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 Cronbach's alpha of 0.88 for a 15-item anxiety scale indicates:

  1. AThe scale predicts future anxiety well
  2. BThe scale has a significant effect on anxiety
  3. CScores are stable over six months
  4. DGood internal consistency of the items
2.Research methods in psychologyHard

Exercise reduces depression, and the effect occurs because exercise improves sleep, which in turn lowers depression. Sleep is best described as a:

  1. AExtraneous variable
  2. BMediator
  3. CControl variable
  4. DModerator
3.Descriptive statisticsMedium
20253035404550

Find the sample standard deviation s of the data: 45, 34, 18, 51, 46, 45.

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

Exam scores are normally distributed with mean 50 and standard deviation 20. Find the z-score z of a score of 70.

μ = 50σ = 20x = 70
z =
5.ProbabilityMedium
xp(x)100.35200.15300.35400.15μ = 23

For the distribution shown, find the standard deviation σ of X.

x: 10, 20, 30, 40p(x): 0.35, 0.15, 0.35, 0.15
σ =
6.ProbabilityHard
xp(x)00.310.1520.2530.3μ

X has the distribution shown, and Y = 2X + 5. Find the standard deviation σY of Y.

x: 0, 1, 2, 3p(x): 0.3, 0.15, 0.25, 0.3Y = 2X + 5
σY =
7.Normal distributionMedium
167.85192P = ?

The heights of adult men are normally distributed with mean 178 cm and standard deviation 7 cm. What fraction lies between 167.85 and 192 cm? Give P as a decimal.

μ = 178σ = 7x1 = 167.85x2 = 192
P =
8.Normal distributionHard
μ = 500x = ?97.5%

The battery lifetimes are normally distributed with mean 500 h and standard deviation 40 h. Find the value x below which 97.5% of them lie (the 97.5th percentile).

μ = 500σ = 40P = 97.5%
x = h
9.Confidence intervalsMedium
n = 16, s = 15LUx̄ = 76E = ?99%

A sample of 16 items has mean 76 and standard deviation s = 15. The population is roughly normal. Find the margin of error E of a 99% confidence interval for the mean.

n = 16x̄ = 76s = 1599%t(15) = 2.947
E =
10.Confidence intervalsHard
μ₁ − μ₂, t(19) = 2.0930LUx̄₁ − x̄₂E = ?95%

Two fertilisers are tested. Plot group 1: 13 plots, mean yield 37.5 kg, standard deviation 1.5 kg. Plot group 2: 8 plots, mean 27 kg, standard deviation 6.5 kg. Assume equal variances (pooled t, df = n₁ + n₂ − 2). Find the margin of error E of the 95% confidence interval for μ₁ − μ₂.

n₁ = 13x̄₁ = 37.5s₁ = 1.5n₂ = 8x̄₂ = 27s₂ = 6.595%t(19) = 2.093
E = kg
11.Hypothesis tests & confidence intervalsMedium
t −1.9901.99

A sample of 81 students has a mean score of 117 and a sample standard deviation of 15. Find the test statistic t for H₀: μ = 120.

μ₀ = 120x̄ = 117s = 15n = 81
t =
12.Hypothesis tests & confidence intervalsHard
−1.961.96z = ?

In a trial, 25 of 100 patients on treatment A recover, and 18 of 100 patients on treatment B. Using the pooled proportion, find the test statistic z for H₀: p₁ = p₂.

x₁ = 25n₁ = 100x₂ = 18n₂ = 100
z =
13.Linear regression & correlationMedium
xy156218852191241042712330130xy15202530100

The table gives 6 paired observations of x = temperature in °C and y = ice creams sold. Find Pearson's correlation coefficient r.

n = 6
r =
14.Linear regression & correlationHard
xy123221318413513xy241520x = 1.5

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. Predict ŷ for x = 1.5.

n = 5x = 1.5
ŷ =
15.Chi-square testsMedium
no defectminormajorTotalday shift43351593night shift31222982Total745744175

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

row total = 82column total = 57N = 175
E =
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 compares observed counts in 9 categories with fully specified expected proportions (no parameters are estimated from the 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: D · Good internal consistency of the items

Why this is the answer

Cronbach's alpha estimates internal consistency; values around 0.7 or above are usually considered acceptable. It does not by itself show validity.

Answer

D · Good internal consistency of the items
2.Research methods in psychology

Answer: B · Mediator

Why this is the answer

A mediator lies on the causal path between the predictor and outcome, explaining how or why the effect occurs.

Answer

B · Mediator
3.Descriptive statistics

Answer: s = 12.06

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 = 5.167² + (−5.833)² + (−21.83)² + 11.17² + 6.167² + 5.167² = 726.8

Standard deviation

s = √(726.8/5) = 12.06

Answer

s = 12.06
4.Descriptive statistics

Answer: z = 1

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 = (70 − 50)/20 = 1

Answer

z = 1
5.Probability

Answer: σ = 11

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 = 10·0.35 + 20·0.15 + 30·0.35 + 40·0.15 = 23

Mean of the squares

E(X²) = 100·0.35 + 400·0.15 + 900·0.35 + 1600·0.15 = 650

Variance

Var(X) = 650 − 23² = 121
σ = √121 = 11

Answer

σ = 11
6.Probability

Answer: σY = 2.406

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 = 1.55, Var(X) = 1.448, σX = 1.203

Linear change

σY = |2|·1.203 = 2.406

Answer

σY = 2.406

The mean does move: E(Y) = 2·1.55 + 5 = 8.1.

7.Normal distribution

Answer: P = 0.9037

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 = (167.85 − 178)/7 = −1.45
z₂ = (192 − 178)/7 = 2

Area between

P = Φ(2) − Φ(−1.45) = 0.9772 − 0.07353 = 0.9037

Answer

P = 0.9037
8.Normal distribution

Answer: x = 578.4 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) = 0.975 → z = 1.96

Back to x

x = 500 + 1.96·40 = 578.4 h

Answer

x = 578.4 h
9.Confidence intervals

Answer: E = 11.05

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 = 2.947

Margin of error

E = 2.947·15/√16 = 11.05

Answer

E = 11.05
10.Confidence intervals

Answer: E = 3.876 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 = (12·1.5² + 7·6.5²)/19 = 16.99
sp = 4.122

Margin of error

E = 2.093·4.122·√(1/13 + 1/8) = 3.876

Answer

E = 3.876 kg
11.Hypothesis tests & confidence intervals

Answer: t = −1.8

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 = 15/√81 = 15/9 = 1.667

Test statistic

t = (117 − 120)/1.667 = −1.8

Answer

t = −1.8
t0.025; df = 80 = 1.99

|t| < 1.99, so H₀ is not rejected at the 5% level (two-sided).

12.Hypothesis tests & confidence intervals

Answer: z = 1.205

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 = 25/100 = 0.25
p̂₂ = 18/100 = 0.18
p̂ = 43/200 = 0.215

Standard error under H₀

SE = √(0.215·0.785·(1/100 + 1/100)) = 0.0581

Test statistic

z = (0.25 − 0.18)/0.0581 = 1.205

Answer

z = 1.205

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

13.Linear regression & correlation

Answer: r = 0.9881

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 = 135/6 = 22.5
ȳ = 595/6 = 99.17

Sums of squares and products

Sxx = Σ(x − x̄)² = 157.5
Sxy = Σ(x − x̄)(y − ȳ) = 700.5
Syy = Σ(y − ȳ)² = 3191

Correlation coefficient

r = 700.5/√(157.5·3191) = 0.9881

Answer

r = 0.9881
14.Linear regression & correlation

Answer: ŷ = 21.8

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 = 88/5 = 17.6

Sums of squares and products

Sxx = Σ(x − x̄)² = 10
Sxy = Σ(x − x̄)(y − ȳ) = −28

Regression line

b = −28/10 = −2.8
a = 17.6 − (−2.8)·3 = 26
ŷ = 26 − 2.8·x

Prediction

ŷ = 26 + (−2.8)·1.5 = 21.8

Answer

ŷ = 21.8
15.Chi-square tests

Answer: E = 26.71

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 = 82·57/175 = 26.71

Answer

E = 26.71
16.Chi-square tests

Answer: χ²crit = 15.51

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

Table

χ²crit(df = 8, α = 0.05) = 15.507

Answer

χ²crit = 15.507

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

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