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 a content analysis, inter-rater reliability is checked by:

  1. AUsing a larger sample of materials
  2. BHaving two researchers code the same material and comparing results
  3. CAsking participants afterwards whether the researcher's coding of their material is correct
  4. DCoding the material twice in one day
2.Research methods in psychologyHard

Patients chosen because they have extremely high anxiety scores show lower scores at a second test, even without treatment. What is the most likely explanation?

  1. AThe Hawthorne effect
  2. BA ceiling effect
  3. CCounterbalancing
  4. DRegression to the mean
3.Descriptive statisticsMedium
101214161820222426

Find the sample standard deviation s of the data: 13, 14, 19, 14, 10, 26, 15.

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

Exam scores are normally distributed with mean 70 and standard deviation 5. Find the z-score z of a score of 75.

μ = 70σ = 5x = 75
z =
5.ProbabilityMedium
red: 3blue: 3

A bag holds 3 red and 3 blue balls. You draw two balls without putting the first one back. What is the probability P that both are red?

3 / 3
P =
6.ProbabilityHard
€p(x)−50.79200.2450.01

A game costs €5 to play. You win €25 with probability 0.2 and €50 with probability 0.01; otherwise you win nothing. The stake is not returned. Find the expected net gain E(G) per game.

stake = €5€25: p = 0.2€50: p = 0.01
E(G) = €
7.Normal distributionMedium
200 mm199.8 mmX + Y?

X is normal with mean 120 mm and standard deviation 0.3 mm; Y is independent and normal with mean 80 mm and standard deviation 0.4 mm. Find the probability P that X + Y is more than 199.8 mm. Round z to two decimals.

μX = 120, σX = 0.3μY = 80, σY = 0.4X + Y > 199.8 mm
P =
8.Normal distributionHard
35455565758595pointstop 25%

The exam scores are normally distributed with mean 65 points and standard deviation 10 points. Drag the cut-off so the shaded top 25% is to the right of it. Find the cut-off x.

μ = 65 pointsσ = 10 points25%
x = points
9.Confidence intervalsMedium
n = 16, s = 3.5LU = ?x̄ = 65E90%

A sample of 16 items has mean 65 and standard deviation s = 3.5. The population is roughly normal. Find the upper limit of the 90% confidence interval for the mean.

n = 16x̄ = 65s = 3.590%t(15) = 1.753
upper limit =
10.Confidence intervalsHard
σ = 39, n = ?x̄ − Ex̄ + Ex̄E = 2.595%

The population standard deviation is 39. 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 2.5?

σ = 39E = 2.595%
n =
11.Hypothesis tests & confidence intervalsMedium
t −2.13102.131

A sample of 16 students has a mean score of 69.5 and a sample standard deviation of 10. Find the test statistic t for H₀: μ = 75.

μ₀ = 75x̄ = 69.5s = 10n = 16
t =
12.Hypothesis tests & confidence intervalsHard
−1.961.96z = ?

In a trial, 90 of 300 patients on treatment A recover, and 34 of 200 patients on treatment B. Using the pooled proportion, find the test statistic z for H₀: p₁ = p₂.

x₁ = 90n₁ = 300x₂ = 34n₂ = 200
z =
13.Linear regression & correlationMedium
xy2941161782510281235xy510102030

The table gives 6 paired observations of x = load in kN and y = extension in mm. Find the coefficient of determination r² of the least-squares line.

n = 6
r² =
14.Linear regression & correlationHard
xy160261362476575685xy246607080x = 3.5

The table gives 6 paired observations of x = hours studied and y = exam score. Use the least-squares line ŷ = a + b·x. Predict ŷ for x = 3.5.

n = 6x = 3.5
ŷ =
15.Chi-square testsMedium
lowmediumhighTotalurban223741100rural334047120Total557788220

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 = 2202 × 3
χ² =
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 11 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: B · Having two researchers code the same material and comparing results

Why this is the answer

If independent coders produce similar codes for the same data (high agreement or correlation), the coding scheme is reliable.

Answer

B · Having two researchers code the same material and comparing results
2.Research methods in psychology

Answer: D · Regression to the mean

Why this is the answer

When people are selected for extreme scores, measurement error and chance mean their later scores tend to move towards the average. Control groups help separate this from treatment effects.

Answer

D · Regression to the mean
3.Descriptive statistics

Answer: s = 5.21

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 = 111/7 = 15.86

Squared deviations

Σ(x − x̄)² = (−2.857)² + (−1.857)² + 3.143² + (−1.857)² + (−5.857)² + 10.14² + (−0.8571)² = 162.9

Standard deviation

s = √(162.9/6) = 5.21

Answer

s = 5.21
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 = (75 − 70)/5 = 1

Answer

z = 1
5.Probability

Answer: P = 0.2

Probability: with equally likely outcomes, P(A) = favourable/total. For independent events multiply: P(A and B) = P(A)·P(B). The complement often helps: P(at least one) = 1 − P(none). Binomial: in n independent trials with success probability p, P(X = k) = C(n, k)·pᵏ·(1 − p)ⁿ⁻ᵏ.

First and second draw

P = 3/6 · 2/5 = 6/30 = 0.2

Answer

P = 0.2
6.Probability

Answer: E(G) = 0.5 €

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: 20 (p = 0.2), 45 (p = 0.01), −5 (p = 0.79)

Expected value

E(G) = 20·0.2 + 45·0.01 − 5·0.79 = 0.5 €

Answer

E(G) = 0.5 €

A positive value means you win money on average.

7.Normal distribution

Answer: P = 0.6554

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 sum

μ = 200 mm, σ = √(0.3² + 0.4²) = 0.5 mm

Standardise

z = (199.8 − 200)/0.5 = −0.4

Upper tail

P = 1 − Φ(−0.4) = 1 − 0.3446 = 0.6554

Answer

P = 0.6554
8.Normal distribution

Answer: x = 71.7 points

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

Φ(z) = 0.75

z from the table

z ≈ 0.67

Back to the original units

x = 65 + 0.67·10 = 71.7 points

Answer

x = 71.7 points
9.Confidence intervals

Answer: upper limit = 66.53

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 = 1.753

Margin of error

E = 1.753·3.5/√16 = 1.534

Interval

65 ± 1.534 → [63.47, 66.53]

Answer

upper limit = 66.53
10.Confidence intervals

Answer: n = 935

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 = (1.96·39/2.5)² = 934.9

Round up

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

n = 935

Answer

n = 935
11.Hypothesis tests & confidence intervals

Answer: t = −2.2

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).

Standard error

SE = 10/√16 = 10/4 = 2.5

Test statistic

t = (69.5 − 75)/2.5 = −2.2

Answer

t = −2.2
t0.025; df = 15 = 2.131

|t| > 2.131, so H₀ is rejected at the 5% level (two-sided).

12.Hypothesis tests & confidence intervals

Answer: z = 3.298

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 = 90/300 = 0.3
p̂₂ = 34/200 = 0.17
p̂ = 124/500 = 0.248

Standard error under H₀

SE = √(0.248·0.752·(1/300 + 1/200)) = 0.03942

Test statistic

z = (0.3 − 0.17)/0.03942 = 3.298

Answer

z = 3.298

|z| > 1.96, so H₀ is rejected at the 5% level (two-sided).

13.Linear regression & correlation

Answer: r² = 0.9798

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 = 125/6 = 20.83

Sums of squares and products

Sxx = Σ(x − x̄)² = 70
Sxy = Σ(x − x̄)(y − ȳ) = 189
Syy = Σ(y − ȳ)² = 520.8

Coefficient of determination

r² = 189²/(70·520.8) = 0.9798

Answer

r² = 0.9798

About 97.98% of the variation in y is explained by the line.

14.Linear regression & correlation

Answer: ŷ = 69.83

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 = 419/6 = 69.83

Sums of squares and products

Sxx = Σ(x − x̄)² = 17.5
Sxy = Σ(x − x̄)(y − ȳ) = 90.5

Regression line

b = 90.5/17.5 = 5.171
a = 69.83 − 5.171·3.5 = 51.73
ŷ = 51.73 + 5.171·x

Prediction

ŷ = 51.73 + 5.171·3.5 = 69.83

Answer

ŷ = 69.83
15.Chi-square tests

Answer: χ² = 0.9154

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

urban: 100·55/220 = 25, 100·77/220 = 35, 100·88/220 = 40
rural: 120·55/220 = 30, 120·77/220 = 42, 120·88/220 = 48

Chi-square statistic

χ² = (22 − 25)²/25 + (37 − 35)²/35 + (41 − 40)²/40 + (33 − 30)²/30 + (40 − 42)²/42 + (47 − 48)²/48
χ² = 0.9154

Answer

χ² = 0.9154

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

16.Chi-square tests

Answer: χ²crit = 18.31

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 = 11 − 1 = 10

Table

χ²crit(df = 10, α = 0.05) = 18.307

Answer

χ²crit = 18.307

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

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