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

Ecological validity refers to:

  1. AWhether a test predicts future behaviour
  2. BHow consistent a test is over time
  3. CWhether the test looks like it measures the right thing
  4. DHow well findings generalise to real-life settings
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. AControl variable
  2. BMediator
  3. CModerator
  4. DExtraneous variable
3.Descriptive statisticsMedium
2025303540455055

Find the sample standard deviation s of the data: 28, 29, 20, 47, 39, 57.

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

Exam scores are normally distributed with mean 50 and standard deviation 5. Find the z-score z of a score of 47.5.

μ = 50σ = 5x = 47.5
z =
5.ProbabilityMedium
xp(x)00.0510.3520.2530.35μ = 1.9

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

x: 0, 1, 2, 3p(x): 0.05, 0.35, 0.25, 0.35
σ =
6.ProbabilityHard
xp(x)00.320.340.0560.35μ

X has the distribution shown, and Y = −2X + 1. Find the standard deviation σY of Y.

x: 0, 2, 4, 6p(x): 0.3, 0.3, 0.05, 0.35Y = −2X + 1
σY =
7.Normal distributionMedium
48 kg45.6 kgT = X₁ + … + X₄?

The masses of boxes are independent and normal with mean 12 kg and standard deviation 0.8 kg. Find the probability P that the total mass of 4 of them is below 45.6 kg. Round z to two decimals.

μ = 12 kgσ = 0.8 kgn = 4< 45.6 kg
P =
8.Normal distributionHard
μ = 100x = ?99%

The IQ scores are normally distributed with mean 100 points and standard deviation 15 points. Find the value x below which 99% of them lie (the 99th percentile).

μ = 100σ = 15P = 99%
x = points
9.Confidence intervalsMedium
μ₁ − μ₂0LU = ?x̄₁ − x̄₂E95%

The battery life of two phone models is compared. Model 1: 80 phones, mean 81 h, known standard deviation 3 h. Model 2: 30 phones, mean 75 h, known standard deviation 17 h. Find the upper limit of the 95% confidence interval for μ₁ − μ₂.

n₁ = 80x̄₁ = 81 hσ₁ = 3 hn₂ = 30x̄₂ = 75 hσ₂ = 17 h95%
upper limit = h
10.Confidence intervalsHard
σ = 10, n = ?x̄ − Ex̄ + Ex̄E = 1.595%

The population standard deviation is 10. How large a sample n do you need so that a 95% confidence interval for the mean has a margin of error of at most 1.5?

σ = 10E = 1.595%
n =
11.Hypothesis tests & confidence intervalsMedium
t −2.13102.131

A sample of 16 students has a mean score of 93 and a sample standard deviation of 20. Find the test statistic t for H₀: μ = 100.

μ₀ = 100x̄ = 93s = 20n = 16
t =
12.Hypothesis tests & confidence intervalsHard
μ₀x̄c

A one-sided z test of H₀: μ = 200 against H₁: μ < 200 is done at α = 0.01 with a sample of 25; the standard deviation is known to be 12.5. In reality the mean is 194.25. Find the probability of a Type II error β in percent.

μ₀ = 200μ = 194.25σ = 12.5n = 25α = 0.01
β = %
13.Linear regression & correlationMedium
xy1578177819872110423112xy152080100

The table gives 5 paired observations of x = temperature in °C and y = ice creams sold. Find the coefficient of determination r² of the least-squares line.

n = 5
r² =
14.Linear regression & correlationHard
xy234861181210141217xy51051015x = 11

The table gives 6 paired observations of x = load in kN and y = extension in mm. Use the least-squares line ŷ = a + b·x. A new observation (11, 13) is made. Find its residual e = y − ŷ.

n = 6(11, 13)
e =
15.Chi-square testsMedium
busbikecarTotalstudents352936100staff253144100Total606080200

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 = 2002 × 3
χ² =
16.Chi-square testsHard
yesnoTotalplant A194160plant B6535100plant C364480Total120120240dfα = 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 = 2403 × 2α = 0.01
χ² =

3Solutions

1.Research methods in psychology

Answer: D · 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

D · How well findings generalise to real-life settings
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 = 13.69

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 = 220/6 = 36.67

Squared deviations

Σ(x − x̄)² = (−8.667)² + (−7.667)² + (−16.67)² + 10.33² + 2.333² + 20.33² = 937.3

Standard deviation

s = √(937.3/5) = 13.69

Answer

s = 13.69
4.Descriptive statistics

Answer: z = −0.5

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 = (47.5 − 50)/5 = −0.5

Answer

z = −0.5
5.Probability

Answer: σ = 0.9434

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 = 0·0.05 + 1·0.35 + 2·0.25 + 3·0.35 = 1.9

Mean of the squares

E(X²) = 0·0.05 + 1·0.35 + 4·0.25 + 9·0.35 = 4.5

Variance

Var(X) = 4.5 − 1.9² = 0.89
σ = √0.89 = 0.9434

Answer

σ = 0.9434
6.Probability

Answer: σY = 4.976

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 = 2.9, Var(X) = 6.19, σX = 2.488

Linear change

σY = |−2|·2.488 = 4.976

Answer

σY = 4.976

The mean does move: E(Y) = −2·2.9 + 1 = −4.8.

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 = 48 kg, σT = 0.8·√4 = 1.6 kg

Standardise

z = (45.6 − 48)/1.6 = −1.5

Probability

P = Φ(−1.5) = 0.06681

Answer

P = 0.06681
8.Normal distribution

Answer: x = 134.9 points

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) = 0.99 → z = 2.326

Back to x

x = 100 + 2.326·15 = 134.89 points

Answer

x = 134.89 points
9.Confidence intervals

Answer: upper limit = 12.12 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 = 81 − 75 = 6

Standard error

SE = √(3²/80 + 17²/30) = 3.122

Margin of error

E = 1.96·3.122 = 6.119

Interval

6 ± 6.119 → [−0.1188, 12.12]

Answer

upper limit = 12.12 h

The interval contains 0, so the data do not show a significant difference.

10.Confidence intervals

Answer: n = 171

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.

Solve for n

n = (1.96·10/1.5)² = 170.7

Round up

A sample size is a whole number, and rounding down would make the margin too large.

n = 171

Answer

n = 171
11.Hypothesis tests & confidence intervals

Answer: t = −1.4

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/√16 = 20/4 = 5

Test statistic

t = (93 − 100)/5 = −1.4

Answer

t = −1.4
t0.025; df = 15 = 2.131

|t| < 2.131, so H₀ is not rejected at the 5% level (two-sided).

12.Hypothesis tests & confidence intervals

Answer: β = 51.04 %

Power of a test: power = 1 − β is the probability of rejecting H₀ when it is false. Find the cutoff x̄c from μ₀ and z_α, then compute the probability of landing beyond x̄c using the true mean: z = (x̄c − μ)/(σ/√n).

Critical sample mean

SE = 12.5/√25 = 2.5
x̄c = 200 − 2.326·2.5 = 194.2

z under the true mean

z = (194.2 − 194.25)/2.5 = −0.026

Type II error

power = P(Z < −0.026) = 48.96%
β = 100% − 48.96% = 51.04%

Answer

β = 51.04%
13.Linear regression & correlation

Answer: r² = 0.9197

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 = 95/5 = 19
ȳ = 459/5 = 91.8

Sums of squares and products

Sxx = Σ(x − x̄)² = 40
Sxy = Σ(x − x̄)(y − ȳ) = 188
Syy = Σ(y − ȳ)² = 960.8

Coefficient of determination

r² = 188²/(40·960.8) = 0.9197

Answer

r² = 0.9197

About 91.97% of the variation in y is explained by the line.

14.Linear regression & correlation

Answer: e = −2.919

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 = 42/6 = 7
ȳ = 65/6 = 10.83

Sums of squares and products

Sxx = Σ(x − x̄)² = 70
Sxy = Σ(x − x̄)(y − ȳ) = 89

Regression line

b = 89/70 = 1.271
a = 10.83 − 1.271·7 = 1.933
ŷ = 1.933 + 1.271·x

Predicted value

ŷ = 1.933 + 1.271·11 = 15.92

Residual

e = 13 − 15.92 = −2.919

Answer

e = −2.919

The point lies below the line.

15.Chi-square tests

Answer: χ² = 2.533

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

students: 100·60/200 = 30, 100·60/200 = 30, 100·80/200 = 40
staff: 100·60/200 = 30, 100·60/200 = 30, 100·80/200 = 40

Chi-square statistic

χ² = (35 − 30)²/30 + (29 − 30)²/30 + (36 − 40)²/40 + (25 − 30)²/30 + (31 − 30)²/30 + (44 − 40)²/40
χ² = 2.533

Answer

χ² = 2.533

df = 2. At α = 0.05 the critical value is 5.991, so H₀ (independence) is not rejected.

16.Chi-square tests

Answer: χ² = 17.87

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: 30, 30
plant B: 50, 50
plant C: 40, 40

Chi-square statistic

χ² = (19 − 30)²/30 + (41 − 30)²/30 + (65 − 50)²/50 + (35 − 50)²/50 + (36 − 40)²/40 + (44 − 40)²/40
χ² = 17.87

Decision

df = 2, χ²crit = 9.210
17.87 > 9.210

Answer

χ² = 17.87

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

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