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 study conducted only with American undergraduates may lack:

  1. APopulation validity
  2. BContent validity
  3. CInternal consistency
  4. DInter-rater reliability
2.Research methods in psychologyHard

A researcher records data at nine time points and tests at the 5% level each time without correction. What is the problem?

  1. AThere is no problem with this approach
  2. BThe effect size falls with each test
  3. CThe chance of at least one false positive rises well above 5%
  4. DThe data become more reliable each time, so the final test is the most accurate
3.Descriptive statisticsMedium
3540455055

Find the sample standard deviation s of the data: 55, 31, 41, 31, 32.

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

Exam scores are normally distributed with mean 70 and standard deviation 10. Find the z-score z of a score of 50.

μ = 70σ = 10x = 50
z =
5.ProbabilityMedium
xp(x)10.3520.230.3540.1μ = 2.2

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

x: 1, 2, 3, 4p(x): 0.35, 0.2, 0.35, 0.1
Var(X) =
6.ProbabilityHard
n = 6, p = 0.1: P(X = 2) = ?k0123456

A test has 6 multiple-choice questions. A student who guesses gets each one right with probability 0.1, independently. What is the probability P of exactly 2 correct answers?

n = 6p = 0.1k = 2
P =
7.Normal distributionMedium
340 kg364 kgT = X₁ + … + X₄?

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 4 of them exceeds 364 kg. Round z to two decimals.

μ = 85 kgσ = 12 kgn = 4> 364 kg
P =
8.Normal distributionHard
μ = 178x = ?90%

The heights of adult men are normally distributed with mean 178 cm and standard deviation 7 cm. Find the value x below which 90% of them lie (the 90th percentile).

μ = 178σ = 7P = 90%
x = cm
9.Confidence intervalsMedium
n = 12, s = 16LUx̄ = 55.5E = ?95%

A sample of 12 items has mean 55.5 and standard deviation s = 16. The population is roughly normal. Find the margin of error E of a 95% confidence interval for the mean.

n = 12x̄ = 55.5s = 1695%t(11) = 2.201
E =
10.Confidence intervalsHard
σ = 14, n = ?x̄ − Ex̄ + Ex̄E = 199%

The population standard deviation is 14. 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 1?

σ = 14E = 199%
n =
11.Hypothesis tests & confidence intervalsMedium
t −1.98401.984

A sample of 100 students has a mean score of 205.8 and a sample standard deviation of 20. Find the test statistic t for H₀: μ = 200.

μ₀ = 200x̄ = 205.8s = 20n = 100
t =
12.Hypothesis tests & confidence intervalsHard
z −1.64501.645

A sample of 64 measurements has mean 100.8. The population standard deviation is 20. Find the lower limit L of the 90% confidence interval for the mean.

x̄ = 100.8σ = 20n = 6490%
L =
13.Linear regression & correlationMedium
xy2941261882410301233xy510102030

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
xy168919101221202514628154xy2025100150ŷ = 96

The table gives 5 paired observations of x = temperature in °C and y = ice creams sold. Use the least-squares line ŷ = a + b·x. For which value x does the line predict ŷ = 96?

n = 5ŷ = 96
x =
15.Chi-square testsMedium
no defectminormajorTotalday shift173053100night shift276924120Total449977220

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 = 2202 × 3
χ² =
16.Chi-square testsHard

A chi-square goodness-of-fit test compares observed counts in 8 categories with fully specified expected proportions (no parameters are estimated from the data). How many degrees of freedom df does the test have?

k = 8
df =

3Solutions

1.Research methods in psychology

Answer: A · Population validity

Why this is the answer

Population validity is a type of external validity. Samples that are mostly young, Western and educated may not represent other people.

Answer

A · Population validity
2.Research methods in psychology

Answer: C · The chance of at least one false positive rises well above 5%

Why this is the answer

With nine independent tests the chance of at least one Type I error is about 1 − 0.95⁹ ≈ 37%. Corrections such as Bonferroni address this.

Answer

C · The chance of at least one false positive rises well above 5%
3.Descriptive statistics

Answer: s = 10.39

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 = 190/5 = 38

Squared deviations

Σ(x − x̄)² = 17² + (−7)² + 3² + (−7)² + (−6)² = 432

Standard deviation

s = √(432/4) = 10.39

Answer

s = 10.39
4.Descriptive statistics

Answer: z = −2

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 = (50 − 70)/10 = −2

Answer

z = −2
5.Probability

Answer: Var(X) = 1.06

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 = 1·0.35 + 2·0.2 + 3·0.35 + 4·0.1 = 2.2

Mean of the squares

E(X²) = 1·0.35 + 4·0.2 + 9·0.35 + 16·0.1 = 5.9

Variance

Var(X) = 5.9 − 2.2² = 1.06

Answer

Var(X) = 1.06
6.Probability

Answer: P = 0.09842

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 = 2) = 15·0.12·0.94 = 15·0.01·0.6561 = 0.09842

Answer

P = 0.09842
7.Normal distribution

Answer: P = 0.1587

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 = 340 kg, σT = 12·√4 = 24 kg

Standardise

z = (364 − 340)/24 = 1

Probability

P = 1 − Φ(1) = 0.1587

Answer

P = 0.1587
8.Normal distribution

Answer: x = 187 cm

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) = 0.9 → z = 1.282

Back to x

x = 178 + 1.282·7 = 186.97 cm

Answer

x = 186.97 cm
9.Confidence intervals

Answer: E = 10.17

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 = 11, t = 2.201

Margin of error

E = 2.201·16/√12 = 10.17

Answer

E = 10.17
10.Confidence intervals

Answer: n = 1301

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 = (2.576·14/1)² = 1301

Round up

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

n = 1301

Answer

n = 1301
11.Hypothesis tests & confidence intervals

Answer: t = 2.9

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).

Standard error

SE = 20/√100 = 20/10 = 2

Test statistic

t = (205.8 − 200)/2 = 2.9

Answer

t = 2.9
t0.025; df = 99 = 1.984

|t| > 1.984, so H₀ is rejected at the 5% level (two-sided).

12.Hypothesis tests & confidence intervals

Answer: L = 96.69

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 = 1.645·20/√64 = 4.112

Interval limit

L = 100.8 − 4.112 = 96.69

Answer

L = 96.69
13.Linear regression & correlation

Answer: r = 0.9945

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 = 126/6 = 21

Sums of squares and products

Sxx = Σ(x − x̄)² = 70
Sxy = Σ(x − x̄)(y − ȳ) = 180
Syy = Σ(y − ȳ)² = 468

Correlation coefficient

r = 180/√(70·468) = 0.9945

Answer

r = 0.9945
14.Linear regression & correlation

Answer: x = 17.54

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 = 110/5 = 22
ȳ = 610/5 = 122

Sums of squares and products

Sxx = Σ(x − x̄)² = 90
Sxy = Σ(x − x̄)(y − ȳ) = 525

Regression line

b = 525/90 = 5.833
a = 122 − 5.833·22 = −6.333
ŷ = −6.333 + 5.833·x

Solve for x

x = (96 − (−6.333))/5.833 = 17.54

Answer

x = 17.54
15.Chi-square tests

Answer: χ² = 26.96

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

day shift: 100·44/220 = 20, 100·99/220 = 45, 100·77/220 = 35
night shift: 120·44/220 = 24, 120·99/220 = 54, 120·77/220 = 42

Chi-square statistic

χ² = (17 − 20)²/20 + (30 − 45)²/45 + (53 − 35)²/35 + (27 − 24)²/24 + (69 − 54)²/54 + (24 − 42)²/42
χ² = 26.96

Answer

χ² = 26.96

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

16.Chi-square tests

Answer: df = 7

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 = 8 − 1 = 7

Answer

df = 7

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