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 Cronbach's alpha of 0.88 for a 15-item anxiety scale indicates:

  1. AScores are stable over six months
  2. BThe scale predicts future anxiety well
  3. CThe scale has a significant effect on anxiety
  4. DGood internal consistency of the items
2.Research methods in psychologyHard

A researcher uses a questionnaire, interviews and observation to study workplace stress, and the findings agree. What is this approach called?

  1. ACounterbalancing
  2. BReplication
  3. CStratification
  4. DTriangulation
3.Descriptive statisticsMedium
253035404550

Find the sample standard deviation s of the data: 53, 37, 21, 35, 28.

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

A quantity is normally distributed with mean 100 and standard deviation 5. What proportion P of values is below 110?

μ = 100σ = 5x = 110
P =
5.ProbabilityMedium
xp(x)00.310.3520.35μ = 1.05

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

x: 0, 1, 2p(x): 0.3, 0.35, 0.35
Var(X) =
6.ProbabilityHard
n = 9, p = 0.1: P(X = 1) = ?k0123456789

A test has 9 multiple-choice questions. A student who guesses gets each one right with probability 0.1, independently. What is the probability P of exactly 1 correct answers?

n = 9p = 0.1k = 1
P =
7.Normal distributionMedium
510 kg480.6 kgT = X₁ + … + X₆?

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 6 of them is below 480.6 kg. Round z to two decimals.

μ = 85 kgσ = 12 kgn = 6< 480.6 kg
P =
8.Normal distributionHard
μ = 178x = ?99%

The heights of adult men are normally distributed with mean 178 cm and standard deviation 7 cm. Find the value x below which 99% of them lie (the 99th percentile).

μ = 178σ = 7P = 99%
x = cm
9.Confidence intervalsMedium
n = 12, s = 3.5LU = ?x̄ = 92E99%

A sample of 12 items has mean 92 and standard deviation s = 3.5. The population is roughly normal. Find the upper limit of the 99% confidence interval for the mean.

n = 12x̄ = 92s = 3.599%t(11) = 3.106
upper limit =
10.Confidence intervalsHard
p = 0.5, n = ?p̂ − Ep̂ + Ep̂E = 0.0495%

Nothing is known in advance about the proportion of defective products. How many products n must be inspected so that a 95% confidence interval for the proportion has a margin of error of at most 0.04?

p = 0.5E = 0.0495%
n =
11.Hypothesis tests & confidence intervalsMedium
t −1.99801.998

A sample of 64 students has a mean score of 101.8 and a sample standard deviation of 20. Find the test statistic t for H₀: μ = 100.

μ₀ = 100x̄ = 101.8s = 20n = 64
t =
12.Hypothesis tests & confidence intervalsHard
z −1.9601.96

A sample of 16 measurements has mean 48.9. The population standard deviation is 15. Find the lower limit L of the 95% confidence interval for the mean.

x̄ = 48.9σ = 15n = 1695%
L =
13.Linear regression & correlationMedium
xy158418110211252414227162xy152025100150

The table gives 5 paired observations of x = temperature in °C and y = ice creams sold. Their sums are given. Find Pearson's correlation coefficient r from these sums.

n = 5Σx = 105Σy = 623Σx² = 2295Σy² = 81189Σxy = 13647
r =
14.Linear regression & correlationHard
xy130225327421522xy242530x = 1.5

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. Predict ŷ for x = 1.5.

n = 5x = 1.5
ŷ =
15.Chi-square testsMedium
no defectminormajorTotalday shift41311789night shift31211264Total725229153

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

row total = 89column total = 52N = 153
E =
16.Chi-square testsHard
yesnoTotalplant A6535100plant B241640plant C114960Total100100200dfα = 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 = 2003 × 2α = 0.01
χ² =

3Solutions

1.Research methods in psychology

Answer: D · Good internal consistency of the items

Why this is the answer

Cronbach's alpha estimates internal consistency; values around 0.7 or above are usually considered acceptable. It does not by itself show validity.

Answer

D · Good internal consistency of the items
2.Research methods in psychology

Answer: D · Triangulation

Why this is the answer

Triangulation uses multiple methods, sources or researchers. Agreement between them increases confidence in the validity of the findings.

Answer

D · Triangulation
3.Descriptive statistics

Answer: s = 11.97

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 = 18.2² + 2.2² + (−13.8)² + 0.2² + (−6.8)² = 572.8

Standard deviation

s = √(572.8/4) = 11.97

Answer

s = 11.97
4.Descriptive statistics

Answer: P = 0.9772

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 = 2

Standard normal table

Φ(2) = 0.9772
P = 0.9772

Answer

P = 0.9772
5.Probability

Answer: Var(X) = 0.6475

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 = 0·0.3 + 1·0.35 + 2·0.35 = 1.05

Mean of the squares

E(X²) = 0·0.3 + 1·0.35 + 4·0.35 = 1.75

Variance

Var(X) = 1.75 − 1.05² = 0.6475

Answer

Var(X) = 0.6475
6.Probability

Answer: P = 0.3874

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 = 1) = 9·0.11·0.98 = 9·0.1·0.4305 = 0.3874

Answer

P = 0.3874
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 = 510 kg, σT = 12·√6 = 29.39 kg

Standardise

z = (480.6 − 510)/29.39 = −1

Probability

P = Φ(−1) = 0.1587

Answer

P = 0.1587
8.Normal distribution

Answer: x = 194.3 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) = 0.99 → z = 2.326

Back to x

x = 178 + 2.326·7 = 194.28 cm

Answer

x = 194.28 cm
9.Confidence intervals

Answer: upper limit = 95.14

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 = 3.106

Margin of error

E = 3.106·3.5/√12 = 3.138

Interval

92 ± 3.138 → [88.86, 95.14]

Answer

upper limit = 95.14
10.Confidence intervals

Answer: n = 601

Confidence interval for a proportion: p̂ ± z·√(p̂(1 − p̂)/n), where p̂ = x/n. To reach a margin E you need n ≥ z²·p(1 − p)/E², rounded up; without a prior estimate use p = 0.5, which gives the largest n.

Solve for n

n = 1.96²·0.5·0.5/0.04² = 600.2

Round up

n = 601

Answer

n = 601
11.Hypothesis tests & confidence intervals

Answer: t = 0.72

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/√64 = 20/8 = 2.5

Test statistic

t = (101.8 − 100)/2.5 = 0.72

Answer

t = 0.72
t0.025; df = 63 = 1.998

|t| < 1.998, so H₀ is not rejected at the 5% level (two-sided).

12.Hypothesis tests & confidence intervals

Answer: L = 41.55

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 = 1.96·15/√16 = 7.35

Interval limit

L = 48.9 − 7.35 = 41.55

Answer

L = 41.55
13.Linear regression & correlation

Answer: r = 0.996

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 = 13647 − 105·623/5 = 564
Sxx = 2295 − 105²/5 = 90
Syy = 81189 − 623²/5 = 3563

Correlation coefficient

r = 564/√(90·3563) = 0.996

Answer

r = 0.996
14.Linear regression & correlation

Answer: ŷ = 28

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 = 125/5 = 25

Sums of squares and products

Sxx = Σ(x − x̄)² = 10
Sxy = Σ(x − x̄)(y − ȳ) = −20

Regression line

b = −20/10 = −2
a = 25 − (−2)·3 = 31
ŷ = 31 − 2·x

Prediction

ŷ = 31 + (−2)·1.5 = 28

Answer

ŷ = 28
15.Chi-square tests

Answer: E = 30.25

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 = 89·52/153 = 30.25

Answer

E = 30.25
16.Chi-square tests

Answer: χ² = 34.67

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

plant A: 50, 50
plant B: 20, 20
plant C: 30, 30

Chi-square statistic

χ² = (65 − 50)²/50 + (35 − 50)²/50 + (24 − 20)²/20 + (16 − 20)²/20 + (11 − 30)²/30 + (49 − 30)²/30
χ² = 34.67

Decision

df = 2, χ²crit = 9.210
34.67 > 9.210

Answer

χ² = 34.67

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

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