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.

New problemsBack to the course

In the print window, choose “Save as PDF” to keep a copy on your device.

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 DV is not measured
  3. CThere are only two conditions, and a true experiment needs at least three levels of the IV
  4. DThe IV is a pre-existing characteristic, so participants cannot be randomly allocated to it
2.Research methods in psychologyHard

What is the main purpose of random allocation to conditions in an experiment?

  1. ATo increase the response rate
  2. BTo ensure participants do not know the aim
  3. CTo spread participant variables evenly so they do not confound the results
  4. DTo make the sample representative of the target population so results generalise
3.Descriptive statisticsMedium
1015202530354045505560

Find the sample standard deviation s of the data: 38, 18, 37, 10, 18, 60.

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

Exam scores are normally distributed with mean 70 and standard deviation 5. Find the z-score z of a score of 76.25.

μ = 70σ = 5x = 76.25
z =
5.ProbabilityMedium
xp(x)00.3510.3520.3μ = 0.95

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

x: 0, 1, 2p(x): 0.35, 0.35, 0.3
σ =
6.ProbabilityHard
xp(x)10.1520.1530.340.250.2μ

X has the distribution shown, and Y = −2X − 3. Find the standard deviation σY of Y.

x: 1, 2, 3, 4, 5p(x): 0.15, 0.15, 0.3, 0.2, 0.2Y = −2X − 3
σY =
7.Normal distributionMedium
−3−2−10123zshaded area to the right of z = 0.7673

Drag z so that the shaded area under the standard normal curve, to the right of z, is 0.7673. Find z to two decimals.

A = 0.7673
z =
8.Normal distributionHard
μ = 178x = ?97.5%

The heights of adult men are normally distributed with mean 178 cm and standard deviation 7 cm. Find the value x below which 97.5% of them lie (the 97.5th percentile).

μ = 178σ = 7P = 97.5%
x = cm
9.Confidence intervalsMedium
μ₁ − μ₂0LUx̄₁ − x̄₂E = ?90%

The battery life of two phone models is compared. Model 1: 100 phones, mean 139 h, known standard deviation 8 h. Model 2: 30 phones, mean 145 h, known standard deviation 10 h. Find the margin of error E of the 90% confidence interval for μ₁ − μ₂.

n₁ = 100x̄₁ = 139 hσ₁ = 8 hn₂ = 30x̄₂ = 145 hσ₂ = 10 h90%
E = h
10.Confidence intervalsHard
σ = 10, n = ?x̄ − Ex̄ + Ex̄E = 1.590%

The population standard deviation is 10. How large a sample n do you need so that a 90% confidence interval for the mean has a margin of error of at most 1.5?

σ = 10E = 1.590%
n =
11.Hypothesis tests & confidence intervalsMedium
−1.961.96z = ?

Two production lines make the same part. Line 1: a sample of 100 parts has mean length 51 mm, and the line's known standard deviation is 16 mm. Line 2: a sample of 30 parts has mean 54.6 mm, known standard deviation 20 mm. Find the test statistic z for H₀: μ₁ = μ₂, taking the difference as x̄₁ − x̄₂ (line 1 minus line 2).

n₁ = 100x̄₁ = 51 mmσ₁ = 16 mmn₂ = 30x̄₂ = 54.6 mmσ₂ = 20 mm
z =
12.Hypothesis tests & confidence intervalsHard
μ₀x̄c

A one-sided z test of H₀: μ = 50 against H₁: μ > 50 is done at α = 0.01 with a sample of 25; the standard deviation is known to be 7.5. In reality the mean is 53.9. Find the power 1 − β of the test in percent.

μ₀ = 50μ = 53.9σ = 7.5n = 25α = 0.01
1 − β = %
13.Linear regression & correlationMedium
xy360468570671782892xy46860708090

The table gives 6 paired observations of x = hours studied and y = exam score. Find the coefficient of determination r² of the least-squares line.

n = 6
r² =
14.Linear regression & correlationHard
xy2741762783110411248xy5102040x = 11

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. Predict ŷ for x = 11.

n = 6x = 11
ŷ =
15.Chi-square testsMedium
lowmediumhighTotalurban28302280rural20184280Total484864160

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 = 1602 × 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 checks whether counts in 8 categories follow a Poisson distribution whose mean is estimated from the same data. Using the table, find the critical value χ²crit at significance level α = 0.01.

α = 0.01
χ²crit =

3Solutions

1.Research methods in psychology

Answer: D · 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

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

Answer: C · To spread participant variables evenly so they do not confound the results

Why this is the answer

Random allocation balances individual differences across conditions, supporting causal conclusions. Random sampling concerns generalisation to a population.

Answer

C · To spread participant variables evenly so they do not confound the results
3.Descriptive statistics

Answer: s = 18.44

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 = 181/6 = 30.17

Squared deviations

Σ(x − x̄)² = 7.833² + (−12.17)² + 6.833² + (−20.17)² + (−12.17)² + 29.83² = 1701

Standard deviation

s = √(1701/5) = 18.44

Answer

s = 18.44
4.Descriptive statistics

Answer: z = 1.25

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 = (76.25 − 70)/5 = 1.25

Answer

z = 1.25
5.Probability

Answer: σ = 0.8047

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 = 0·0.35 + 1·0.35 + 2·0.3 = 0.95

Mean of the squares

E(X²) = 0·0.35 + 1·0.35 + 4·0.3 = 1.55

Variance

Var(X) = 1.55 − 0.95² = 0.6475
σ = √0.6475 = 0.8047

Answer

σ = 0.8047
6.Probability

Answer: σY = 2.629

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 = 3.15, Var(X) = 1.728, σX = 1.314

Linear change

σY = |−2|·1.314 = 2.629

Answer

σY = 2.629

The mean does move: E(Y) = −2·3.15 − 3 = −9.3.

7.Normal distribution

Answer: z = −0.73

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 − 0.7673 = 0.2327

Read the z-table backwards

Find 0.2327 inside the table and read off z.

z = −0.73

Answer

z = −0.73
8.Normal distribution

Answer: x = 191.7 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) = 0.975 → z = 1.96

Back to x

x = 178 + 1.96·7 = 191.72 cm

Answer

x = 191.72 cm
9.Confidence intervals

Answer: E = 3.279 h

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.

Point estimate

x̄₁ − x̄₂ = 139 − 145 = −6

Standard error

SE = √(8²/100 + 10²/30) = 1.993

Margin of error

E = 1.645·1.993 = 3.279

Answer

E = 3.279 h

The interval does not contain 0, so the means differ significantly.

10.Confidence intervals

Answer: n = 121

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 = (1.645·10/1.5)² = 120.3

Round up

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

n = 121

Answer

n = 121
11.Hypothesis tests & confidence intervals

Answer: z = −0.903

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.

Standard error of the difference

SE = √(16²/100 + 20²/30) = √15.89 = 3.987

Test statistic

z = (51 − 54.6)/3.987 = −0.903

Answer

z = −0.903

|z| < 1.96, so H₀ is not rejected at the 5% level (two-sided).

12.Hypothesis tests & confidence intervals

Answer: 1 − β = 60.8 %

Power of a test: power = 1 − β is the probability of rejecting H₀ when it is false. Find the cutoff x̄c from μ₀ and z_α, then compute the probability of landing beyond x̄c using the true mean: z = (x̄c − μ)/(σ/√n).

Critical sample mean

SE = 7.5/√25 = 1.5
x̄c = 50 + 2.326·1.5 = 53.49

z under the true mean

z = (53.49 − 53.9)/1.5 = −0.274

Power

power = P(Z > −0.274) = 60.8%
1 − β = 60.8%

Answer

1 − β = 60.8%
13.Linear regression & correlation

Answer: r² = 0.9129

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 = 33/6 = 5.5
ȳ = 443/6 = 73.83

Sums of squares and products

Sxx = Σ(x − x̄)² = 17.5
Sxy = Σ(x − x̄)(y − ȳ) = 101.5
Syy = Σ(y − ȳ)² = 644.8

Coefficient of determination

r² = 101.5²/(17.5·644.8) = 0.9129

Answer

r² = 0.9129

About 91.29% of the variation in y is explained by the line.

14.Linear regression & correlation

Answer: ŷ = 44.56

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 = 171/6 = 28.5

Sums of squares and products

Sxx = Σ(x − x̄)² = 70
Sxy = Σ(x − x̄)(y − ȳ) = 281

Regression line

b = 281/70 = 4.014
a = 28.5 − 4.014·7 = 0.4
ŷ = 0.4 + 4.014·x

Prediction

ŷ = 0.4 + 4.014·11 = 44.56

Answer

ŷ = 44.56
15.Chi-square tests

Answer: χ² = 10.58

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: 80·48/160 = 24, 80·48/160 = 24, 80·64/160 = 32
rural: 80·48/160 = 24, 80·48/160 = 24, 80·64/160 = 32

Chi-square statistic

χ² = (28 − 24)²/24 + (30 − 24)²/24 + (22 − 32)²/32 + (20 − 24)²/24 + (18 − 24)²/24 + (42 − 32)²/32
χ² = 10.58

Answer

χ² = 10.58

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

16.Chi-square tests

Answer: χ²crit = 16.81

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 − 1 = 6

Table

χ²crit(df = 6, α = 0.01) = 16.812

Answer

χ²crit = 16.812

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

ExamProblems · examproblems.com/courses/research-methods-psychology