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

The standard error of the mean becomes smaller when:

  1. AThe sample size increases
  2. BThe mean increases
  3. CThe sample size decreases
  4. DThe variable is measured at ordinal level
2.Research methods in psychologyHard

Two observers record aggressive acts in a playground and their counts correlate at r = 0.45. What should the researcher do?

  1. AAccept the data because any positive correlation is enough
  2. BUse only the higher of the two counts
  3. CSwitch to a one-tailed hypothesis
  4. DRefine the behaviour categories and train observers, then recheck agreement
3.Descriptive statisticsMedium
3540455055

Find the sample standard deviation s of the data: 56, 31, 45, 49, 41.

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

Exam scores are normally distributed with mean 50 and standard deviation 20. Find the z-score z of a score of 20.

μ = 50σ = 20x = 20
z =
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
€p(x)−100.75200.24900.05

A game costs €10 to play. You win €30 with probability 0.2 and €500 with probability 0.05; otherwise you win nothing. The stake is not returned. Find the expected net gain E(G) per game.

stake = €10€30: p = 0.2€500: p = 0.05
E(G) = €
7.Normal distributionMedium
−3−2−10123zshaded area between −z and z = 0.8098

Drag z so that the shaded area between −z and z under the standard normal curve is 0.8098. Find z to two decimals.

A = 0.8098
z =
8.Normal distributionHard
380420460500540580620htop 85%

The battery lifetimes are normally distributed with mean 500 h and standard deviation 40 h. Drag the cut-off so the shaded top 85% is to the right of it. Find the cut-off x.

μ = 500 hσ = 40 h85%
x = h
9.Confidence intervalsMedium
μ₁ − μ₂0LUx̄₁ − x̄₂E = ?95%

The battery life of two phone models is compared. Model 1: 30 phones, mean 98 h, known standard deviation 8 h. Model 2: 100 phones, mean 91 h, known standard deviation 9 h. Find the margin of error E of the 95% confidence interval for μ₁ − μ₂.

n₁ = 30x̄₁ = 98 hσ₁ = 8 hn₂ = 100x̄₂ = 91 hσ₂ = 9 h95%
E = h
10.Confidence intervalsHard
μ₁ − μ₂, t(24) = 2.7970L = ?Ux̄₁ − x̄₂E99%

Two fertilisers are tested. Plot group 1: 5 plots, mean yield 13 kg, standard deviation 10 kg. Plot group 2: 21 plots, mean 5 kg, standard deviation 7 kg. Assume equal variances (pooled t, df = n₁ + n₂ − 2). Find the lower limit of the 99% confidence interval for μ₁ − μ₂.

n₁ = 5x̄₁ = 13s₁ = 10n₂ = 21x̄₂ = 5s₂ = 799%t(24) = 2.797
lower limit = kg
11.Hypothesis tests & confidence intervalsMedium
2.04p = ?

The mean of a population is claimed to be 120; the population standard deviation is known to be 16. A random sample of 64 observations has a mean of 124.08. Test H₀: μ = 120 against H₁: μ > 120. Find the p-value p.

μ₀ = 120σ = 16n = 64x̄ = 124.08H₁: μ > μ₀
p =
12.Hypothesis tests & confidence intervalsHard
μ₀x̄c

A one-sided z test of H₀: μ = 50 against H₁: μ > 50 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 57. Find the power 1 − β of the test in percent.

μ₀ = 50μ = 57σ = 12.5n = 25α = 0.01
1 − β = %
13.Linear regression & correlationMedium
xy2841161481910281233xy510102030

The table gives 6 paired observations of x = load in kN and y = extension in mm. Find Pearson's correlation coefficient r.

n = 6
r =
14.Linear regression & correlationHard
xy12122031641151067xy246101520ŷ = 8

The table gives 6 paired observations of x = car age in years and y = price in k€. Use the least-squares line ŷ = a + b·x. For which value x does the line predict ŷ = 8?

n = 6ŷ = 8
x =
15.Chi-square testsMedium
improvednot improvedTotalnew drug5545100placebo491160Total10456160

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 test of independence is carried out on a contingency table with 3 rows and 6 columns (not counting the totals). How many degrees of freedom df does the test have?

3 × 6
df =

3Solutions

1.Research methods in psychology

Answer: A · The sample size increases

Why this is the answer

The standard error estimates how much sample means vary around the population mean. Larger samples give more precise estimates.

Answer

A · The sample size increases
2.Research methods in psychology

Answer: D · Refine the behaviour categories and train observers, then recheck agreement

Why this is the answer

Low inter-observer reliability means the categories are applied inconsistently. Clearer operational definitions and practice sessions should raise agreement before main data collection.

Answer

D · Refine the behaviour categories and train observers, then recheck agreement
3.Descriptive statistics

Answer: s = 9.317

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 = 222/5 = 44.4

Squared deviations

Σ(x − x̄)² = 11.6² + (−13.4)² + 0.6² + 4.6² + (−3.4)² = 347.2

Standard deviation

s = √(347.2/4) = 9.317

Answer

s = 9.317
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 = (20 − 50)/20 = −1.5

Answer

z = −1.5
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: E(G) = 21 €

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), 490 (p = 0.05), −10 (p = 0.75)

Expected value

E(G) = 20·0.2 + 490·0.05 − 10·0.75 = 21 €

Answer

E(G) = 21 €

A positive value means you win money on average.

7.Normal distribution

Answer: z = 1.31

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 of z

Φ(z) = (1 + 0.8098)/2 = 0.9049

Read the z-table backwards

Find 0.9049 inside the table and read off z.

z = 1.31

Answer

z = 1.31
8.Normal distribution

Answer: x = 458 h

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) = 0.15

z from the table

z ≈ −1.04

Back to the original units

x = 500 − 1.04·40 = 458 h

Answer

x = 458 h
9.Confidence intervals

Answer: E = 3.363 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 = 98 − 91 = 7

Standard error

SE = √(8²/30 + 9²/100) = 1.716

Margin of error

E = 1.96·1.716 = 3.363

Answer

E = 3.363 h

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

10.Confidence intervals

Answer: lower limit = −2.554 kg

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.

Pooled standard deviation

sp² = (4·10² + 20·7²)/24 = 57.5
sp = 7.583

Margin of error

E = 2.797·7.583·√(1/5 + 1/21) = 10.55

Interval

8 ± 10.55 → [−2.554, 18.55]

Answer

lower limit = −2.554 kg
11.Hypothesis tests & confidence intervals

Answer: p = 0.02068

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 = 16/√64 = 2
z = (124.08 − 120)/2 = 2.04

Tail area

Φ(2.04) = 0.9793
1 − Φ(2.04) = 0.0207

p-value

p = 0.02068

Answer

p = 0.02068

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

12.Hypothesis tests & confidence intervals

Answer: 1 − β = 68.23 %

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 = 50 + 2.326·2.5 = 55.81

z under the true mean

z = (55.81 − 57)/2.5 = −0.474

Power

power = P(Z > −0.474) = 68.23%
1 − β = 68.23%

Answer

1 − β = 68.23%
13.Linear regression & correlation

Answer: r = 0.9805

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 = 113/6 = 18.83

Sums of squares and products

Sxx = Σ(x − x̄)² = 70
Sxy = Σ(x − x̄)(y − ȳ) = 181
Syy = Σ(y − ȳ)² = 486.8

Correlation coefficient

r = 181/√(70·486.8) = 0.9805

Answer

r = 0.9805
14.Linear regression & correlation

Answer: x = 5.556

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 = 85/6 = 14.17

Sums of squares and products

Sxx = Σ(x − x̄)² = 17.5
Sxy = Σ(x − x̄)(y − ȳ) = −52.5

Regression line

b = −52.5/17.5 = −3
a = 14.17 − (−3)·3.5 = 24.67
ŷ = 24.67 − 3·x

Solve for x

x = (8 − 24.67)/(−3) = 5.556

Answer

x = 5.556
15.Chi-square tests

Answer: χ² = 11.72

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: 100·104/160 = 65, 100·56/160 = 35
placebo: 60·104/160 = 39, 60·56/160 = 21

Chi-square statistic

χ² = (55 − 65)²/65 + (45 − 35)²/35 + (49 − 39)²/39 + (11 − 21)²/21
χ² = 11.72

Answer

χ² = 11.72

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

16.Chi-square tests

Answer: df = 10

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 = (3 − 1)·(6 − 1) = 10

Answer

df = 10

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