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

A test is so easy that almost all participants score near the maximum. This is called:

  1. AA floor effect
  2. BAn order effect
  3. CA practice effect
  4. DA ceiling effect
2.Research methods in psychologyHard

Interpretative phenomenological analysis (IPA) mainly aims to:

  1. ATest causal hypotheses with large samples
  2. BExplore how individuals make sense of their own lived experiences
  3. CCount the frequency of words in texts
  4. DMeasure reaction times accurately
3.Descriptive statisticsMedium
152025303540455055

Find the sample standard deviation s of the data: 57, 46, 56, 49, 13, 57.

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

A quantity is normally distributed with mean 500 and standard deviation 25. What proportion P of values is below 525?

μ = 500σ = 25x = 525
P =
5.ProbabilityMedium
P(at least one six) = ??1?2?3?4P(six) = 1/6 on each roll

A fair die is rolled 4 times. What is the probability P of getting at least one six?

n = 4
P =
6.ProbabilityHard
€p(x)−10.7530.290.05

A game costs €1 to play. You win €4 with probability 0.2 and €10 with probability 0.05; otherwise you win nothing. The stake is not returned. Find the expected net gain E(G) per game.

stake = €1€4: p = 0.2€10: p = 0.05
E(G) = €
7.Normal distributionMedium
−3−2−10123zshaded area to the right of z = 0.7357

Drag z so that the shaded area under the standard normal curve, to the right of z, is 0.7357. Find z to two decimals.

A = 0.7357
z =
8.Normal distributionHard
35455565758595pointsbottom 10%

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

μ = 65 pointsσ = 10 points10%
x = points
9.Confidence intervalsMedium
n = 9, s = 18.5L = ?Ux̄ = 78E99%

A sample of 9 items has mean 78 and standard deviation s = 18.5. The population is roughly normal. Find the lower limit of the 99% confidence interval for the mean.

n = 9x̄ = 78s = 18.599%t(8) = 3.355
lower limit =
10.Confidence intervalsHard
σ = 16, n = ?x̄ − Ex̄ + Ex̄E = 290%

The population standard deviation is 16. 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 2?

σ = 16E = 290%
n =
11.Hypothesis tests & confidence intervalsMedium
t −2.06402.064

A sample of 25 students has a mean score of 45.84 and a sample standard deviation of 8. Find the test statistic t for H₀: μ = 50.

μ₀ = 50x̄ = 45.84s = 8n = 25
t =
12.Hypothesis tests & confidence intervalsHard
−1.961.96z = ?

In a trial, 120 of 300 patients on treatment A recover, and 94 of 200 patients on treatment B. Using the pooled proportion, find the test statistic z for H₀: p₁ = p₂.

x₁ = 120n₁ = 300x₂ = 94n₂ = 200
z =
13.Linear regression & correlationMedium
xy1413495607719791190xy510406080

The table gives 6 paired observations of x = hours studied and y = exam score. Their sums are given. Find Pearson's correlation coefficient r from these sums.

n = 6Σx = 36Σy = 390Σx² = 286Σy² = 27064Σxy = 2686
r =
14.Linear regression & correlationHard
xy350460568673777881xy46850607080x = 6.5

The table gives 6 paired observations of x = advertising spend in k€ and y = sales in k€. Use the least-squares line ŷ = a + b·x. Predict ŷ for x = 6.5.

n = 6x = 6.5
ŷ =
15.Chi-square testsMedium
lowmediumhighTotalurban14252160rural313851120Total456372180

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 = 1802 × 3
χ² =
16.Chi-square testsHard
passfailTotalline 15941100line 2928100line 38911100Total24060300dfα = 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 = 0.05 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 = 3003 × 2α = 0.05
χ² =

3Solutions

1.Research methods in psychology

Answer: D · A ceiling effect

Why this is the answer

Ceiling effects hide differences between conditions because no one can score higher. Floor effects are the reverse, when a task is too hard.

Answer

D · A ceiling effect
2.Research methods in psychology

Answer: B · Explore how individuals make sense of their own lived experiences

Why this is the answer

IPA studies small, homogeneous samples in depth, examining how participants interpret significant experiences, with the researcher interpreting their sense-making.

Answer

B · Explore how individuals make sense of their own lived experiences
3.Descriptive statistics

Answer: s = 16.97

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 = 278/6 = 46.33

Squared deviations

Σ(x − x̄)² = 10.67² + (−0.3333)² + 9.667² + 2.667² + (−33.33)² + 10.67² = 1439

Standard deviation

s = √(1439/5) = 16.97

Answer

s = 16.97
4.Descriptive statistics

Answer: P = 0.8413

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 = 1

Standard normal table

Φ(1) = 0.8413
P = 0.8413

Answer

P = 0.8413
5.Probability

Answer: P = 0.5177

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.

Complement

P(no six) = (5/6)4 = 0.4823
P = 1 − 0.4823 = 0.5177

Answer

P = 0.5177
6.Probability

Answer: E(G) = 0.3 €

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: 3 (p = 0.2), 9 (p = 0.05), −1 (p = 0.75)

Expected value

E(G) = 3·0.2 + 9·0.05 − 1·0.75 = 0.3 €

Answer

E(G) = 0.3 €

A positive value means you win money on average.

7.Normal distribution

Answer: z = −0.63

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 − 0.7357 = 0.2643

Read the z-table backwards

Find 0.2643 inside the table and read off z.

z = −0.63

Answer

z = −0.63
8.Normal distribution

Answer: x = 52.2 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) = 0.1

z from the table

z ≈ −1.28

Back to the original units

x = 65 − 1.28·10 = 52.2 points

Answer

x = 52.2 points
9.Confidence intervals

Answer: lower limit = 57.31

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 = 8, t = 3.355

Margin of error

E = 3.355·18.5/√9 = 20.69

Interval

78 ± 20.69 → [57.31, 98.69]

Answer

lower limit = 57.31
10.Confidence intervals

Answer: n = 174

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 = (1.645·16/2)² = 173.2

Round up

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

n = 174

Answer

n = 174
11.Hypothesis tests & confidence intervals

Answer: t = −2.6

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/√25 = 8/5 = 1.6

Test statistic

t = (45.84 − 50)/1.6 = −2.6

Answer

t = −2.6
t0.025; df = 24 = 2.064

|t| > 2.064, so H₀ is rejected at the 5% level (two-sided).

12.Hypothesis tests & confidence intervals

Answer: z = −1.55

Hypothesis test: the test statistic = (estimate − value under H₀)/(standard error under H₀). The p-value is the probability, assuming H₀ is true, of a statistic at least as extreme as the one observed: the area in one tail for a one-sided H₁, twice the tail area for a two-sided H₁. Reject H₀ when p ≤ α.

Sample proportions

p̂₁ = 120/300 = 0.4
p̂₂ = 94/200 = 0.47
p̂ = 214/500 = 0.428

Standard error under H₀

SE = √(0.428·0.572·(1/300 + 1/200)) = 0.04517

Test statistic

z = (0.4 − 0.47)/0.04517 = −1.55

Answer

z = −1.55

|z| < 1.96, so H₀ is not rejected at the 5% level (two-sided).

13.Linear regression & correlation

Answer: r = 0.9989

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 = 2686 − 36·390/6 = 346
Sxx = 286 − 36²/6 = 70
Syy = 27064 − 390²/6 = 1714

Correlation coefficient

r = 346/√(70·1714) = 0.9989

Answer

r = 0.9989
14.Linear regression & correlation

Answer: ŷ = 74.2

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 = 33/6 = 5.5
ȳ = 409/6 = 68.17

Sums of squares and products

Sxx = Σ(x − x̄)² = 17.5
Sxy = Σ(x − x̄)(y − ȳ) = 105.5

Regression line

b = 105.5/17.5 = 6.029
a = 68.17 − 6.029·5.5 = 35.01
ŷ = 35.01 + 6.029·x

Prediction

ŷ = 35.01 + 6.029·6.5 = 74.2

Answer

ŷ = 74.2
15.Chi-square tests

Answer: χ² = 1.805

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: 60·45/180 = 15, 60·63/180 = 21, 60·72/180 = 24
rural: 120·45/180 = 30, 120·63/180 = 42, 120·72/180 = 48

Chi-square statistic

χ² = (14 − 15)²/15 + (25 − 21)²/21 + (21 − 24)²/24 + (31 − 30)²/30 + (38 − 42)²/42 + (51 − 48)²/48
χ² = 1.805

Answer

χ² = 1.805

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

16.Chi-square tests

Answer: χ² = 41.63

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

line 1: 80, 20
line 2: 80, 20
line 3: 80, 20

Chi-square statistic

χ² = (59 − 80)²/80 + (41 − 20)²/20 + (92 − 80)²/80 + (8 − 20)²/20 + (89 − 80)²/80 + (11 − 20)²/20
χ² = 41.63

Decision

df = 2, χ²crit = 5.991
41.63 > 5.991

Answer

χ² = 41.63

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

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