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. AThe study takes place in a laboratory
  2. BThe IV is a pre-existing characteristic, so participants cannot be randomly allocated to it
  3. CThe DV is not measured
  4. DThere are only two conditions, and a true experiment needs at least three levels of the IV
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?

  1. AA Type I error
  2. BA floor effect
  3. COnly a main effect of age
  4. DAn interaction between drug and age
3.Descriptive statisticsMedium
20253035404550

Find the sample standard deviation s of the data: 27, 52, 44, 45, 18.

n = 5
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 85.

μ = 100σ = 10x = 85
z =
5.ProbabilityMedium
xp(x)100.25200.2300.25400.3μ = 26

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

x: 10, 20, 30, 40p(x): 0.25, 0.2, 0.25, 0.3
Var(X) =
6.ProbabilityHard
n = 11, p = 0.25: P(X = 4) = ?k01234567891011

A test has 11 multiple-choice questions. A student who guesses gets each one right with probability 0.25, independently. What is the probability P of exactly 4 correct answers?

n = 11p = 0.25k = 4
P =
7.Normal distributionMedium
−3−2−10123zshaded area to the right of z = 0.0351

Drag z so that the shaded area under the standard normal curve, to the right of z, is 0.0351. Find z to two decimals.

A = 0.0351
z =
8.Normal distributionHard
35455565758595pointstop 75%

The exam scores are normally distributed with mean 65 points and standard deviation 10 points. Drag the cut-off so the shaded top 75% is to the right of it. Find the cut-off x.

μ = 65 pointsσ = 10 points75%
x = points
9.Confidence intervalsMedium
n = 9, s = 16LUx̄ = 30E = ?90%

A sample of 9 items has mean 30 and standard deviation s = 16. The population is roughly normal. Find the margin of error E of a 90% confidence interval for the mean.

n = 9x̄ = 30s = 1690%t(8) = 1.86
E =
10.Confidence intervalsHard
n = 30Lσ = ?s = 10.395%

The diameters of 30 machined shafts have a sample standard deviation of 10.3 µm. Assuming the diameters are normally distributed, find the upper limit σ of the 95% confidence interval for the population standard deviation. With df = 29, 2.5% of the χ² distribution lies below 16.047 and 2.5% lies above 45.722.

n = 30s = 10.3 µmχ² = 16.047, 45.722
σ = µm
11.Hypothesis tests & confidence intervalsMedium
−2.0452.045t = ?df = 29

Two teaching methods are compared. Group 1: 11 students, mean score 107, standard deviation 5. Group 2: 20 students, mean score 119, standard deviation 11. Assume equal population variances and use the pooled two-sample t test. Find the test statistic t for H₀: μ₁ = μ₂, taking the difference as x̄₁ − x̄₂ (group 1 minus group 2).

n₁ = 11x̄₁ = 107s₁ = 5n₂ = 20x̄₂ = 119s₂ = 11
t =
12.Hypothesis tests & confidence intervalsHard
σ = 8, n = ?x̄ − Ex̄ + Ex̄E = 1.599%

How large must a sample be so that the 99% confidence interval for a mean has a margin of error of at most 1.5? The standard deviation is 8. Find the smallest sample size n.

σ = 8E = 1.599%
n =
13.Linear regression & correlationMedium
xy141350554763969xy540506070

The table gives 5 paired observations of x = hours studied and y = exam score. Their sums are given. Find Pearson's correlation coefficient r from these sums.

n = 5Σx = 25Σy = 277Σx² = 165Σy² = 15827Σxy = 1523
r =
14.Linear regression & correlationHard
xy155817671964218623812593xy1520256080x = 24

The table gives 6 paired observations of x = temperature in °C and y = ice creams sold. Use the least-squares line ŷ = a + b·x. A new observation (24, 86) is made. Find its residual e = y − ŷ.

n = 6(24, 86)
e =
15.Chi-square testsMedium
no defectminormajorTotalday shift12242460night shift333948120Total456372180

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
dfα = 0.05α = 0.0113.8416.63525.9919.21037.81511.34549.48813.277511.07015.086612.59216.812714.06718.475815.50720.090916.91921.6661018.30723.209

A chi-square goodness-of-fit test compares observed counts in 7 categories with fully specified expected proportions (no parameters are estimated from the data). Using the table, find the critical value χ²crit at significance level α = 0.05.

α = 0.05
χ²crit =

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: D · 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

D · An interaction between drug and age
3.Descriptive statistics

Answer: s = 14.13

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 = 186/5 = 37.2

Squared deviations

Σ(x − x̄)² = (−10.2)² + 14.8² + 6.8² + 7.8² + (−19.2)² = 798.8

Standard deviation

s = √(798.8/4) = 14.13

Answer

s = 14.13
4.Descriptive statistics

Answer: z = −1.5

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 = (85 − 100)/10 = −1.5

Answer

z = −1.5
5.Probability

Answer: Var(X) = 134

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 = 10·0.25 + 20·0.2 + 30·0.25 + 40·0.3 = 26

Mean of the squares

E(X²) = 100·0.25 + 400·0.2 + 900·0.25 + 1600·0.3 = 810

Variance

Var(X) = 810 − 26² = 134

Answer

Var(X) = 134
6.Probability

Answer: P = 0.1721

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 = 4) = 330·0.254·0.757 = 330·0.003906·0.1335 = 0.1721

Answer

P = 0.1721
7.Normal distribution

Answer: z = 1.81

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 of z

Φ(z) = 1 − 0.0351 = 0.9649

Read the z-table backwards

Find 0.9649 inside the table and read off z.

z = 1.81

Answer

z = 1.81
8.Normal distribution

Answer: x = 58.3 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) = 0.25

z from the table

z ≈ −0.67

Back to the original units

x = 65 − 0.67·10 = 58.3 points

Answer

x = 58.3 points
9.Confidence intervals

Answer: E = 9.92

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 = 1.86·16/√9 = 9.92

Answer

E = 9.92
10.Confidence intervals

Answer: σ = 13.85 µm

Interval for a variance: (n − 1)s²/σ² has a χ² distribution with n − 1 degrees of freedom. The 95% interval for σ² runs from (n − 1)s² divided by the upper χ² value (2.5% above it) to (n − 1)s² divided by the lower χ² value (2.5% below it); the interval for σ is the square root of both limits. It is not symmetric around s.

Sum of squares

(n − 1)s² = 29·10.3² = 3077

Limits for σ²

3077/45.722 = 67.29
3077/16.047 = 191.7

Limit for σ

σ = √191.7 = 13.85 µm

Answer

σ = 13.85 µm

95% interval for σ: [8.203, 13.85] µm.

11.Hypothesis tests & confidence intervals

Answer: t = −3.41

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.

Pooled variance

sp² = (10·5² + 19·11²)/29 = 87.9
sp = 9.375

Standard error

SE = 9.375·√(1/11 + 1/20) = 3.519

Test statistic

t = (107 − 119)/3.519 = −3.41

Answer

t = −3.41

With df = 29, the two-sided 5% critical value is 2.045; |t| is larger, so H₀ is rejected.

12.Hypothesis tests & confidence intervals

Answer: n = 189

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 = (2.576·8/1.5)² = 188.8

Round up

n = 189

Answer

n = 189
13.Linear regression & correlation

Answer: r = 0.9947

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 = 1523 − 25·277/5 = 138
Sxx = 165 − 25²/5 = 40
Syy = 15827 − 277²/5 = 481.2

Correlation coefficient

r = 138/√(40·481.2) = 0.9947

Answer

r = 0.9947
14.Linear regression & correlation

Answer: e = −2.49

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 = 120/6 = 20
ȳ = 449/6 = 74.83

Sums of squares and products

Sxx = Σ(x − x̄)² = 70
Sxy = Σ(x − x̄)(y − ȳ) = 239

Regression line

b = 239/70 = 3.414
a = 74.83 − 3.414·20 = 6.548
ŷ = 6.548 + 3.414·x

Predicted value

ŷ = 6.548 + 3.414·24 = 88.49

Residual

e = 86 − 88.49 = −2.49

Answer

e = −2.49

The point lies below the line.

15.Chi-square tests

Answer: χ² = 1.543

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: 60·45/180 = 15, 60·63/180 = 21, 60·72/180 = 24
night shift: 120·45/180 = 30, 120·63/180 = 42, 120·72/180 = 48

Chi-square statistic

χ² = (12 − 15)²/15 + (24 − 21)²/21 + (24 − 24)²/24 + (33 − 30)²/30 + (39 − 42)²/42 + (48 − 48)²/48
χ² = 1.543

Answer

χ² = 1.543

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

16.Chi-square tests

Answer: χ²crit = 12.59

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 = 7 − 1 = 6

Table

χ²crit(df = 6, α = 0.05) = 12.592

Answer

χ²crit = 12.592

Reject H₀ if the computed χ² is larger than this value.

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