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 new depression questionnaire predicts which students will seek treatment over the next year. This supports which type of validity?

  1. AConcurrent validity
  2. BContent validity
  3. CPredictive validity
  4. DFace validity
2.Research methods in psychologyHard

After analysing the data, a researcher writes the introduction as if an unexpected finding had been predicted all along. This is called:

  1. ADebriefing
  2. BOperationalisation
  3. CCounterbalancing
  4. DHARKing (hypothesising after the results are known)
3.Descriptive statisticsMedium
3035404550

Find the sample standard deviation s of the data: 35, 51, 51, 33, 41, 28.

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

A quantity is normally distributed with mean 60 and standard deviation 5. What proportion P of values is above 57.5?

μ = 60σ = 5x = 57.5
P =
5.ProbabilityMedium
red: 5blue: 8

A bag holds 5 red and 8 blue balls. You draw two balls without putting the first one back. What is the probability P that both are red?

5 / 8
P =
6.ProbabilityHard
€p(x)−20.7560.2980.05

A game costs €2 to play. You win €8 with probability 0.2 and €100 with probability 0.05; otherwise you win nothing. The stake is not returned. Find the expected net gain E(G) per game.

stake = €2€8: p = 0.2€100: p = 0.05
E(G) = €
7.Normal distributionMedium
502534P = ?

The battery lifetimes are normally distributed with mean 500 h and standard deviation 40 h. What fraction lies between 502 and 534 h? Give P as a decimal.

μ = 500σ = 40x1 = 502x2 = 534
P =
8.Normal distributionHard
μ = 250x = ?95%

The bag weights are normally distributed with mean 250 g and standard deviation 5 g. Find the value x below which 95% of them lie (the 95th percentile).

μ = 250σ = 5P = 95%
x = g
9.Confidence intervalsMedium
n = 16, s = 15.5L = ?Ux̄ = 111E90%

A sample of 16 items has mean 111 and standard deviation s = 15.5. The population is roughly normal. Find the lower limit of the 90% confidence interval for the mean.

n = 16x̄ = 111s = 15.590%t(15) = 1.753
lower limit =
10.Confidence intervalsHard
n = 16Lσ = ?s = 2.995%

The diameters of 16 machined shafts have a sample standard deviation of 2.9 µm. Assuming the diameters are normally distributed, find the upper limit σ of the 95% confidence interval for the population standard deviation. With df = 15, 2.5% of the χ² distribution lies below 6.262 and 2.5% lies above 27.488.

n = 16s = 2.9 µmχ² = 6.262, 27.488
σ = µm
11.Hypothesis tests & confidence intervalsMedium
t −2.0302.03

A sample of 36 students has a mean score of 491.3 and a sample standard deviation of 20. Find the test statistic t for H₀: μ = 500.

μ₀ = 500x̄ = 491.3s = 20n = 36
t =
12.Hypothesis tests & confidence intervalsHard
σ = 20, 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 20. Find the smallest sample size n.

σ = 20E = 399%
n =
13.Linear regression & correlationMedium
xy3495577739801190xy5106080

The table gives 5 paired observations of x = hours studied and y = exam score. Find the coefficient of determination r² of the least-squares line.

n = 5
r² =
14.Linear regression & correlationHard
xy12422031741451568xy24610152025ŷ = 15

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. For which value x does the line predict ŷ = 15?

n = 6ŷ = 15
x =
15.Chi-square testsMedium
busbikecarTotalstudents32372695staff393045114Total716771209

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

row total = 95column total = 67N = 209
E =
16.Chi-square testsHard

A chi-square goodness-of-fit test compares observed counts in 4 categories with fully specified expected proportions (no parameters are estimated from the data). How many degrees of freedom df does the test have?

k = 4
df =

3Solutions

1.Research methods in psychology

Answer: C · Predictive validity

Why this is the answer

Predictive validity is shown when scores predict a later criterion. Concurrent validity uses a criterion measured at the same time.

Answer

C · Predictive validity
2.Research methods in psychology

Answer: D · HARKing (hypothesising after the results are known)

Why this is the answer

HARKing presents exploratory results as confirmatory, making findings look more robust than they are. It contributes to the replication crisis.

Answer

D · HARKing (hypothesising after the results are known)
3.Descriptive statistics

Answer: s = 9.6

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.

Mean

x̄ = 239/6 = 39.83

Squared deviations

Σ(x − x̄)² = (−4.833)² + 11.17² + 11.17² + (−6.833)² + 1.167² + (−11.83)² = 460.8

Standard deviation

s = √(460.8/5) = 9.6

Answer

s = 9.6
4.Descriptive statistics

Answer: P = 0.6915

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 = −0.5

Standard normal table

Φ(−0.5) = 0.3085
P = 1 − 0.3085 = 0.6915

Answer

P = 0.6915
5.Probability

Answer: P = 0.1282

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 = 5/13 · 4/12 = 20/156 = 0.1282

Answer

P = 0.1282
6.Probability

Answer: E(G) = 4.6 €

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: 6 (p = 0.2), 98 (p = 0.05), −2 (p = 0.75)

Expected value

E(G) = 6·0.2 + 98·0.05 − 2·0.75 = 4.6 €

Answer

E(G) = 4.6 €

A positive value means you win money on average.

7.Normal distribution

Answer: P = 0.2824

Normal distribution: convert a value x to a z-score, z = (x − μ)/σ, which counts standard deviations from the mean. A z-table (or Φ(z)) gives the area to the left of z, which is P(X < x). The area to the right is 1 − Φ(z), and the area between two values is the difference of their two areas.

z-scores

z₁ = (502 − 500)/40 = 0.05
z₂ = (534 − 500)/40 = 0.85

Area between

P = Φ(0.85) − Φ(0.05) = 0.8023 − 0.5199 = 0.2824

Answer

P = 0.2824
8.Normal distribution

Answer: x = 258.2 g

Normal distribution: convert a value x to a z-score, z = (x − μ)/σ, which counts standard deviations from the mean. A z-table (or Φ(z)) gives the area to the left of z, which is P(X < x). The area to the right is 1 − Φ(z), and the area between two values is the difference of their two areas.

z for the percentile

Φ(z) = 0.95 → z = 1.645

Back to x

x = 250 + 1.645·5 = 258.22 g

Answer

x = 258.22 g
9.Confidence intervals

Answer: lower limit = 104.2

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 = 15, t = 1.753

Margin of error

E = 1.753·15.5/√16 = 6.793

Interval

111 ± 6.793 → [104.2, 117.8]

Answer

lower limit = 104.2
10.Confidence intervals

Answer: σ = 4.488 µm

Interval for a variance: (n − 1)s²/σ² has a χ² distribution with n − 1 degrees of freedom. The 95% interval for σ² runs from (n − 1)s² divided by the upper χ² value (2.5% above it) to (n − 1)s² divided by the lower χ² value (2.5% below it); the interval for σ is the square root of both limits. It is not symmetric around s.

Sum of squares

(n − 1)s² = 15·2.9² = 126.1

Limits for σ²

126.1/27.488 = 4.589
126.1/6.262 = 20.15

Limit for σ

σ = √20.15 = 4.488 µm

Answer

σ = 4.488 µm

95% interval for σ: [2.142, 4.488] µm.

11.Hypothesis tests & confidence intervals

Answer: t = −2.61

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 = 20/√36 = 20/6 = 3.333

Test statistic

t = (491.3 − 500)/3.333 = −2.61

Answer

t = −2.61
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: n = 295

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 = (2.576·20/3)² = 294.9

Round up

n = 295

Answer

n = 295
13.Linear regression & correlation

Answer: r² = 0.9854

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 = 35/5 = 7
ȳ = 349/5 = 69.8

Sums of squares and products

Sxx = Σ(x − x̄)² = 40
Sxy = Σ(x − x̄)(y − ȳ) = 210
Syy = Σ(y − ȳ)² = 1119

Coefficient of determination

r² = 210²/(40·1119) = 0.9854

Answer

r² = 0.9854

About 98.54% of the variation in y is explained by the line.

14.Linear regression & correlation

Answer: x = 3.976

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 = 98/6 = 16.33

Sums of squares and products

Sxx = Σ(x − x̄)² = 17.5
Sxy = Σ(x − x̄)(y − ȳ) = −49

Regression line

b = −49/17.5 = −2.8
a = 16.33 − (−2.8)·3.5 = 26.13
ŷ = 26.13 − 2.8·x

Solve for x

x = (15 − 26.13)/(−2.8) = 3.976

Answer

x = 3.976
15.Chi-square tests

Answer: E = 30.45

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 = 95·67/209 = 30.45

Answer

E = 30.45
16.Chi-square tests

Answer: df = 3

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 = 3

Answer

df = 3

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