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 normal distribution, approximately what percentage of scores lie within one standard deviation of the mean?

  1. A99.7%
  2. B95%
  3. C50%
  4. D68%
2.Research methods in psychologyHard

A psychology department recruits participants only from introductory psychology courses. Critics say samples are "WEIRD". What does this refer to?

  1. AWestern, Educated, Industrialised, Rich and Democratic samples that may not generalise
  2. BSamples recruited using weighted random methods
  3. CSamples that withdraw early from studies
  4. DSamples that are unusually strange in their behaviour
3.Descriptive statisticsMedium
3035404550

Find the sample standard deviation s of the data: 40, 45, 35, 26, 26, 52.

n = 6
s =
4.Descriptive statisticsHard
z −2−1012

Exam scores are normally distributed with mean 100 and standard deviation 10. Find the z-score z of a score of 104.

μ = 100σ = 10x = 104
z =
5.ProbabilityMedium
xp(x)00.2510.220.230.35μ = 1.65

For the distribution shown, find the variance Var(X) of X.

x: 0, 1, 2, 3p(x): 0.25, 0.2, 0.2, 0.35
Var(X) =
6.ProbabilityHard
n = 9, p = 0.3: P(X ≤ 1) = ?k0123456789

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

n = 9p = 0.3
P =
7.Normal distributionMedium
120 kg122.5 kgT = X₁ + … + X₁₀?

The masses of boxes are independent and normal with mean 12 kg and standard deviation 0.8 kg. Find the probability P that the total mass of 10 of them is below 122.5 kg. Round z to two decimals.

μ = 12 kgσ = 0.8 kgn = 10< 122.5 kg
P =
8.Normal distributionHard
μ = 250x = ?99%

The bag weights are normally distributed with mean 250 g and standard deviation 5 g. Find the value x below which 99% of them lie (the 99th percentile).

μ = 250σ = 5P = 99%
x = g
9.Confidence intervalsMedium
n = 6, sd​ = 7.40L = ?Ud̄ = 5.5E95%

The blood pressure of 6 patients is measured before and after a treatment. The differences d = before − after (mmHg) have mean 5.5 and standard deviation 7.4. Find the lower limit of the 95% confidence interval for the mean difference.

n = 6d̄ = 5.5sd = 7.495%t(5) = 2.571
lower limit = mmHg
10.Confidence intervalsHard
σ = 31, n = ?x̄ − Ex̄ + Ex̄E = 0.599%

The population standard deviation is 31. 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 0.5?

σ = 31E = 0.599%
n =
11.Hypothesis tests & confidence intervalsMedium
t −2.06402.064

A sample of 25 students has a mean score of 51.68 and a sample standard deviation of 12. Find the test statistic t for H₀: μ = 50.

μ₀ = 50x̄ = 51.68s = 12n = 25
t =
12.Hypothesis tests & confidence intervalsHard
z −2.57602.576

A sample of 25 measurements has mean 117.8. The population standard deviation is 8. Find the upper limit U of the 99% confidence interval for the mean.

x̄ = 117.8σ = 8n = 2599%
U =
13.Linear regression & correlationMedium
xy234561281110151219xy5105101520

The table gives 6 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 = 6Σx = 42Σy = 65Σx² = 364Σy² = 885Σxy = 564
r =
14.Linear regression & correlationHard
xy264861081110161219xy5105101520ŷ = 17

The table gives 6 paired observations of x = load in kN and y = extension in mm. Use the least-squares line ŷ = a + b·x. For which value x does the line predict ŷ = 17?

n = 6ŷ = 17
x =
15.Chi-square testsMedium
busbikecarTotalstudents34101256staff12402678Total465038134

The contingency table shows observed counts. For a chi-square test of independence, find the expected count E for the cell “students, bus”.

row total = 56column total = 46N = 134
E =
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: D · 68%

Why this is the answer

About 68% of scores fall within ±1 SD, about 95% within ±2 SD and about 99.7% within ±3 SD.

Answer

D · 68%
2.Research methods in psychology

Answer: A · Western, Educated, Industrialised, Rich and Democratic samples that may not generalise

Why this is the answer

Henrich and colleagues noted that most psychology participants come from WEIRD societies, which are unrepresentative of humanity, threatening population validity.

Answer

A · Western, Educated, Industrialised, Rich and Democratic samples that may not generalise
3.Descriptive statistics

Answer: s = 10.42

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 = 224/6 = 37.33

Squared deviations

Σ(x − x̄)² = 2.667² + 7.667² + (−2.333)² + (−11.33)² + (−11.33)² + 14.67² = 543.3

Standard deviation

s = √(543.3/5) = 10.42

Answer

s = 10.42
4.Descriptive statistics

Answer: z = 0.4

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 = (104 − 100)/10 = 0.4

Answer

z = 0.4
5.Probability

Answer: Var(X) = 1.428

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 + 1·0.2 + 2·0.2 + 3·0.35 = 1.65

Mean of the squares

E(X²) = 0·0.25 + 1·0.2 + 4·0.2 + 9·0.35 = 4.15

Variance

Var(X) = 4.15 − 1.65² = 1.428

Answer

Var(X) = 1.428
6.Probability

Answer: P = 0.196

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) = 0.79 = 0.04035
P(X = 1) = 9·0.3·0.78 = 0.1556

Add them

P = 0.04035 + 0.1556 = 0.196

Answer

P = 0.196
7.Normal distribution

Answer: P = 0.8389

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 = 120 kg, σT = 0.8·√10 = 2.53 kg

Standardise

z = (122.5 − 120)/2.53 = 0.99

Probability

P = Φ(0.99) = 0.8389

Answer

P = 0.8389
8.Normal distribution

Answer: x = 261.6 g

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) = 0.99 → z = 2.326

Back to x

x = 250 + 2.326·5 = 261.63 g

Answer

x = 261.63 g
9.Confidence intervals

Answer: lower limit = −2.267 mmHg

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.

Critical value

df = 5, t = 2.571

Margin of error

E = 2.571·7.4/√6 = 7.767

Interval

5.5 ± 7.767 → [−2.267, 13.27]

Answer

lower limit = −2.267 mmHg
10.Confidence intervals

Answer: n = 25508

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 = (2.576·31/0.5)² = 25508

Round up

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

n = 25508

Answer

n = 25508
11.Hypothesis tests & confidence intervals

Answer: t = 0.7

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 = 12/√25 = 12/5 = 2.4

Test statistic

t = (51.68 − 50)/2.4 = 0.7

Answer

t = 0.7
t0.025; df = 24 = 2.064

|t| < 2.064, so H₀ is not rejected at the 5% level (two-sided).

12.Hypothesis tests & confidence intervals

Answer: U = 121.9

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 = 2.576·8/√25 = 4.122

Interval limit

U = 117.8 + 4.122 = 121.9

Answer

U = 121.9
13.Linear regression & correlation

Answer: r = 0.9688

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 = 564 − 42·65/6 = 109
Sxx = 364 − 42²/6 = 70
Syy = 885 − 65²/6 = 180.8

Correlation coefficient

r = 109/√(70·180.8) = 0.9688

Answer

r = 0.9688
14.Linear regression & correlation

Answer: x = 11.15

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 = 70/6 = 11.67

Sums of squares and products

Sxx = Σ(x − x̄)² = 70
Sxy = Σ(x − x̄)(y − ȳ) = 90

Regression line

b = 90/70 = 1.286
a = 11.67 − 1.286·7 = 2.667
ŷ = 2.667 + 1.286·x

Solve for x

x = (17 − 2.667)/1.286 = 11.15

Answer

x = 11.15
15.Chi-square tests

Answer: E = 19.22

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 count

E = 56·46/134 = 19.22

Answer

E = 19.22
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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