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 behaviour only at fixed time intervals
  2. BChoosing participants at random
  3. CRecording only events the participants report
  4. DRecording every time a specified behaviour occurs
2.Research methods in psychologyHard

What is the main purpose of random allocation to conditions in an experiment?

  1. ATo make the sample representative of the target population so results generalise
  2. BTo increase the response rate
  3. CTo spread participant variables evenly so they do not confound the results
  4. DTo ensure participants do not know the aim
3.Descriptive statisticsMedium
3234363840424446485052

Find the sample standard deviation s of the data: 42, 47, 32, 52, 35, 38.

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

Exam scores are normally distributed with mean 60 and standard deviation 10. Find the z-score z of a score of 75.

μ = 60σ = 10x = 75
z =
5.ProbabilityMedium
xp(x)10.220.330.0540.250.25μ = 3

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

x: 1, 2, 3, 4, 5p(x): 0.2, 0.3, 0.05, 0.2, 0.25
Var(X) =
6.ProbabilityHard
xp(x)00.310.120.0530.2540.3μ

X has the distribution shown, and Y = 5X + 4. Find the standard deviation σY of Y.

x: 0, 1, 2, 3, 4p(x): 0.3, 0.1, 0.05, 0.25, 0.3Y = 5X + 4
σY =
7.Normal distributionMedium
1020 kg1082 kgT = X₁ + … + X₁₂?

The masses of passengers with luggage are independent and normal with mean 85 kg and standard deviation 12 kg. Find the probability P that the total mass of 12 of them exceeds 1082 kg. Round z to two decimals.

μ = 85 kgσ = 12 kgn = 12> 1082 kg
P =
8.Normal distributionHard
μ = 500x = ?10%

The battery lifetimes are normally distributed with mean 500 h and standard deviation 40 h. Find the value x below which 10% of them lie (the 10th percentile).

μ = 500σ = 40P = 10%
x = h
9.Confidence intervalsMedium
n = 25, s = 18.5LUx̄ = 94E = ?99%

A sample of 25 items has mean 94 and standard deviation s = 18.5. The population is roughly normal. Find the margin of error E of a 99% confidence interval for the mean.

n = 25x̄ = 94s = 18.599%t(24) = 2.797
E =
10.Confidence intervalsHard
μ₁ − μ₂, t(14) = 2.9770LUx̄₁ − x̄₂E = ?99%

Two fertilisers are tested. Plot group 1: 7 plots, mean yield 94 kg, standard deviation 9 kg. Plot group 2: 9 plots, mean 84 kg, standard deviation 7.5 kg. Assume equal variances (pooled t, df = n₁ + n₂ − 2). Find the margin of error E of the 99% confidence interval for μ₁ − μ₂.

n₁ = 7x̄₁ = 94s₁ = 9n₂ = 9x̄₂ = 84s₂ = 7.599%t(14) = 2.977
E = kg
11.Hypothesis tests & confidence intervalsMedium
−1.62p = ?

The mean of a population is claimed to be 50; the population standard deviation is known to be 24. A random sample of 144 observations has a mean of 46.76. Test H₀: μ = 50 against H₁: μ < 50. Find the p-value p.

μ₀ = 50σ = 24n = 144x̄ = 46.76H₁: μ < μ₀
p =
12.Hypothesis tests & confidence intervalsHard
z −2.57602.576

A sample of 36 measurements has mean 51.6. The population standard deviation is 12. Find the lower limit L of the 99% confidence interval for the mean.

x̄ = 51.6σ = 12n = 3699%
L =
13.Linear regression & correlationMedium
xy127224321418515614xy246152025

The table gives 6 paired observations of x = car age in years and y = price in k€. Find Pearson's correlation coefficient r.

n = 6
r =
14.Linear regression & correlationHard
xy128222320419515xy24152025x = 4.5

The table gives 5 paired observations of x = car age in years and y = price in k€. Use the least-squares line ŷ = a + b·x. A new observation (4.5, 11) is made. Find its residual e = y − ŷ.

n = 5(4.5, 11)
e =
15.Chi-square testsMedium
busbikecarTotalstudents14134168staff253640101Total394981169

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

row total = 101column total = 49N = 169
E =
16.Chi-square testsHard

A chi-square goodness-of-fit test compares observed counts in 9 categories with fully specified expected proportions (no parameters are estimated from the data). How many degrees of freedom df does the test have?

k = 9
df =

3Solutions

1.Research methods in psychology

Answer: D · 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

D · Recording every time a specified behaviour occurs
2.Research methods in psychology

Answer: C · To spread participant variables evenly so they do not confound the results

Why this is the answer

Random allocation balances individual differences across conditions, supporting causal conclusions. Random sampling concerns generalisation to a population.

Answer

C · To spread participant variables evenly so they do not confound the results
3.Descriptive statistics

Answer: s = 7.537

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 = 246/6 = 41

Squared deviations

Σ(x − x̄)² = 1² + 6² + (−9)² + 11² + (−6)² + (−3)² = 284

Standard deviation

s = √(284/5) = 7.537

Answer

s = 7.537
4.Descriptive statistics

Answer: z = 1.5

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 = (75 − 60)/10 = 1.5

Answer

z = 1.5
5.Probability

Answer: Var(X) = 2.3

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.

Mean

μ = 1·0.2 + 2·0.3 + 3·0.05 + 4·0.2 + 5·0.25 = 3

Mean of the squares

E(X²) = 1·0.2 + 4·0.3 + 9·0.05 + 16·0.2 + 25·0.25 = 11.3

Variance

Var(X) = 11.3 − 3² = 2.3

Answer

Var(X) = 2.3
6.Probability

Answer: σY = 8.258

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.

Spread of X

μ = 2.15, Var(X) = 2.728, σX = 1.652

Linear change

σY = |5|·1.652 = 8.258

Answer

σY = 8.258

The mean does move: E(Y) = 5·2.15 + 4 = 14.75.

7.Normal distribution

Answer: P = 0.06681

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 total

μT = 1020 kg, σT = 12·√12 = 41.57 kg

Standardise

z = (1082 − 1020)/41.57 = 1.5

Probability

P = 1 − Φ(1.5) = 0.06681

Answer

P = 0.06681
8.Normal distribution

Answer: x = 448.7 h

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) = 0.1 → z = −1.282

Back to x

x = 500 + −1.282·40 = 448.74 h

Answer

x = 448.74 h
9.Confidence intervals

Answer: E = 10.35

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 = 24, t = 2.797

Margin of error

E = 2.797·18.5/√25 = 10.35

Answer

E = 10.35
10.Confidence intervals

Answer: E = 12.27 kg

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.

Pooled standard deviation

sp² = (6·9² + 8·7.5²)/14 = 66.86
sp = 8.177

Margin of error

E = 2.977·8.177·√(1/7 + 1/9) = 12.27

Answer

E = 12.27 kg
11.Hypothesis tests & confidence intervals

Answer: p = 0.05262

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.

Test statistic

SE = σ/√n = 24/√144 = 2
z = (46.76 − 50)/2 = −1.62

Tail area

Φ(1.62) = 0.9474
1 − Φ(1.62) = 0.0526

p-value

p = 0.05262

Answer

p = 0.05262

p > 0.05, so H₀ is not rejected at the 5% level.

12.Hypothesis tests & confidence intervals

Answer: L = 46.45

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/√36 = 5.152

Interval limit

L = 51.6 − 5.152 = 46.45

Answer

L = 46.45
13.Linear regression & correlation

Answer: r = −0.9927

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 = 119/6 = 19.83

Sums of squares and products

Sxx = Σ(x − x̄)² = 17.5
Sxy = Σ(x − x̄)(y − ȳ) = −47.5
Syy = Σ(y − ȳ)² = 130.8

Correlation coefficient

r = −47.5/√(17.5·130.8) = −0.9927

Answer

r = −0.9927
14.Linear regression & correlation

Answer: e = −5.45

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 = 15/5 = 3
ȳ = 104/5 = 20.8

Sums of squares and products

Sxx = Σ(x − x̄)² = 10
Sxy = Σ(x − x̄)(y − ȳ) = −29

Regression line

b = −29/10 = −2.9
a = 20.8 − (−2.9)·3 = 29.5
ŷ = 29.5 − 2.9·x

Predicted value

ŷ = 29.5 + (−2.9)·4.5 = 16.45

Residual

e = 11 − 16.45 = −5.45

Answer

e = −5.45

The point lies below the line.

15.Chi-square tests

Answer: E = 29.28

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 = 101·49/169 = 29.28

Answer

E = 29.28
16.Chi-square tests

Answer: df = 8

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 = 9 − 1 = 8

Answer

df = 8

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