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

To obtain a sample with the same proportions of first-, second- and third-year students as the university population, a researcher should use:

  1. AStratified sampling
  2. BVolunteer sampling
  3. CSnowball sampling
  4. DOpportunity sampling
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. AModerator
  2. BMediator
  3. CControl variable
  4. DExtraneous variable
3.Descriptive statisticsMedium
20253035404550

Find the sample standard deviation s of the data: 19, 45, 53, 50, 47, 28, 17.

n = 7
s =
4.Descriptive statisticsHard
z −2−1012

Exam scores are normally distributed with mean 500 and standard deviation 50. Find the z-score z of a score of 460.

μ = 500σ = 50x = 460
z =
5.ProbabilityMedium
xp(x)10.3520.230.240.25μ = 2.35

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

x: 1, 2, 3, 4p(x): 0.35, 0.2, 0.2, 0.25
Var(X) =
6.ProbabilityHard
€p(x)−30.78120.21470.02

A game costs €3 to play. You win €15 with probability 0.2 and €150 with probability 0.02; otherwise you win nothing. The stake is not returned. Find the expected net gain E(G) per game.

stake = €3€15: p = 0.2€150: p = 0.02
E(G) = €
7.Normal distributionMedium
−3−2−10123zshaded area to the right of z = 0.1685

Drag z so that the shaded area under the standard normal curve, to the right of z, is 0.1685. Find z to two decimals.

A = 0.1685
z =
8.Normal distributionHard
10 cm0X − Y?

The heights X of men and Y of women in a group are independent and normal: X has mean 180 cm and standard deviation 10 cm, Y has mean 170 cm and standard deviation 12 cm. Find the probability P that X is greater than Y. Round z to two decimals.

μX = 180 cm, σX = 10 cmμY = 170 cm, σY = 12 cm
P =
9.Confidence intervalsMedium
n = 30, s = 12L = ?Ux̄ = 42.5E95%

A sample of 30 items has mean 42.5 and standard deviation s = 12. The population is roughly normal. Find the lower limit of the 95% confidence interval for the mean.

n = 30x̄ = 42.5s = 1295%t(29) = 2.045
lower limit =
10.Confidence intervalsHard
n = 15σ = ?Us = 0.695%

The diameters of 15 machined shafts have a sample standard deviation of 0.6 µm. Assuming the diameters are normally distributed, find the lower limit σ of the 95% confidence interval for the population standard deviation. With df = 14, 2.5% of the χ² distribution lies below 5.629 and 2.5% lies above 26.119.

n = 15s = 0.6 µmχ² = 5.629, 26.119
σ = µm
11.Hypothesis tests & confidence intervalsMedium
−2.4472.447t = ?df = 6

7 employees take a typing test before and after a training course. For each person the difference d = after − before (words per minute) is recorded. The mean difference is 2.04 and the standard deviation of the differences is 6. Find the paired test statistic t for H₀: μd = 0.

n = 7d̄ = 2.04sd = 6
t =
12.Hypothesis tests & confidence intervalsHard
z −1.9601.96

A sample of 100 measurements has mean 98.6. The population standard deviation is 8. Find the upper limit U of the 95% confidence interval for the mean.

x̄ = 98.6σ = 8n = 10095%
U =
13.Linear regression & correlationMedium
xy131235332440539xy243540

The table gives 5 paired observations of x = advertising spend in k€ and y = sales in k€. Find Pearson's correlation coefficient r.

n = 5
r =
14.Linear regression & correlationHard
xy158255360467564678xy246607080x = 2.5

The table gives 6 paired observations of x = hours studied and y = exam score. Use the least-squares line ŷ = a + b·x. A new observation (2.5, 64) is made. Find its residual e = y − ŷ.

n = 6(2.5, 64)
e =
15.Chi-square testsMedium
no defectminormajorTotalday shift11131943night shift22173271Total333051114

The contingency table shows observed counts. For a chi-square test of independence, find the expected count E for the cell “day shift, no defect”.

row total = 43column total = 33N = 114
E =
16.Chi-square testsHard
passfailTotalplant A235780plant B354580plant C384280Total96144240dfα = 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 = 2403 × 2α = 0.05
χ² =

3Solutions

1.Research methods in psychology

Answer: A · Stratified sampling

Why this is the answer

Stratified sampling identifies subgroups (strata) and randomly samples from each in proportion to their size in the population, making the sample more representative.

Answer

A · Stratified sampling
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 = 15.24

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 = 259/7 = 37

Squared deviations

Σ(x − x̄)² = (−18)² + 8² + 16² + 13² + 10² + (−9)² + (−20)² = 1394

Standard deviation

s = √(1394/6) = 15.24

Answer

s = 15.24
4.Descriptive statistics

Answer: z = −0.8

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 = (460 − 500)/50 = −0.8

Answer

z = −0.8
5.Probability

Answer: Var(X) = 1.427

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 = 1·0.35 + 2·0.2 + 3·0.2 + 4·0.25 = 2.35

Mean of the squares

E(X²) = 1·0.35 + 4·0.2 + 9·0.2 + 16·0.25 = 6.95

Variance

Var(X) = 6.95 − 2.35² = 1.427

Answer

Var(X) = 1.427
6.Probability

Answer: E(G) = 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: 12 (p = 0.2), 147 (p = 0.02), −3 (p = 0.78)

Expected value

E(G) = 12·0.2 + 147·0.02 − 3·0.78 = 3 €

Answer

E(G) = 3 €

A positive value means you win money on average.

7.Normal distribution

Answer: z = 0.96

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 − 0.1685 = 0.8315

Read the z-table backwards

Find 0.8315 inside the table and read off z.

z = 0.96

Answer

z = 0.96
8.Normal distribution

Answer: P = 0.7389

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 difference

μ = 180 − 170 = 10 cm, σ = 15.62 cm

Standardise

z = (0 − 10)/15.62 = −0.64

Probability

P = 1 − Φ(−0.64) = 0.7389

Answer

P = 0.7389
9.Confidence intervals

Answer: lower limit = 38.02

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 = 29, t = 2.045

Margin of error

E = 2.045·12/√30 = 4.48

Interval

42.5 ± 4.48 → [38.02, 46.98]

Answer

lower limit = 38.02
10.Confidence intervals

Answer: σ = 0.4393 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 = 14·0.6² = 5.04

Limits for σ²

5.04/26.119 = 0.193
5.04/5.629 = 0.8954

Limit for σ

σ = √0.193 = 0.4393 µm

Answer

σ = 0.4393 µm

95% interval for σ: [0.4393, 0.9462] µm.

11.Hypothesis tests & confidence intervals

Answer: t = 0.8996

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 of the mean difference

SE = 6/√7 = 2.268

Test statistic

t = 2.04/2.268 = 0.8996

Answer

t = 0.8996

With df = 6, the two-sided 5% critical value is 2.447; |t| is smaller, so H₀ is not rejected.

12.Hypothesis tests & confidence intervals

Answer: U = 100.2

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 = 1.96·8/√100 = 1.568

Interval limit

U = 98.6 + 1.568 = 100.2

Answer

U = 100.2
13.Linear regression & correlation

Answer: r = 0.8224

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 = 177/5 = 35.4

Sums of squares and products

Sxx = Σ(x − x̄)² = 10
Sxy = Σ(x − x̄)(y − ȳ) = 21
Syy = Σ(y − ȳ)² = 65.2

Correlation coefficient

r = 21/√(10·65.2) = 0.8224

Answer

r = 0.8224
14.Linear regression & correlation

Answer: e = 4.162

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 = 382/6 = 63.67

Sums of squares and products

Sxx = Σ(x − x̄)² = 17.5
Sxy = Σ(x − x̄)(y − ȳ) = 67

Regression line

b = 67/17.5 = 3.829
a = 63.67 − 3.829·3.5 = 50.27
ŷ = 50.27 + 3.829·x

Predicted value

ŷ = 50.27 + 3.829·2.5 = 59.84

Residual

e = 64 − 59.84 = 4.162

Answer

e = 4.162

The point lies above the line.

15.Chi-square tests

Answer: E = 12.45

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 = 43·33/114 = 12.45

Answer

E = 12.45
16.Chi-square tests

Answer: χ² = 6.563

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

plant A: 32, 48
plant B: 32, 48
plant C: 32, 48

Chi-square statistic

χ² = (23 − 32)²/32 + (57 − 48)²/48 + (35 − 32)²/32 + (45 − 48)²/48 + (38 − 32)²/32 + (42 − 48)²/48
χ² = 6.563

Decision

df = 2, χ²crit = 5.991
6.563 > 5.991

Answer

χ² = 6.563

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

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