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 repeated measures design, order effects can be controlled by:

  1. ARandom sampling
  2. BSingle-blind procedures
  3. CUsing a larger sample only
  4. DCounterbalancing
2.Research methods in psychologyHard

A 2 × 3 factorial ANOVA shows that a drug improves memory in young adults but not in older adults. What has been found?

  1. AA floor effect
  2. BA Type I error
  3. COnly a main effect of age
  4. DAn interaction between drug and age
3.Descriptive statisticsMedium
1520253035404550

Find the sample standard deviation s of the data: 39, 28, 51, 11, 45.

n = 5
s =
4.Descriptive statisticsHard
z −2−1012

Exam scores are normally distributed with mean 100 and standard deviation 8. Find the z-score z of a score of 88.

μ = 100σ = 8x = 88
z =
5.ProbabilityMedium
xp(x)00.2510.3520.0530.35μ = 1.5

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

x: 0, 1, 2, 3p(x): 0.25, 0.35, 0.05, 0.35
σ =
6.ProbabilityHard
n = 8, p = 0.3: P(X ≤ 1) = ?k012345678

8 components are tested, and each one fails with probability 0.3, independently. What is the probability P that at most one fails?

n = 8p = 0.3
P =
7.Normal distributionMedium
300 kg303.5 kgT = X₁ + … + X₁₂?

The masses of bags of cement are independent and normal with mean 25 kg and standard deviation 0.5 kg. Find the probability P that the total mass of 12 of them is below 303.5 kg. Round z to two decimals.

μ = 25 kgσ = 0.5 kgn = 12< 303.5 kg
P =
8.Normal distributionHard
μ = 500x = ?10%

The battery lifetimes are normally distributed with mean 500 h and standard deviation 40 h. Find the value x below which 10% of them lie (the 10th percentile).

μ = 500σ = 40P = 10%
x = h
9.Confidence intervalsMedium
μ₁ − μ₂0LUx̄₁ − x̄₂E = ?99%

The battery life of two phone models is compared. Model 1: 50 phones, mean 43 h, known standard deviation 14 h. Model 2: 100 phones, mean 31.5 h, known standard deviation 14 h. Find the margin of error E of the 99% confidence interval for μ₁ − μ₂.

n₁ = 50x̄₁ = 43 hσ₁ = 14 hn₂ = 100x̄₂ = 31.5 hσ₂ = 14 h99%
E = h
10.Confidence intervalsHard
σ = 8, n = ?x̄ − Ex̄ + Ex̄E = 290%

The population standard deviation is 8. 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 2?

σ = 8E = 290%
n =
11.Hypothesis tests & confidence intervalsMedium
−2.0452.045t = ?df = 29

Two teaching methods are compared. Group 1: 15 students, mean score 107, standard deviation 7. Group 2: 16 students, mean score 110.6, standard deviation 3. Assume equal population variances and use the pooled two-sample t test. Find the test statistic t for H₀: μ₁ = μ₂, taking the difference as x̄₁ − x̄₂ (group 1 minus group 2).

n₁ = 15x̄₁ = 107s₁ = 7n₂ = 16x̄₂ = 110.6s₂ = 3
t =
12.Hypothesis tests & confidence intervalsHard
−1.961.96z = ?

In a trial, 24 of 80 patients on treatment A recover, and 80 of 200 patients on treatment B. Using the pooled proportion, find the test statistic z for H₀: p₁ = p₂.

x₁ = 24n₁ = 80x₂ = 80n₂ = 200
z =
13.Linear regression & correlationMedium
xy3485647669781177xy51050607080

The table gives 5 paired observations of x = hours studied and y = exam score. Their sums are given. Find Pearson's correlation coefficient r from these sums.

n = 5Σx = 35Σy = 333Σx² = 285Σy² = 22769Σxy = 2475
r =
14.Linear regression & correlationHard
xy244355458566667773xy246506070ŷ = 55

The table gives 6 paired observations of x = hours studied and y = exam score. Use the least-squares line ŷ = a + b·x. For which value x does the line predict ŷ = 55?

n = 6ŷ = 55
x =
15.Chi-square testsMedium
yesnoTotalmen184260women303060Total4872120

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 = 1202 × 2
χ² =
16.Chi-square testsHard
passfailTotalage 18–306040100age 31–50374380age 51+991100Total19684280dfα = 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 = 2803 × 2α = 0.01
χ² =

3Solutions

1.Research methods in psychology

Answer: D · Counterbalancing

Why this is the answer

Counterbalancing distributes practice and fatigue effects evenly across conditions.

Answer

D · Counterbalancing
2.Research methods in psychology

Answer: D · An interaction between drug and age

Why this is the answer

An interaction occurs when the effect of one factor differs across levels of another. Here the drug effect depends on age group.

Answer

D · An interaction between drug and age
3.Descriptive statistics

Answer: s = 15.79

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 = 174/5 = 34.8

Squared deviations

Σ(x − x̄)² = 4.2² + (−6.8)² + 16.2² + (−23.8)² + 10.2² = 996.8

Standard deviation

s = √(996.8/4) = 15.79

Answer

s = 15.79
4.Descriptive statistics

Answer: z = −1.5

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 = (88 − 100)/8 = −1.5

Answer

z = −1.5
5.Probability

Answer: σ = 1.204

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 + 1·0.35 + 2·0.05 + 3·0.35 = 1.5

Mean of the squares

E(X²) = 0·0.25 + 1·0.35 + 4·0.05 + 9·0.35 = 3.7

Variance

Var(X) = 3.7 − 1.5² = 1.45
σ = √1.45 = 1.204

Answer

σ = 1.204
6.Probability

Answer: P = 0.2553

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.

Zero and one failure

P(X = 0) = 0.78 = 0.05765
P(X = 1) = 8·0.3·0.77 = 0.1977

Add them

P = 0.05765 + 0.1977 = 0.2553

Answer

P = 0.2553
7.Normal distribution

Answer: P = 0.9783

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 = 300 kg, σT = 0.5·√12 = 1.732 kg

Standardise

z = (303.5 − 300)/1.732 = 2.02

Probability

P = Φ(2.02) = 0.9783

Answer

P = 0.9783
8.Normal distribution

Answer: x = 448.7 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) = 0.1 → z = −1.282

Back to x

x = 500 + −1.282·40 = 448.74 h

Answer

x = 448.74 h
9.Confidence intervals

Answer: E = 6.246 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 = 43 − 31.5 = 11.5

Standard error

SE = √(14²/50 + 14²/100) = 2.425

Margin of error

E = 2.576·2.425 = 6.246

Answer

E = 6.246 h

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

10.Confidence intervals

Answer: n = 44

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 = (1.645·8/2)² = 43.3

Round up

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

n = 44

Answer

n = 44
11.Hypothesis tests & confidence intervals

Answer: t = −1.883

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.

Pooled variance

sp² = (14·7² + 15·3²)/29 = 28.31
sp = 5.321

Standard error

SE = 5.321·√(1/15 + 1/16) = 1.912

Test statistic

t = (107 − 110.6)/1.912 = −1.883

Answer

t = −1.883

With df = 29, the two-sided 5% critical value is 2.045; |t| is smaller, so H₀ is not rejected.

12.Hypothesis tests & confidence intervals

Answer: z = −1.564

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 = 24/80 = 0.3
p̂₂ = 80/200 = 0.4
p̂ = 104/280 = 0.3714

Standard error under H₀

SE = √(0.3714·0.6286·(1/80 + 1/200)) = 0.06392

Test statistic

z = (0.3 − 0.4)/0.06392 = −1.564

Answer

z = −1.564

|z| < 1.96, so H₀ is not rejected at the 5% level (two-sided).

13.Linear regression & correlation

Answer: r = 0.9364

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 = 2475 − 35·333/5 = 144
Sxx = 285 − 35²/5 = 40
Syy = 22769 − 333²/5 = 591.2

Correlation coefficient

r = 144/√(40·591.2) = 0.9364

Answer

r = 0.9364
14.Linear regression & correlation

Answer: x = 3.481

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 = 27/6 = 4.5
ȳ = 363/6 = 60.5

Sums of squares and products

Sxx = Σ(x − x̄)² = 17.5
Sxy = Σ(x − x̄)(y − ȳ) = 94.5

Regression line

b = 94.5/17.5 = 5.4
a = 60.5 − 5.4·4.5 = 36.2
ŷ = 36.2 + 5.4·x

Solve for x

x = (55 − 36.2)/5.4 = 3.481

Answer

x = 3.481
15.Chi-square tests

Answer: χ² = 5

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: 60·48/120 = 24, 60·72/120 = 36
women: 60·48/120 = 24, 60·72/120 = 36

Chi-square statistic

χ² = (18 − 24)²/24 + (42 − 36)²/36 + (30 − 24)²/24 + (30 − 36)²/36
χ² = 5

Answer

χ² = 5

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

16.Chi-square tests

Answer: χ² = 66.3

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: 70, 30
age 31–50: 56, 24
age 51+: 70, 30

Chi-square statistic

χ² = (60 − 70)²/70 + (40 − 30)²/30 + (37 − 56)²/56 + (43 − 24)²/24 + (99 − 70)²/70 + (1 − 30)²/30
χ² = 66.3

Decision

df = 2, χ²crit = 9.210
66.3 > 9.210

Answer

χ² = 66.3

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

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