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 researcher studies the effect of gender on memory scores. Why is this a quasi-experiment rather than a true experiment?

  1. AThere are only two conditions, and a true experiment needs at least three levels of the IV
  2. BThe IV is a pre-existing characteristic, so participants cannot be randomly allocated to it
  3. CThe DV is not measured
  4. DThe study takes place in a laboratory
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. CRefine the behaviour categories and train observers, then recheck agreement
  4. DSwitch to a one-tailed hypothesis
3.Descriptive statisticsMedium
384042444648505254

Find the sample standard deviation s of the data: 44, 39, 37, 55, 41, 46.

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

A quantity is normally distributed with mean 170 and standard deviation 5. What proportion P of values is above 173?

μ = 170σ = 5x = 173
P =
5.ProbabilityMedium
xp(x)10.220.330.1540.35μ = 2.65

For the distribution shown, find the standard deviation σ of X.

x: 1, 2, 3, 4p(x): 0.2, 0.3, 0.15, 0.35
σ =
6.ProbabilityHard
xp(x)00.210.320.2530.25μ

X has the distribution shown, and Y = 5X − 3. Find the standard deviation σY of Y.

x: 0, 1, 2, 3p(x): 0.2, 0.3, 0.25, 0.25Y = 5X − 3
σY =
7.Normal distributionMedium
765 kg747 kgT = X₁ + … + X₉?

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 9 of them exceeds 747 kg. Round z to two decimals.

μ = 85 kgσ = 12 kgn = 9> 747 kg
P =
8.Normal distributionHard
μ = 178x = ?25%

The heights of adult men are normally distributed with mean 178 cm and standard deviation 7 cm. Find the value x below which 25% of them lie (the 25th percentile).

μ = 178σ = 7P = 25%
x = cm
9.Confidence intervalsMedium
n = 8, s = 15.5L = ?Ux̄ = 39.5E95%

A sample of 8 items has mean 39.5 and standard deviation s = 15.5. The population is roughly normal. Find the lower limit of the 95% confidence interval for the mean.

n = 8x̄ = 39.5s = 15.595%t(7) = 2.365
lower limit =
10.Confidence intervalsHard
σ = 20, n = ?x̄ − Ex̄ + Ex̄E = 295%

The population standard deviation is 20. How large a sample n do you need so that a 95% confidence interval for the mean has a margin of error of at most 2?

σ = 20E = 295%
n =
11.Hypothesis tests & confidence intervalsMedium
−2.142.14p = ?

The mean of a population is claimed to be 500; the population standard deviation is known to be 25. A random sample of 100 observations has a mean of 494.65. Test H₀: μ = 500 against H₁: μ ≠ 500. Find the p-value p.

μ₀ = 500σ = 25n = 100x̄ = 494.65H₁: μ ≠ μ₀
p =
12.Hypothesis tests & confidence intervalsHard
σ = 15, n = ?x̄ − Ex̄ + Ex̄E = 199%

How large must a sample be so that the 99% confidence interval for a mean has a margin of error of at most 1? The standard deviation is 15. Find the smallest sample size n.

σ = 15E = 199%
n =
13.Linear regression & correlationMedium
xy18742199241022712630136xy20253080100120140

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

n = 5
r =
14.Linear regression & correlationHard
xy140242343448554666xy246405060ŷ = 50

The table gives 6 paired observations of x = advertising spend in k€ and y = sales in k€. Use the least-squares line ŷ = a + b·x. For which value x does the line predict ŷ = 50?

n = 6ŷ = 50
x =
15.Chi-square testsMedium
lowmediumhighTotalurban10151540rural253441100Total354956140

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 = 1402 × 3
χ² =
16.Chi-square testsHard
yesnoTotalline 1245680line 2433780line 33763100Total104156260dfα = 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 = 2603 × 2α = 0.01
χ² =

3Solutions

1.Research methods in psychology

Answer: B · The IV is a pre-existing characteristic, so participants cannot be randomly allocated to it

Why this is the answer

In a quasi-experiment the IV is a naturally occurring difference between people. Without random allocation, other differences between groups may confound the results.

Answer

B · The IV is a pre-existing characteristic, so participants cannot be randomly allocated to it
2.Research methods in psychology

Answer: C · 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

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

Answer: s = 6.439

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 = 262/6 = 43.67

Squared deviations

Σ(x − x̄)² = 0.3333² + (−4.667)² + (−6.667)² + 11.33² + (−2.667)² + 2.333² = 207.3

Standard deviation

s = √(207.3/5) = 6.439

Answer

s = 6.439
4.Descriptive statistics

Answer: P = 0.2743

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 = 0.6

Standard normal table

Φ(0.6) = 0.7257
P = 1 − 0.7257 = 0.2743

Answer

P = 0.2743
5.Probability

Answer: σ = 1.152

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 = 1·0.2 + 2·0.3 + 3·0.15 + 4·0.35 = 2.65

Mean of the squares

E(X²) = 1·0.2 + 4·0.3 + 9·0.15 + 16·0.35 = 8.35

Variance

Var(X) = 8.35 − 2.65² = 1.327
σ = √1.327 = 1.152

Answer

σ = 1.152
6.Probability

Answer: σY = 5.356

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.

Spread of X

μ = 1.55, Var(X) = 1.147, σX = 1.071

Linear change

σY = |5|·1.071 = 5.356

Answer

σY = 5.356

The mean does move: E(Y) = 5·1.55 − 3 = 4.75.

7.Normal distribution

Answer: P = 0.6915

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 = 765 kg, σT = 12·√9 = 36 kg

Standardise

z = (747 − 765)/36 = −0.5

Probability

P = 1 − Φ(−0.5) = 0.6915

Answer

P = 0.6915
8.Normal distribution

Answer: x = 173.3 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) = 0.25 → z = −0.6745

Back to x

x = 178 + −0.6745·7 = 173.28 cm

Answer

x = 173.28 cm
9.Confidence intervals

Answer: lower limit = 26.54

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 = 7, t = 2.365

Margin of error

E = 2.365·15.5/√8 = 12.96

Interval

39.5 ± 12.96 → [26.54, 52.46]

Answer

lower limit = 26.54
10.Confidence intervals

Answer: n = 385

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 = (1.96·20/2)² = 384.2

Round up

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

n = 385

Answer

n = 385
11.Hypothesis tests & confidence intervals

Answer: p = 0.03235

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 = 25/√100 = 2.5
z = (494.65 − 500)/2.5 = −2.14

Tail area

Φ(2.14) = 0.9838
1 − Φ(2.14) = 0.0162

p-value

p = 2·0.0162 = 0.03235

Answer

p = 0.03235

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

12.Hypothesis tests & confidence intervals

Answer: n = 1494

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).

Solve for n

n = (2.576·15/1)² = 1493

Round up

n = 1494

Answer

n = 1494
13.Linear regression & correlation

Answer: r = 0.979

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 = 120/5 = 24
ȳ = 537/5 = 107.4

Sums of squares and products

Sxx = Σ(x − x̄)² = 90
Sxy = Σ(x − x̄)(y − ȳ) = 453
Syy = Σ(y − ȳ)² = 2379

Correlation coefficient

r = 453/√(90·2379) = 0.979

Answer

r = 0.979
14.Linear regression & correlation

Answer: x = 3.739

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 = 293/6 = 48.83

Sums of squares and products

Sxx = Σ(x − x̄)² = 17.5
Sxy = Σ(x − x̄)(y − ȳ) = 85.5

Regression line

b = 85.5/17.5 = 4.886
a = 48.83 − 4.886·3.5 = 31.73
ŷ = 31.73 + 4.886·x

Solve for x

x = (50 − 31.73)/4.886 = 3.739

Answer

x = 3.739
15.Chi-square tests

Answer: χ² = 0.1875

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

urban: 40·35/140 = 10, 40·49/140 = 14, 40·56/140 = 16
rural: 100·35/140 = 25, 100·49/140 = 35, 100·56/140 = 40

Chi-square statistic

χ² = (10 − 10)²/10 + (15 − 14)²/14 + (15 − 16)²/16 + (25 − 25)²/25 + (34 − 35)²/35 + (41 − 40)²/40
χ² = 0.1875

Answer

χ² = 0.1875

df = 2. At α = 0.05 the critical value is 5.991, so H₀ (independence) is not rejected.

16.Chi-square tests

Answer: χ² = 10.01

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

line 1: 32, 48
line 2: 32, 48
line 3: 40, 60

Chi-square statistic

χ² = (24 − 32)²/32 + (56 − 48)²/48 + (43 − 32)²/32 + (37 − 48)²/48 + (37 − 40)²/40 + (63 − 60)²/60
χ² = 10.01

Decision

df = 2, χ²crit = 9.210
10.01 > 9.210

Answer

χ² = 10.01

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

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