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

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

Increasing the sample size in a study most directly:

  1. ARemoves all bias from the sample
  2. BIncreases statistical power to detect a real effect
  3. CGuarantees a statistically significant result if any effect exists
  4. DIncreases the effect size
3.Descriptive statisticsMedium
1520253035404550

Find the sample standard deviation s of the data: 49, 51, 34, 24, 51, 39, 15.

n = 7
s =
4.Descriptive statisticsHard
z −2−1012

A quantity is normally distributed with mean 50 and standard deviation 20. What proportion P of values is below 58?

μ = 50σ = 20x = 58
P =
5.ProbabilityMedium
P(at least one six) = ??1?2?3?4?5?6?7?8P(six) = 1/6 on each roll

A fair die is rolled 8 times. What is the probability P of getting at least one six?

n = 8
P =
6.ProbabilityHard
xp(x)00.3520.340.0560.3μ

X has the distribution shown, and Y = 5X + 20. Find the standard deviation σY of Y.

x: 0, 2, 4, 6p(x): 0.35, 0.3, 0.05, 0.3Y = 5X + 20
σY =
7.Normal distributionMedium
247.75257.5P = ?

The bag weights are normally distributed with mean 250 g and standard deviation 5 g. What fraction lies between 247.75 and 257.5 g? Give P as a decimal.

μ = 250σ = 5x1 = 247.75x2 = 257.5
P =
8.Normal distributionHard
557085100115130145pointstop 25%

The IQ scores are normally distributed with mean 100 points and standard deviation 15 points. Drag the cut-off so the shaded top 25% is to the right of it. Find the cut-off x.

μ = 100 pointsσ = 15 points25%
x = points
9.Confidence intervalsMedium
n = 11, sd​ = 6.90L = ?Ud̄ = −4.1E99%

The blood pressure of 11 patients is measured before and after a treatment. The differences d = before − after (mmHg) have mean −4.1 and standard deviation 6.9. Find the lower limit of the 99% confidence interval for the mean difference.

n = 11d̄ = −4.1sd = 6.999%t(10) = 3.169
lower limit = mmHg
10.Confidence intervalsHard
σ = 31, n = ?x̄ − Ex̄ + Ex̄E = 499%

The population standard deviation is 31. 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 4?

σ = 31E = 499%
n =
11.Hypothesis tests & confidence intervalsMedium
t −2.13102.131

A sample of 16 students has a mean score of 114.8 and a sample standard deviation of 8. Find the test statistic t for H₀: μ = 120.

μ₀ = 120x̄ = 114.8s = 8n = 16
t =
12.Hypothesis tests & confidence intervalsHard
H₀H₁μ₀x̄cμ₀ + 4α = 0.01βσ = 18, power 95%, n = ?

A one-sided z test at α = 0.01 must detect a shift of the mean of 4 units with a power of 95%. The standard deviation is 18. Find the minimum sample size n.

α = 0.01power = 95%δ = 4σ = 18zα = 2.326zβ = 1.645
n =
13.Linear regression & correlationMedium
xy144250355457563xy245060

The table gives 5 paired observations of x = advertising spend in k€ and y = sales in k€. Find the coefficient of determination r² of the least-squares line.

n = 5
r² =
14.Linear regression & correlationHard
xy340450549659766xy46405060ŷ = 50

The table gives 5 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 = 5ŷ = 50
x =
15.Chi-square testsMedium
no defectminormajorTotalday shift10311960night shift224137100Total327256160

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

A chi-square goodness-of-fit test compares observed counts in 11 categories with fully specified expected proportions (no parameters are estimated from the data). How many degrees of freedom df does the test have?

k = 11
df =

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: B · Increases statistical power to detect a real effect

Why this is the answer

Larger samples reduce standard error, increasing power. They do not change the true effect size or remove systematic bias.

Answer

B · Increases statistical power to detect a real effect
3.Descriptive statistics

Answer: s = 14.14

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 = 263/7 = 37.57

Squared deviations

Σ(x − x̄)² = 11.43² + 13.43² + (−3.571)² + (−13.57)² + 13.43² + 1.429² + (−22.57)² = 1200

Standard deviation

s = √(1200/6) = 14.14

Answer

s = 14.14
4.Descriptive statistics

Answer: P = 0.6554

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 = 0.4

Standard normal table

Φ(0.4) = 0.6554
P = 0.6554

Answer

P = 0.6554
5.Probability

Answer: P = 0.7674

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.

Complement

P(no six) = (5/6)8 = 0.2326
P = 1 − 0.2326 = 0.7674

Answer

P = 0.7674
6.Probability

Answer: σY = 12.29

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 = 2.6, Var(X) = 6.04, σX = 2.458

Linear change

σY = |5|·2.458 = 12.29

Answer

σY = 12.29

The mean does move: E(Y) = 5·2.6 + 20 = 33.

7.Normal distribution

Answer: P = 0.6068

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-scores

z₁ = (247.75 − 250)/5 = −0.45
z₂ = (257.5 − 250)/5 = 1.5

Area between

P = Φ(1.5) − Φ(−0.45) = 0.9332 − 0.3264 = 0.6068

Answer

P = 0.6068
8.Normal distribution

Answer: x = 110.1 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) = 0.75

z from the table

z ≈ 0.67

Back to the original units

x = 100 + 0.67·15 = 110.1 points

Answer

x = 110.1 points
9.Confidence intervals

Answer: lower limit = −10.69 mmHg

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.

Critical value

df = 10, t = 3.169

Margin of error

E = 3.169·6.9/√11 = 6.593

Interval

−4.1 ± 6.593 → [−10.69, 2.493]

Answer

lower limit = −10.69 mmHg
10.Confidence intervals

Answer: n = 399

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 = (2.576·31/4)² = 398.6

Round up

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

n = 399

Answer

n = 399
11.Hypothesis tests & confidence intervals

Answer: t = −2.6

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 = 8/√16 = 8/4 = 2

Test statistic

t = (114.8 − 120)/2 = −2.6

Answer

t = −2.6
t0.025; df = 15 = 2.131

|t| > 2.131, so H₀ is rejected at the 5% level (two-sided).

12.Hypothesis tests & confidence intervals

Answer: n = 320

Sample size for a test: to detect a true shift δ with power 1 − β in a one-sided z test at level α, the standard error must satisfy δ = (z_α + z_β)·σ/√n, so n = ((z_α + z_β)·σ/δ)², rounded up.

Solve for n

n = ((2.326 + 1.645)·18/4)² = 319.3

Round up

n = 320

Answer

n = 320
13.Linear regression & correlation

Answer: r² = 0.9792

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 = 269/5 = 53.8

Sums of squares and products

Sxx = Σ(x − x̄)² = 10
Sxy = Σ(x − x̄)(y − ȳ) = 45
Syy = Σ(y − ȳ)² = 206.8

Coefficient of determination

r² = 45²/(10·206.8) = 0.9792

Answer

r² = 0.9792

About 97.92% of the variation in y is explained by the line.

14.Linear regression & correlation

Answer: x = 4.541

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 = 25/5 = 5
ȳ = 264/5 = 52.8

Sums of squares and products

Sxx = Σ(x − x̄)² = 10
Sxy = Σ(x − x̄)(y − ȳ) = 61

Regression line

b = 61/10 = 6.1
a = 52.8 − 6.1·5 = 22.3
ŷ = 22.3 + 6.1·x

Solve for x

x = (50 − 22.3)/6.1 = 4.541

Answer

x = 4.541
15.Chi-square tests

Answer: χ² = 1.786

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: 60·32/160 = 12, 60·72/160 = 27, 60·56/160 = 21
night shift: 100·32/160 = 20, 100·72/160 = 45, 100·56/160 = 35

Chi-square statistic

χ² = (10 − 12)²/12 + (31 − 27)²/27 + (19 − 21)²/21 + (22 − 20)²/20 + (41 − 45)²/45 + (37 − 35)²/35
χ² = 1.786

Answer

χ² = 1.786

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

16.Chi-square tests

Answer: df = 10

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 = 11 − 1 = 10

Answer

df = 10

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