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)·p^k(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

Convert the value to a z-score: z = (x − μ)/σ.
Look up Φ(z), the area to the left of z.
For the area to the right use 1 − Φ(z); for an interval subtract the two left areas.
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

Decide between z (population σ known) and t (only the sample s, with n − 1 degrees of freedom).
Find the critical value for the confidence level.
Margin of error: E = critical value × standard deviation/√n.
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

State H₀ and H₁ and decide whether the test is one- or two-sided.
Calculate the test statistic: (x̄ − μ₀)/(σ/√n), or with s and t.
Compare with the critical value, or find the p-value.
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: S_xx = Σ(x − x̄)² = Σx² − (Σx)²/n, S_xy = Σxy − ΣxΣy/n
Slope: b = S_xy/S_xx
Intercept: a = ȳ − b·x̄
Correlation: r = S_xy/√(S_xxS_yy)
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 > χ²_crit

2Practice problems

1.Research methods in psychologyMedium

Which design uses different participants in each condition?

AIndependent groups design
BMatched pairs design with the same people
CWithin-subjects design
DRepeated measures design

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?

AAn interaction between drug and age
BA floor effect
CA Type I error
DOnly a main effect of age

3.Descriptive statisticsMedium

Find the sample standard deviation s of the data: 19, 34, 16, 42, 11.

n = 5

s =

4.Descriptive statisticsHard

Exam scores are normally distributed with mean 500 and standard deviation 60. Find the z-score z of a score of 536.

μ = 500σ = 60x = 536

z =

5.ProbabilityMedium

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

A test has 11 multiple-choice questions. A student who guesses gets each one right with probability 0.5, independently. What is the probability P of exactly 1 correct answers?

n = 11p = 0.5k = 1

P =

7.Normal distributionMedium

The battery lifetimes are normally distributed with mean 500 h and standard deviation 40 h. What fraction lies between 500 and 548 h? Give P as a decimal.

μ = 500σ = 40x₁ = 500x₂ = 548

P =

8.Normal distributionHard

The IQ scores are normally distributed with mean 100 points and standard deviation 15 points. Drag the cut-off so the shaded bottom 15% is to the left of it. Find the cut-off x.

μ = 100 pointsσ = 15 points15%

x = points

9.Confidence intervalsMedium

A sample of 9 items has mean 119 and standard deviation s = 12. The population is roughly normal. Find the lower limit of the 90% confidence interval for the mean.

n = 9x̄ = 119s = 1290%t(8) = 1.86

lower limit =

10.Confidence intervalsHard

The population standard deviation is 21. How large a sample n do you need so that a 99% confidence interval for the mean has a margin of error of at most 2?

σ = 21E = 299%

n =

11.Hypothesis tests & confidence intervalsMedium

A sample of 49 students has a mean score of 78 and a sample standard deviation of 10. Find the test statistic t for H₀: μ = 75.

μ₀ = 75x̄ = 78s = 10n = 49

t =

12.Hypothesis tests & confidence intervalsHard

How large must a sample be so that the 95% confidence interval for a mean has a margin of error of at most 2.5? The standard deviation is 12. Find the smallest sample size n.

σ = 12E = 2.595%

n =

13.Linear regression & correlationMedium

The table gives 5 paired observations of x = car age in years and y = price in k€. Find Pearson's correlation coefficient r.

n = 5

r =

14.Linear regression & correlationHard

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. For which value x does the line predict ŷ = 19?

n = 5ŷ = 19

x =

15.Chi-square testsMedium

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 = 1802 × 3

χ² =

16.Chi-square testsHard

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 = 2603 × 2α = 0.01

χ² =

3Solutions

1.Research methods in psychology

Answer: A · Independent groups design

Why this is the answer

In an independent groups design each condition has different participants, avoiding order effects but risking participant variables.

Answer

A · Independent groups design

2.Research methods in psychology

Answer: A · 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

A · An interaction between drug and age

3.Descriptive statistics

Answer: s = 13.05

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 = 122/5 = 24.4

Squared deviations

Σ(x − x̄)² = (−5.4)² + 9.6² + (−8.4)² + 17.6² + (−13.4)² = 681.2

Standard deviation

s = √(681.2/4) = 13.05

Answer

s = 13.05

4.Descriptive statistics

Answer: z = 0.6

Standardise

z = (536 − 500)/60 = 0.6

Answer

z = 0.6

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: P = 0.005371

Binomial formula

P(X = 1) = 11·0.5¹·0.5¹⁰ = 11·0.5·0.0009766 = 0.005371

Answer

P = 0.005371

7.Normal distribution

Answer: P = 0.3849

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-scores

z₁ = (500 − 500)/40 = 0

z₂ = (548 − 500)/40 = 1.2

Area between

P = Φ(1.2) − Φ(0) = 0.8849 − 0.5 = 0.3849

Answer

P = 0.3849

8.Normal distribution

Answer: x = 84.4 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) = 0.15

z from the table

z ≈ −1.04

Back to the original units

x = 100 − 1.04·15 = 84.4 points

Answer

x = 84.4 points

9.Confidence intervals

Answer: lower limit = 111.6

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 = 8, t = 1.86

Margin of error

E = 1.86·12/√9 = 7.44

Interval

119 ± 7.44 → [111.6, 126.4]

Answer

lower limit = 111.6

10.Confidence intervals

Answer: n = 732

Solve for n

n = (2.576·21/2)² = 731.6

Round up

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

n = 732

Answer

n = 732

11.Hypothesis tests & confidence intervals

Answer: t = 2.1

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/√49 = 10/7 = 1.429

Test statistic

t = (78 − 75)/1.429 = 2.1

Answer

t = 2.1

t_{0.025; df = 48} = 2.011

|t| > 2.011, so H₀ is rejected at the 5% level (two-sided).

12.Hypothesis tests & confidence intervals

Answer: n = 89

Solve for n

n = (1.96·12/2.5)² = 88.51

Round up

n = 89

Answer

n = 89

13.Linear regression & correlation

Answer: r = −0.9847

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 = 93/5 = 18.6

Sums of squares and products

S_xx = Σ(x − x̄)² = 10

S_xy = Σ(x − x̄)(y − ȳ) = −37

S_yy = Σ(y − ȳ)² = 141.2

Correlation coefficient

r = −37/√(10·141.2) = −0.9847

Answer

r = −0.9847

14.Linear regression & correlation

Answer: x = 3.6

Means

x̄ = 15/5 = 3

ȳ = 98/5 = 19.6

Sums of squares and products

S_xx = Σ(x − x̄)² = 10

S_xy = Σ(x − x̄)(y − ȳ) = −10

Regression line

b = −10/10 = −1

a = 19.6 − (−1)·3 = 22.6

ŷ = 22.6 − 1·x

Solve for x

x = (19 − 22.6)/(−1) = 3.6

Answer

x = 3.6

15.Chi-square tests

Answer: χ² = 8

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

students: 120·27/180 = 18, 120·72/180 = 48, 120·81/180 = 54

staff: 60·27/180 = 9, 60·72/180 = 24, 60·81/180 = 27

Chi-square statistic

χ² = (12 − 18)²/18 + (48 − 48)²/48 + (60 − 54)²/54 + (15 − 9)²/9 + (24 − 24)²/24 + (21 − 27)²/27

χ² = 8

Answer

χ² = 8

df = 2. At α = 0.05 the critical value is 5.991, so H₀ (independence) is rejected.

16.Chi-square tests

Answer: χ² = 12.11

Expected counts

plant A: 50, 50

plant B: 50, 50

plant C: 30, 30

Chi-square statistic

χ² = (40 − 50)²/50 + (60 − 50)²/50 + (49 − 50)²/50 + (51 − 50)²/50 + (41 − 30)²/30 + (19 − 30)²/30

χ² = 12.11

Decision

df = 2, χ²_crit = 9.210

12.11 > 9.210

Answer

χ² = 12.11

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

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