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

In observational research, event sampling means:

  1. ARecording every time a specified behaviour occurs
  2. BRecording behaviour only at fixed time intervals
  3. CRecording only events the participants report
  4. DChoosing participants at random
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
38404244464850525456

Find the sample standard deviation s of the data: 47, 42, 38, 43, 57.

n = 5
s =
4.Descriptive statisticsHard
z −2−1012

A quantity is normally distributed with mean 50 and standard deviation 10. What proportion P of values is below 65?

μ = 50σ = 10x = 65
P =
5.ProbabilityMedium
n = 7k = 2N = ?

In how many ways N can a committee of 2 people be chosen from a group of 7? The order does not matter.

n = 7k = 2
N =
6.ProbabilityHard
€p(x)−20.7980.2180.01

A game costs €2 to play. You win €10 with probability 0.2 and €20 with probability 0.01; otherwise you win nothing. The stake is not returned. Find the expected net gain E(G) per game.

stake = €2€10: p = 0.2€20: p = 0.01
E(G) = €
7.Normal distributionMedium
32 min34.5 minX + Y?

X is normal with mean 12 min and standard deviation 1.5 min; Y is independent and normal with mean 20 min and standard deviation 2 min. Find the probability P that X + Y is more than 34.5 min. Round z to two decimals.

μX = 12, σX = 1.5μY = 20, σY = 2X + Y > 34.5 min
P =
8.Normal distributionHard
235240245250255260265gtop 70%

The bag weights are normally distributed with mean 250 g and standard deviation 5 g. Drag the cut-off so the shaded top 70% is to the right of it. Find the cut-off x.

μ = 250 gσ = 5 g70%
x = g
9.Confidence intervalsMedium
n = 15, sd​ = 7.10LU = ?d̄ = 6.7E95%

The blood pressure of 15 patients is measured before and after a treatment. The differences d = before − after (mmHg) have mean 6.7 and standard deviation 7.1. Find the upper limit of the 95% confidence interval for the mean difference.

n = 15d̄ = 6.7sd = 7.195%t(14) = 2.145
upper limit = mmHg
10.Confidence intervalsHard
p = 0.5, n = ?p̂ − Ep̂ + Ep̂E = 0.0299%

Nothing is known in advance about the proportion of defective products. How many products n must be inspected so that a 99% confidence interval for the proportion has a margin of error of at most 0.02?

p = 0.5E = 0.0299%
n =
11.Hypothesis tests & confidence intervalsMedium
t −2.0302.03

A sample of 36 students has a mean score of 97.42 and a sample standard deviation of 5. Find the test statistic t for H₀: μ = 100.

μ₀ = 100x̄ = 97.42s = 5n = 36
t =
12.Hypothesis tests & confidence intervalsHard
μ₀x̄c

A one-sided z test of H₀: μ = 50 against H₁: μ > 50 is done at α = 0.05 with a sample of 36; the standard deviation is known to be 9. In reality the mean is 53.45. Find the power 1 − β of the test in percent.

μ₀ = 50μ = 53.45σ = 9n = 36α = 0.05
1 − β = %
13.Linear regression & correlationMedium
xy1863218124912711830121xy2025306080100120

The table gives 5 paired observations of x = temperature in °C and y = ice creams sold. Find the coefficient of determination r² of the least-squares line.

n = 5
r² =
14.Linear regression & correlationHard
xy1252203184165969xy24610152025x = 4.5

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. Predict ŷ for x = 4.5.

n = 6x = 4.5
ŷ =
15.Chi-square testsMedium
no defectminormajorTotalday shift12183060night shift16451980Total286349140

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
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 goodness-of-fit test checks whether counts in 4 categories follow a Poisson distribution whose mean is estimated from the same data. Using the table, find the critical value χ²crit at significance level α = 0.05.

α = 0.05
χ²crit =

3Solutions

1.Research methods in psychology

Answer: A · Recording every time a specified behaviour occurs

Why this is the answer

Event sampling counts each occurrence of target behaviours. Time sampling records what happens at set intervals, such as every 30 seconds.

Answer

A · Recording every time a specified behaviour occurs
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 = 7.232

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 = 227/5 = 45.4

Squared deviations

Σ(x − x̄)² = 1.6² + (−3.4)² + (−7.4)² + (−2.4)² + 11.6² = 209.2

Standard deviation

s = √(209.2/4) = 7.232

Answer

s = 7.232
4.Descriptive statistics

Answer: P = 0.9332

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 = 1.5

Standard normal table

Φ(1.5) = 0.9332
P = 0.9332

Answer

P = 0.9332
5.Probability

Answer: N = 21

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.

Combinations

C(7, 2) = (7·6)/(2·1) = 21

Answer

N = 21
6.Probability

Answer: E(G) = 0.2 €

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: 8 (p = 0.2), 18 (p = 0.01), −2 (p = 0.79)

Expected value

E(G) = 8·0.2 + 18·0.01 − 2·0.79 = 0.2 €

Answer

E(G) = 0.2 €

A positive value means you win money on average.

7.Normal distribution

Answer: P = 0.1587

Sums and differences of normal variables: if X and Y are independent and normal, X + Y and X − Y are normal too. Means add or subtract: E(X ± Y) = μX ± μY. Variances always add, also for a difference: Var(X ± Y) = σX² + σY², so σ = √(σX² + σY²). Standard deviations never add directly. The total of n independent items with the same distribution has mean n·μ and standard deviation σ·√n. Then standardise as usual: z = (x − mean)/σ and read Φ(z).

Distribution of the sum

μ = 32 min, σ = √(1.5² + 2²) = 2.5 min

Standardise

z = (34.5 − 32)/2.5 = 1

Upper tail

P = 1 − Φ(1) = 1 − 0.8413 = 0.1587

Answer

P = 0.1587
8.Normal distribution

Answer: x = 247.4 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) = 0.3

z from the table

z ≈ −0.52

Back to the original units

x = 250 − 0.52·5 = 247.4 g

Answer

x = 247.4 g
9.Confidence intervals

Answer: upper limit = 10.63 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 = 14, t = 2.145

Margin of error

E = 2.145·7.1/√15 = 3.932

Interval

6.7 ± 3.932 → [2.768, 10.63]

Answer

upper limit = 10.63 mmHg
10.Confidence intervals

Answer: n = 4148

Confidence interval for a proportion: p̂ ± z·√(p̂(1 − p̂)/n), where p̂ = x/n. To reach a margin E you need n ≥ z²·p(1 − p)/E², rounded up; without a prior estimate use p = 0.5, which gives the largest n.

Solve for n

n = 2.576²·0.5·0.5/0.02² = 4147

Round up

n = 4148

Answer

n = 4148
11.Hypothesis tests & confidence intervals

Answer: t = −3.096

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/√36 = 5/6 = 0.8333

Test statistic

t = (97.42 − 100)/0.8333 = −3.096

Answer

t = −3.096
t0.025; df = 35 = 2.03

|t| > 2.03, so H₀ is rejected at the 5% level (two-sided).

12.Hypothesis tests & confidence intervals

Answer: 1 − β = 74.38 %

Power of a test: power = 1 − β is the probability of rejecting H₀ when it is false. Find the cutoff x̄c from μ₀ and z_α, then compute the probability of landing beyond x̄c using the true mean: z = (x̄c − μ)/(σ/√n).

Critical sample mean

SE = 9/√36 = 1.5
x̄c = 50 + 1.645·1.5 = 52.47

z under the true mean

z = (52.47 − 53.45)/1.5 = −0.655

Power

power = P(Z > −0.655) = 74.38%
1 − β = 74.38%

Answer

1 − β = 74.38%
13.Linear regression & correlation

Answer: r² = 0.9591

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 = 120/5 = 24
ȳ = 474/5 = 94.8

Sums of squares and products

Sxx = Σ(x − x̄)² = 90
Sxy = Σ(x − x̄)(y − ȳ) = 459
Syy = Σ(y − ȳ)² = 2441

Coefficient of determination

r² = 459²/(90·2441) = 0.9591

Answer

r² = 0.9591

About 95.91% of the variation in y is explained by the line.

14.Linear regression & correlation

Answer: ŷ = 12.88

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 = 97/6 = 16.17

Sums of squares and products

Sxx = Σ(x − x̄)² = 17.5
Sxy = Σ(x − x̄)(y − ȳ) = −57.5

Regression line

b = −57.5/17.5 = −3.286
a = 16.17 − (−3.286)·3.5 = 27.67
ŷ = 27.67 − 3.286·x

Prediction

ŷ = 27.67 + (−3.286)·4.5 = 12.88

Answer

ŷ = 12.88
15.Chi-square tests

Answer: χ² = 12

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

day shift: 60·28/140 = 12, 60·63/140 = 27, 60·49/140 = 21
night shift: 80·28/140 = 16, 80·63/140 = 36, 80·49/140 = 28

Chi-square statistic

χ² = (12 − 12)²/12 + (18 − 27)²/27 + (30 − 21)²/21 + (16 − 16)²/16 + (45 − 36)²/36 + (19 − 28)²/28
χ² = 12

Answer

χ² = 12

df = 2. At α = 0.05 the critical value is 5.991, so H₀ (independence) is rejected.

16.Chi-square tests

Answer: χ²crit = 5.991

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 − 1 = 2

Table

χ²crit(df = 2, α = 0.05) = 5.991

Answer

χ²crit = 5.991

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

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