 to play. You win €12 with probability 0.2 and €150 with probability 0.05; otherwise you win nothing. The stake is not returned. Find the expected net gain E(G) per game.

stake = €3€12: p = 0.2€150: p = 0.05
E(G) = €
7.Normal distributionMedium
−3−2−10123zshaded area to the right of z = 0.9854

Drag z so that the shaded area under the standard normal curve, to the right of z, is 0.9854. Find z to two decimals.

A = 0.9854
z =
8.Normal distributionHard
35455565758595pointstop 75%

The exam scores are normally distributed with mean 65 points and standard deviation 10 points. Drag the cut-off so the shaded top 75% is to the right of it. Find the cut-off x.

μ = 65 pointsσ = 10 points75%
x = points
9.Confidence intervalsMedium
μ₁ − μ₂0LU = ?x̄₁ − x̄₂E90%

The battery life of two phone models is compared. Model 1: 100 phones, mean 78 h, known standard deviation 10 h. Model 2: 50 phones, mean 70 h, known standard deviation 5 h. Find the upper limit of the 90% confidence interval for μ₁ − μ₂.

n₁ = 100x̄₁ = 78 hσ₁ = 10 hn₂ = 50x̄₂ = 70 hσ₂ = 5 h90%
upper limit = h
10.Confidence intervalsHard
σ = 19, n = ?x̄ − Ex̄ + Ex̄E = 499%

The population standard deviation is 19. 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 = 19E = 499%
n =
11.Hypothesis tests & confidence intervalsMedium
t −1.99801.998

A sample of 64 students has a mean score of 499.1 and a sample standard deviation of 8. Find the test statistic t for H₀: μ = 500.

μ₀ = 500x̄ = 499.1s = 8n = 64
t =
12.Hypothesis tests & confidence intervalsHard
σ = 8, n = ?x̄ − Ex̄ + Ex̄E = 399%

How large must a sample be so that the 99% confidence interval for a mean has a margin of error of at most 3? The standard deviation is 8. Find the smallest sample size n.

σ = 8E = 399%
n =
13.Linear regression & correlationMedium
xy284196268351043xy5102040

The table gives 5 paired observations of x = load in kN and y = extension in mm. Their sums are given. Find Pearson's correlation coefficient r from these sums.

n = 5Σx = 30Σy = 131Σx² = 220Σy² = 4175Σxy = 958
r =
14.Linear regression & correlationHard
xy18872010022982411126120xy2025100120x = 21

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. A new observation (21, 96) is made. Find its residual e = y − ŷ.

n = 5(21, 96)
e =
15.Chi-square testsMedium
lowmediumhighTotalurban27372185rural12252057Total396241142

The contingency table shows observed counts. For a chi-square test of independence, find the expected count E for the cell “rural, low”.

row total = 57column total = 39N = 142
E =
16.Chi-square testsHard

A chi-square test of independence is carried out on a contingency table with 2 rows and 6 columns (not counting the totals). How many degrees of freedom df does the test have?

2 × 6
df =

3Solutions

1.Research methods in psychology

Answer: A · Content validity

Why this is the answer

Content validity is the extent to which a measure covers all relevant aspects of the construct, usually judged by experts.

Answer

A · Content validity
2.Research methods in psychology

Answer: D · Wilcoxon signed-rank test

Why this is the answer

The Wilcoxon signed-rank test is for paired ordinal data; Mann–Whitney U is its counterpart for independent groups.

Answer

D · Wilcoxon signed-rank test
3.Descriptive statistics

Answer: s = 15.04

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 = 241/6 = 40.17

Squared deviations

Σ(x − x̄)² = 2.833² + 11.83² + 16.83² + 3.833² + (−23.17)² + (−12.17)² = 1131

Standard deviation

s = √(1131/5) = 15.04

Answer

s = 15.04
4.Descriptive statistics

Answer: z = 0.6

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 = (112 − 100)/20 = 0.6

Answer

z = 0.6
5.Probability

Answer: Var(X) = 94

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 = 10·0.2 + 20·0.35 + 30·0.3 + 40·0.15 = 24

Mean of the squares

E(X²) = 100·0.2 + 400·0.35 + 900·0.3 + 1600·0.15 = 670

Variance

Var(X) = 670 − 24² = 94

Answer

Var(X) = 94
6.Probability

Answer: E(G) = 6.9 €

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: 9 (p = 0.2), 147 (p = 0.05), −3 (p = 0.75)

Expected value

E(G) = 9·0.2 + 147·0.05 − 3·0.75 = 6.9 €

Answer

E(G) = 6.9 €

A positive value means you win money on average.

7.Normal distribution

Answer: z = −2.18

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 − 0.9854 = 0.0146

Read the z-table backwards

Find 0.0146 inside the table and read off z.

z = −2.18

Answer

z = −2.18
8.Normal distribution

Answer: x = 58.3 points

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) = 0.25

z from the table

z ≈ −0.67

Back to the original units

x = 65 − 0.67·10 = 58.3 points

Answer

x = 58.3 points
9.Confidence intervals

Answer: upper limit = 10.01 h

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.

Point estimate

x̄₁ − x̄₂ = 78 − 70 = 8

Standard error

SE = √(10²/100 + 5²/50) = 1.225

Margin of error

E = 1.645·1.225 = 2.015

Interval

8 ± 2.015 → [5.985, 10.01]

Answer

upper limit = 10.01 h

The interval does not contain 0, so the means differ significantly.

10.Confidence intervals

Answer: n = 150

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 = (2.576·19/4)² = 149.7

Round up

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

n = 150

Answer

n = 150
11.Hypothesis tests & confidence intervals

Answer: t = −0.9

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 = 8/√64 = 8/8 = 1

Test statistic

t = (499.1 − 500)/1 = −0.9

Answer

t = −0.9
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 = 48

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).

Solve for n

n = (2.576·8/3)² = 47.19

Round up

n = 48

Answer

n = 48
13.Linear regression & correlation

Answer: r = 0.9978

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 = 958 − 30·131/5 = 172
Sxx = 220 − 30²/5 = 40
Syy = 4175 − 131²/5 = 742.8

Correlation coefficient

r = 172/√(40·742.8) = 0.9978

Answer

r = 0.9978
14.Linear regression & correlation

Answer: e = −3.35

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 = 110/5 = 22
ȳ = 516/5 = 103.2

Sums of squares and products

Sxx = Σ(x − x̄)² = 40
Sxy = Σ(x − x̄)(y − ȳ) = 154

Regression line

b = 154/40 = 3.85
a = 103.2 − 3.85·22 = 18.5
ŷ = 18.5 + 3.85·x

Predicted value

ŷ = 18.5 + 3.85·21 = 99.35

Residual

e = 96 − 99.35 = −3.35

Answer

e = −3.35

The point lies below the line.

15.Chi-square tests

Answer: E = 15.65

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 count

E = 57·39/142 = 15.65

Answer

E = 15.65
16.Chi-square tests

Answer: df = 5

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 = (2 − 1)·(6 − 1) = 5

Answer

df = 5

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