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

Internal validity refers to:

  1. AWhether participants found the study interesting
  2. BWhether the observed effect was really caused by the IV rather than other factors
  3. CWhether the findings generalise beyond the study to other populations, settings and times
  4. DWhether a test gives the same results over time
2.Research methods in psychologyHard

A researcher uses a questionnaire, interviews and observation to study workplace stress, and the findings agree. What is this approach called?

  1. ATriangulation
  2. BCounterbalancing
  3. CStratification
  4. DReplication
3.Descriptive statisticsMedium
152025303540455055

Find the sample standard deviation s of the data: 32, 56, 38, 34, 28, 15, 18.

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

A quantity is normally distributed with mean 70 and standard deviation 20. What proportion P of values is below 54?

μ = 70σ = 20x = 54
P =
5.ProbabilityMedium
P(at least one six) = ??1?2?3?4?5?6P(six) = 1/6 on each roll

A fair die is rolled 6 times. What is the probability P of getting at least one six?

n = 6
P =
6.ProbabilityHard
n = 7, p = 0.4: P(X = 1) = ?k01234567

A test has 7 multiple-choice questions. A student who guesses gets each one right with probability 0.4, independently. What is the probability P of exactly 1 correct answers?

n = 7p = 0.4k = 1
P =
7.Normal distributionMedium
750 g759.8 gX + Y?

X is normal with mean 250 g and standard deviation 4 g; Y is independent and normal with mean 500 g and standard deviation 9 g. Find the probability P that X + Y is more than 759.8 g. Round z to two decimals.

μX = 250, σX = 4μY = 500, σY = 9X + Y > 759.8 g
P =
8.Normal distributionHard
0.05 mm0D − d?

Hole diameters D are normal with mean 25.05 mm and standard deviation 0.02 mm. Shaft diameters d are independent and normal with mean 25 mm and standard deviation 0.02 mm. A randomly chosen shaft fits when D − d > 0. Find the probability P that it fits. Round z to two decimals.

hole: μ = 25.05 mm, σ = 0.02 mmshaft: μ = 25 mm, σ = 0.02 mm
P =
9.Confidence intervalsMedium
n = 16, s = 5L = ?Ux̄ = 39E90%

A sample of 16 items has mean 39 and standard deviation s = 5. The population is roughly normal. Find the lower limit of the 90% confidence interval for the mean.

n = 16x̄ = 39s = 590%t(15) = 1.753
lower limit =
10.Confidence intervalsHard
n = 21Lσ = ?s = 0.695%

The diameters of 21 machined shafts have a sample standard deviation of 0.6 µm. Assuming the diameters are normally distributed, find the upper limit σ of the 95% confidence interval for the population standard deviation. With df = 20, 2.5% of the χ² distribution lies below 9.591 and 2.5% lies above 34.17.

n = 21s = 0.6 µmχ² = 9.591, 34.17
σ = µm
11.Hypothesis tests & confidence intervalsMedium
−2.22p = ?

The mean of a population is claimed to be 120; the population standard deviation is known to be 10. A random sample of 25 observations has a mean of 115.56. Test H₀: μ = 120 against H₁: μ < 120. Find the p-value p.

μ₀ = 120σ = 10n = 25x̄ = 115.56H₁: μ < μ₀
p =
12.Hypothesis tests & confidence intervalsHard
z −1.64501.645

A sample of 25 measurements has mean 493.5. The population standard deviation is 12. Find the lower limit L of the 90% confidence interval for the mean.

x̄ = 493.5σ = 12n = 2590%
L =
13.Linear regression & correlationMedium
xy274106138211024xy5101020

The table gives 5 paired observations of x = load in kN and y = extension in mm. Their sums are given. Find Pearson's correlation coefficient r from these sums.

n = 5Σx = 30Σy = 75Σx² = 220Σy² = 1335Σxy = 540
r =
14.Linear regression & correlationHard
xy124223321417514xy24152025x = 4.5

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 = 4.5.

n = 5x = 4.5
ŷ =
15.Chi-square testsMedium
yesnoTotalmen392160women4159100Total8080160

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 × 2
χ² =
16.Chi-square testsHard

A chi-square goodness-of-fit test checks whether counts in 6 categories follow a Poisson distribution whose mean is estimated from the same data. How many degrees of freedom df does the test have?

k = 6
df =

3Solutions

1.Research methods in psychology

Answer: B · Whether the observed effect was really caused by the IV rather than other factors

Why this is the answer

High internal validity means alternative explanations, such as confounds, demand characteristics or experimenter bias, have been ruled out. External validity concerns generalisation.

Answer

B · Whether the observed effect was really caused by the IV rather than other factors
2.Research methods in psychology

Answer: A · Triangulation

Why this is the answer

Triangulation uses multiple methods, sources or researchers. Agreement between them increases confidence in the validity of the findings.

Answer

A · Triangulation
3.Descriptive statistics

Answer: s = 13.64

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 = 221/7 = 31.57

Squared deviations

Σ(x − x̄)² = 0.4286² + 24.43² + 6.429² + 2.429² + (−3.571)² + (−16.57)² + (−13.57)² = 1116

Standard deviation

s = √(1116/6) = 13.64

Answer

s = 13.64
4.Descriptive statistics

Answer: P = 0.2119

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 = −0.8

Standard normal table

Φ(−0.8) = 0.2119
P = 0.2119

Answer

P = 0.2119
5.Probability

Answer: P = 0.6651

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.

Complement

P(no six) = (5/6)6 = 0.3349
P = 1 − 0.3349 = 0.6651

Answer

P = 0.6651
6.Probability

Answer: P = 0.1306

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.

Binomial formula

P(X = 1) = 7·0.41·0.66 = 7·0.4·0.04666 = 0.1306

Answer

P = 0.1306
7.Normal distribution

Answer: P = 0.1587

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 = 750 g, σ = √(4² + 9²) = 9.849 g

Standardise

z = (759.8 − 750)/9.849 = 1

Upper tail

P = 1 − Φ(1) = 1 − 0.8413 = 0.1587

Answer

P = 0.1587
8.Normal distribution

Answer: P = 0.9616

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 = 25.05 − 25 = 0.05 mm, σ = 0.02828 mm

Standardise

z = (0 − 0.05)/0.02828 = −1.77

Probability

P = 1 − Φ(−1.77) = 0.9616

Answer

P = 0.9616
9.Confidence intervals

Answer: lower limit = 36.81

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 = 1.753·5/√16 = 2.191

Interval

39 ± 2.191 → [36.81, 41.19]

Answer

lower limit = 36.81
10.Confidence intervals

Answer: σ = 0.8664 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 = 20·0.6² = 7.2

Limits for σ²

7.2/34.17 = 0.2107
7.2/9.591 = 0.7507

Limit for σ

σ = √0.7507 = 0.8664 µm

Answer

σ = 0.8664 µm

95% interval for σ: [0.459, 0.8664] µm.

11.Hypothesis tests & confidence intervals

Answer: p = 0.01321

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.

Test statistic

SE = σ/√n = 10/√25 = 2
z = (115.56 − 120)/2 = −2.22

Tail area

Φ(2.22) = 0.9868
1 − Φ(2.22) = 0.0132

p-value

p = 0.01321

Answer

p = 0.01321

p ≤ 0.05, so H₀ is rejected at the 5% level.

12.Hypothesis tests & confidence intervals

Answer: L = 489.6

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 = 1.645·12/√25 = 3.948

Interval limit

L = 493.5 − 3.948 = 489.6

Answer

L = 489.6
13.Linear regression & correlation

Answer: r = 0.982

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 = 540 − 30·75/5 = 90
Sxx = 220 − 30²/5 = 40
Syy = 1335 − 75²/5 = 210

Correlation coefficient

r = 90/√(40·210) = 0.982

Answer

r = 0.982
14.Linear regression & correlation

Answer: ŷ = 15.9

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 = 99/5 = 19.8

Sums of squares and products

Sxx = Σ(x − x̄)² = 10
Sxy = Σ(x − x̄)(y − ȳ) = −26

Regression line

b = −26/10 = −2.6
a = 19.8 − (−2.6)·3 = 27.6
ŷ = 27.6 − 2.6·x

Prediction

ŷ = 27.6 + (−2.6)·4.5 = 15.9

Answer

ŷ = 15.9
15.Chi-square tests

Answer: χ² = 8.64

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

men: 60·80/160 = 30, 60·80/160 = 30
women: 100·80/160 = 50, 100·80/160 = 50

Chi-square statistic

χ² = (39 − 30)²/30 + (21 − 30)²/30 + (41 − 50)²/50 + (59 − 50)²/50
χ² = 8.64

Answer

χ² = 8.64

df = 1. At α = 0.05 the critical value is 3.841, so H₀ (independence) is rejected.

16.Chi-square tests

Answer: df = 4

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 = 6 − 1 − 1 = 4

Answer

df = 4

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