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

A cross-sectional study compares 20-, 40- and 60-year-olds on technology use at the same time. Which problem may confound the age comparison?

  1. APractice effects
  2. BCohort effects
  3. COrder effects
  4. DFloor effects
2.Research methods in psychologyHard

A researcher manipulates mood by showing a sad film, then asks participants to rate how sad they feel before the main task. Why is this rating included?

  1. AAs the main dependent variable of the study
  2. BTo debrief participants about the aims
  3. CTo counterbalance the order of conditions
  4. DAs a manipulation check to confirm the IV worked as intended
3.Descriptive statisticsMedium
152025303540455055

Find the sample standard deviation s of the data: 42, 47, 11, 50, 16, 11, 57.

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

Exam scores are normally distributed with mean 70 and standard deviation 20. Find the z-score z of a score of 54.

μ = 70σ = 20x = 54
z =
5.ProbabilityMedium
xp(x)10.120.2530.3540.3μ = 2.85

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

x: 1, 2, 3, 4p(x): 0.1, 0.25, 0.35, 0.3
σ =
6.ProbabilityHard
n = 12, p = 0.4: P(X = 4) = ?k0123456789101112

A test has 12 multiple-choice questions. A student who guesses gets each one right with probability 0.4, independently. What is the probability P of exactly 4 correct answers?

n = 12p = 0.4k = 4
P =
7.Normal distributionMedium
510 kg554.1 kgT = X₁ + … + X₆?

The masses of passengers with luggage are independent and normal with mean 85 kg and standard deviation 12 kg. Find the probability P that the total mass of 6 of them is below 554.1 kg. Round z to two decimals.

μ = 85 kgσ = 12 kgn = 6< 554.1 kg
P =
8.Normal distributionHard
μ = 500x = ?97.5%

The battery lifetimes are normally distributed with mean 500 h and standard deviation 40 h. Find the value x below which 97.5% of them lie (the 97.5th percentile).

μ = 500σ = 40P = 97.5%
x = h
9.Confidence intervalsMedium
x̄E₁ = 1199%, n = 25x̄E₂ = ?99%, n = 50

A 99% confidence interval for a mean, based on 25 observations and a known σ, has a margin of error of 11. What is the margin of error E₂ if the same study uses 50 observations instead?

99%n₁ = 25E₁ = 11n₂ = 50
E₂ =
10.Confidence intervalsHard
σ = 35, n = ?x̄ − Ex̄ + Ex̄E = 1.590%

The population standard deviation is 35. How large a sample n do you need so that a 90% confidence interval for the mean has a margin of error of at most 1.5?

σ = 35E = 1.590%
n =
11.Hypothesis tests & confidence intervalsMedium
2.12p = ?

The mean of a population is claimed to be 500; the population standard deviation is known to be 40. A random sample of 100 observations has a mean of 508.48. Test H₀: μ = 500 against H₁: μ > 500. Find the p-value p.

μ₀ = 500σ = 40n = 100x̄ = 508.48H₁: μ > μ₀
p =
12.Hypothesis tests & confidence intervalsHard
−1.961.96z = ?

In a trial, 36 of 80 patients on treatment A recover, and 78 of 250 patients on treatment B. Using the pooled proportion, find the test statistic z for H₀: p₁ = p₂.

x₁ = 36n₁ = 80x₂ = 78n₂ = 250
z =
13.Linear regression & correlationMedium
xy2104186288361042xy51010203040

The table gives 5 paired observations of x = load in kN and y = extension in mm. Find the coefficient of determination r² of the least-squares line.

n = 5
r² =
14.Linear regression & correlationHard
xy274176218271038xy51010203040x = 9

The table gives 5 paired observations of x = load in kN and y = extension in mm. Use the least-squares line ŷ = a + b·x. A new observation (9, 31) is made. Find its residual e = y − ŷ.

n = 5(9, 31)
e =
15.Chi-square testsMedium
improvednot improvedTotalnew drug38240placebo6040100Total9842140

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 = 1402 × 2
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 checks whether counts in 5 categories follow a Poisson distribution whose mean is estimated from the same data. Using the table, find the critical value χ²crit at significance level α = 0.01.

α = 0.01
χ²crit =

3Solutions

1.Research methods in psychology

Answer: B · Cohort effects

Why this is the answer

Different age groups were raised in different historical conditions, so differences may reflect generation (cohort) rather than ageing itself.

Answer

B · Cohort effects
2.Research methods in psychology

Answer: D · As a manipulation check to confirm the IV worked as intended

Why this is the answer

A manipulation check verifies that the IV produced the intended psychological state. Without it, a null result could mean the manipulation failed.

Answer

D · As a manipulation check to confirm the IV worked as intended
3.Descriptive statistics

Answer: s = 19.99

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 = 234/7 = 33.43

Squared deviations

Σ(x − x̄)² = 8.571² + 13.57² + (−22.43)² + 16.57² + (−17.43)² + (−22.43)² + 23.57² = 2398

Standard deviation

s = √(2398/6) = 19.99

Answer

s = 19.99
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 = (54 − 70)/20 = −0.8

Answer

z = −0.8
5.Probability

Answer: σ = 0.9631

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 = 1·0.1 + 2·0.25 + 3·0.35 + 4·0.3 = 2.85

Mean of the squares

E(X²) = 1·0.1 + 4·0.25 + 9·0.35 + 16·0.3 = 9.05

Variance

Var(X) = 9.05 − 2.85² = 0.9275
σ = √0.9275 = 0.9631

Answer

σ = 0.9631
6.Probability

Answer: P = 0.2128

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 = 4) = 495·0.44·0.68 = 495·0.0256·0.0168 = 0.2128

Answer

P = 0.2128
7.Normal distribution

Answer: P = 0.9332

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 = 510 kg, σT = 12·√6 = 29.39 kg

Standardise

z = (554.1 − 510)/29.39 = 1.5

Probability

P = Φ(1.5) = 0.9332

Answer

P = 0.9332
8.Normal distribution

Answer: x = 578.4 h

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 = 500 + 1.96·40 = 578.4 h

Answer

x = 578.4 h
9.Confidence intervals

Answer: E₂ = 7.778

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.

Ratio of margins

E₂/E₁ = (2.576/2.576)·√(25/50) = 0.7071

New margin

E₂ = 11·0.7071 = 7.778

Answer

E₂ = 7.778
10.Confidence intervals

Answer: n = 1474

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 = (1.645·35/1.5)² = 1473

Round up

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

n = 1474

Answer

n = 1474
11.Hypothesis tests & confidence intervals

Answer: p = 0.017

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 = 40/√100 = 4
z = (508.48 − 500)/4 = 2.12

Tail area

Φ(2.12) = 0.9830
1 − Φ(2.12) = 0.0170

p-value

p = 0.017

Answer

p = 0.017

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

12.Hypothesis tests & confidence intervals

Answer: z = 2.259

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 = 36/80 = 0.45
p̂₂ = 78/250 = 0.312
p̂ = 114/330 = 0.3455

Standard error under H₀

SE = √(0.3455·0.6545·(1/80 + 1/250)) = 0.06108

Test statistic

z = (0.45 − 0.312)/0.06108 = 2.259

Answer

z = 2.259

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

13.Linear regression & correlation

Answer: r² = 0.9935

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 = 134/5 = 26.8

Sums of squares and products

Sxx = Σ(x − x̄)² = 40
Sxy = Σ(x − x̄)(y − ȳ) = 164
Syy = Σ(y − ȳ)² = 676.8

Coefficient of determination

r² = 164²/(40·676.8) = 0.9935

Answer

r² = 0.9935

About 99.35% of the variation in y is explained by the line.

14.Linear regression & correlation

Answer: e = −1.8

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 = 110/5 = 22

Sums of squares and products

Sxx = Σ(x − x̄)² = 40
Sxy = Σ(x − x̄)(y − ȳ) = 144

Regression line

b = 144/40 = 3.6
a = 22 − 3.6·6 = 0.4
ŷ = 0.4 + 3.6·x

Predicted value

ŷ = 0.4 + 3.6·9 = 32.8

Residual

e = 31 − 32.8 = −1.8

Answer

e = −1.8

The point lies below the line.

15.Chi-square tests

Answer: χ² = 16.67

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

new drug: 40·98/140 = 28, 40·42/140 = 12
placebo: 100·98/140 = 70, 100·42/140 = 30

Chi-square statistic

χ² = (38 − 28)²/28 + (2 − 12)²/12 + (60 − 70)²/70 + (40 − 30)²/30
χ² = 16.67

Answer

χ² = 16.67

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

16.Chi-square tests

Answer: χ²crit = 11.35

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 = 5 − 1 − 1 = 3

Table

χ²crit(df = 3, α = 0.01) = 11.345

Answer

χ²crit = 11.345

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

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