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.

New problemsBack to the course

In the print window, choose “Save as PDF” to keep a copy on your device.

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

Ecological validity refers to:

  1. AHow consistent a test is over time
  2. BWhether the test looks like it measures the right thing
  3. CHow well findings generalise to real-life settings
  4. DWhether a test predicts future behaviour
2.Research methods in psychologyHard

A study finds a correlation of r = 0.05 between height and IQ in only professional basketball players, though it is higher in the general population. What best explains the weak correlation?

  1. ARestriction of range
  2. BDemand characteristics
  3. CA ceiling effect in IQ tests
  4. DRegression to the mean
3.Descriptive statisticsMedium
152025303540

Find the sample standard deviation s of the data: 29, 24, 44, 11, 25, 17, 33.

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

A quantity is normally distributed with mean 500 and standard deviation 100. What proportion P of values is above 600?

μ = 500σ = 100x = 600
P =
5.ProbabilityMedium
xp(x)00.2510.220.1530.3540.05μ = 1.75

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

x: 0, 1, 2, 3, 4p(x): 0.25, 0.2, 0.15, 0.35, 0.05
σ =
6.ProbabilityHard
xp(x)10.120.3530.2540.2550.05μ

X has the distribution shown, and Y = −2X + 20. Find the standard deviation σY of Y.

x: 1, 2, 3, 4, 5p(x): 0.1, 0.35, 0.25, 0.25, 0.05Y = −2X + 20
σY =
7.Normal distributionMedium
−3−2−10123zshaded area to the right of z = 0.0239

Drag z so that the shaded area under the standard normal curve, to the right of z, is 0.0239. Find z to two decimals.

A = 0.0239
z =
8.Normal distributionHard
μ = 250x = ?97.5%

The bag weights are normally distributed with mean 250 g and standard deviation 5 g. Find the value x below which 97.5% of them lie (the 97.5th percentile).

μ = 250σ = 5P = 97.5%
x = g
9.Confidence intervalsMedium
μ₁ − μ₂0LUx̄₁ − x̄₂E = ?99%

The battery life of two phone models is compared. Model 1: 30 phones, mean 104.5 h, known standard deviation 8 h. Model 2: 50 phones, mean 119.5 h, known standard deviation 20 h. Find the margin of error E of the 99% confidence interval for μ₁ − μ₂.

n₁ = 30x̄₁ = 104.5 hσ₁ = 8 hn₂ = 50x̄₂ = 119.5 hσ₂ = 20 h99%
E = h
10.Confidence intervalsHard
μ₁ − μ₂, t(20) = 2.8450LU = ?x̄₁ − x̄₂E99%

Two fertilisers are tested. Plot group 1: 14 plots, mean yield 22.5 kg, standard deviation 4 kg. Plot group 2: 8 plots, mean 29 kg, standard deviation 1 kg. Assume equal variances (pooled t, df = n₁ + n₂ − 2). Find the upper limit of the 99% confidence interval for μ₁ − μ₂.

n₁ = 14x̄₁ = 22.5s₁ = 4n₂ = 8x̄₂ = 29s₂ = 199%t(20) = 2.845
upper limit = kg
11.Hypothesis tests & confidence intervalsMedium
−2.5712.571t = ?df = 5

6 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 −4.86 and the standard deviation of the differences is 7. Find the paired test statistic t for H₀: μd = 0.

n = 6d̄ = −4.86sd = 7
t =
12.Hypothesis tests & confidence intervalsHard
z −1.64501.645

A sample of 25 measurements has mean 75.3. The population standard deviation is 5. Find the upper limit U of the 90% confidence interval for the mean.

x̄ = 75.3σ = 5n = 2590%
U =
13.Linear regression & correlationMedium
xy147316901892201042212124136xy15202580100120140

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

n = 6Σx = 114Σy = 616Σx² = 2236Σy² = 65846Σxy = 12124
r =
14.Linear regression & correlationHard
xy2474586738891098xy5106080100x = 7

The table gives 5 paired observations of x = advertising spend in k€ and y = sales in k€. Use the least-squares line ŷ = a + b·x. A new observation (7, 84) is made. Find its residual e = y − ŷ.

n = 5(7, 84)
e =
15.Chi-square testsMedium
no defectminormajorTotalday shift6122240night shift266034120Total327256160

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 = 1602 × 3
χ² =
16.Chi-square testsHard
passfailTotalage 18–30681280age 31–50261440age 51+7129100Total16555220dfα = 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 = 0.01 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 = 2203 × 2α = 0.01
χ² =

3Solutions

1.Research methods in psychology

Answer: C · How well findings generalise to real-life settings

Why this is the answer

Ecological validity concerns whether results apply to everyday situations. Highly controlled lab studies often score low on it.

Answer

C · How well findings generalise to real-life settings
2.Research methods in psychology

Answer: A · Restriction of range

Why this is the answer

When a sample covers only a narrow range of a variable, correlations involving it are reduced. Basketball players are all tall, so height varies little.

Answer

A · Restriction of range
3.Descriptive statistics

Answer: s = 10.75

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 = 183/7 = 26.14

Squared deviations

Σ(x − x̄)² = 2.857² + (−2.143)² + 17.86² + (−15.14)² + (−1.143)² + (−9.143)² + 6.857² = 692.9

Standard deviation

s = √(692.9/6) = 10.75

Answer

s = 10.75
4.Descriptive statistics

Answer: P = 0.1587

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

Standard normal table

Φ(1) = 0.8413
P = 1 − 0.8413 = 0.1587

Answer

P = 0.1587
5.Probability

Answer: σ = 1.299

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 = 0·0.25 + 1·0.2 + 2·0.15 + 3·0.35 + 4·0.05 = 1.75

Mean of the squares

E(X²) = 0·0.25 + 1·0.2 + 4·0.15 + 9·0.35 + 16·0.05 = 4.75

Variance

Var(X) = 4.75 − 1.75² = 1.688
σ = √1.688 = 1.299

Answer

σ = 1.299
6.Probability

Answer: σY = 2.154

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 = 2.8, Var(X) = 1.16, σX = 1.077

Linear change

σY = |−2|·1.077 = 2.154

Answer

σY = 2.154

The mean does move: E(Y) = −2·2.8 + 20 = 14.4.

7.Normal distribution

Answer: z = 1.98

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 − 0.0239 = 0.9761

Read the z-table backwards

Find 0.9761 inside the table and read off z.

z = 1.98

Answer

z = 1.98
8.Normal distribution

Answer: x = 259.8 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) = 0.975 → z = 1.96

Back to x

x = 250 + 1.96·5 = 259.8 g

Answer

x = 259.8 g
9.Confidence intervals

Answer: E = 8.2 h

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.

Point estimate

x̄₁ − x̄₂ = 104.5 − 119.5 = −15

Standard error

SE = √(8²/30 + 20²/50) = 3.183

Margin of error

E = 2.576·3.183 = 8.2

Answer

E = 8.2 h

The interval does not contain 0, so the means differ significantly.

10.Confidence intervals

Answer: upper limit = −2.366 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 = (13·4² + 7·1²)/20 = 10.75
sp = 3.279

Margin of error

E = 2.845·3.279·√(1/14 + 1/8) = 4.134

Interval

−6.5 ± 4.134 → [−10.63, −2.366]

Answer

upper limit = −2.366 kg
11.Hypothesis tests & confidence intervals

Answer: t = −1.701

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 = 7/√6 = 2.858

Test statistic

t = −4.86/2.858 = −1.701

Answer

t = −1.701

With df = 5, the two-sided 5% critical value is 2.571; |t| is smaller, so H₀ is not rejected.

12.Hypothesis tests & confidence intervals

Answer: U = 76.94

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 = 1.645·5/√25 = 1.645

Interval limit

U = 75.3 + 1.645 = 76.94

Answer

U = 76.94
13.Linear regression & correlation

Answer: r = 0.9839

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.

Centred sums

Sxy = 12124 − 114·616/6 = 420
Sxx = 2236 − 114²/6 = 70
Syy = 65846 − 616²/6 = 2603

Correlation coefficient

r = 420/√(70·2603) = 0.9839

Answer

r = 0.9839
14.Linear regression & correlation

Answer: e = 4.35

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 = 30/5 = 6
ȳ = 365/5 = 73

Sums of squares and products

Sxx = Σ(x − x̄)² = 40
Sxy = Σ(x − x̄)(y − ȳ) = 266

Regression line

b = 266/40 = 6.65
a = 73 − 6.65·6 = 33.1
ŷ = 33.1 + 6.65·x

Predicted value

ŷ = 33.1 + 6.65·7 = 79.65

Residual

e = 84 − 79.65 = 4.35

Answer

e = 4.35

The point lies above the line.

15.Chi-square tests

Answer: χ² = 9.429

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: 40·32/160 = 8, 40·72/160 = 18, 40·56/160 = 14
night shift: 120·32/160 = 24, 120·72/160 = 54, 120·56/160 = 42

Chi-square statistic

χ² = (6 − 8)²/8 + (12 − 18)²/18 + (22 − 14)²/14 + (26 − 24)²/24 + (60 − 54)²/54 + (34 − 42)²/42
χ² = 9.429

Answer

χ² = 9.429

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

16.Chi-square tests

Answer: χ² = 7.253

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

age 18–30: 60, 20
age 31–50: 30, 10
age 51+: 75, 25

Chi-square statistic

χ² = (68 − 60)²/60 + (12 − 20)²/20 + (26 − 30)²/30 + (14 − 10)²/10 + (71 − 75)²/75 + (29 − 25)²/25
χ² = 7.253

Decision

df = 2, χ²crit = 9.210
7.253 < 9.210

Answer

χ² = 7.253

χ² is below the critical value, so H₀ is not rejected: there is no significant evidence of a relationship.

ExamProblems · examproblems.com/courses/research-methods-psychology