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 problems Back 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)·p^k(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

Convert the value to a z-score: z = (x − μ)/σ.
Look up Φ(z), the area to the left of z.
For the area to the right use 1 − Φ(z); for an interval subtract the two left areas.
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

Decide between z (population σ known) and t (only the sample s, with n − 1 degrees of freedom).
Find the critical value for the confidence level.
Margin of error: E = critical value × standard deviation/√n.
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

State H₀ and H₁ and decide whether the test is one- or two-sided.
Calculate the test statistic: (x̄ − μ₀)/(σ/√n), or with s and t.
Compare with the critical value, or find the p-value.
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: S_xx = Σ(x − x̄)² = Σx² − (Σx)²/n, S_xy = Σxy − ΣxΣy/n
Slope: b = S_xy/S_xx
Intercept: a = ȳ − b·x̄
Correlation: r = S_xy/√(S_xxS_yy)
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 > χ²_crit

2Practice problems

1.Research methods in psychologyMedium

In a perfectly normal distribution:

AMost scores lie in the tails
BThe mode is the lowest score
CThe mean is always larger than the median
DThe mean, median and mode are equal

2.Research methods in psychologyHard

A self-report scale finds that people who agree with many statements score high regardless of content. Which response bias is this, and how can it be reduced?

AOrder effects; use counterbalancing of participants
BAcquiescence bias; include reverse-worded items
CSocial desirability bias; use a larger sample
DDemand characteristics; use a double-blind design

3.Descriptive statisticsMedium

Find the sample standard deviation s of the data: 48, 47, 15, 35, 28, 17, 12.

n = 7

s =

4.Descriptive statisticsHard

A quantity is normally distributed with mean 500 and standard deviation 50. What proportion P of values is below 475?

μ = 500σ = 50x = 475

P =

5.ProbabilityMedium

A bag holds 5 red and 3 blue balls. You draw two balls without putting the first one back. What is the probability P that both are red?

5 / 3

P =

6.ProbabilityHard

A test has 11 multiple-choice questions. A student who guesses gets each one right with probability 0.1, independently. What is the probability P of exactly 4 correct answers?

n = 11p = 0.1k = 4

P =

7.Normal distributionMedium

Drag z so that the shaded area under the standard normal curve, to the right of z, is 0.6736. Find z to two decimals.

A = 0.6736

z =

8.Normal distributionHard

The delivery times are normally distributed with mean 32 min and standard deviation 6 min. Find the value x below which 5% of them lie (the 5th percentile).

μ = 32σ = 6P = 5%

x = min

9.Confidence intervalsMedium

The blood pressure of 12 patients is measured before and after a treatment. The differences d = before − after (mmHg) have mean 8.3 and standard deviation 1.9. Find the upper limit of the 90% confidence interval for the mean difference.

n = 12d̄ = 8.3s_d = 1.990%t(11) = 1.796

upper limit = mmHg

10.Confidence intervalsHard

Two fertilisers are tested. Plot group 1: 5 plots, mean yield 22 kg, standard deviation 10 kg. Plot group 2: 7 plots, mean 23.5 kg, standard deviation 4 kg. Assume equal variances (pooled t, df = n₁ + n₂ − 2). Find the upper limit of the 95% confidence interval for μ₁ − μ₂.

n₁ = 5x̄₁ = 22s₁ = 10n₂ = 7x̄₂ = 23.5s₂ = 495%t(10) = 2.228

upper limit = kg

11.Hypothesis tests & confidence intervalsMedium

A sample of 16 students has a mean score of 106 and a sample standard deviation of 20. Find the test statistic t for H₀: μ = 100.

μ₀ = 100x̄ = 106s = 20n = 16

t =

12.Hypothesis tests & confidence intervalsHard

A sample of 49 measurements has mean 118.3. The population standard deviation is 5. Find the upper limit U of the 90% confidence interval for the mean.

x̄ = 118.3σ = 5n = 4990%

U =

13.Linear regression & correlationMedium

The table gives 6 paired observations of x = advertising spend in k€ and y = sales in k€. Their sums are given. Find Pearson's correlation coefficient r from these sums.

n = 6Σx = 48Σy = 473Σx² = 454Σy² = 39259Σxy = 4153

r =

14.Linear regression & correlationHard

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. For which value x does the line predict ŷ = 97?

n = 6ŷ = 97

x =

15.Chi-square testsMedium

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 × 2

χ² =

16.Chi-square testsHard

A chi-square goodness-of-fit test checks whether counts in 7 categories follow a Poisson distribution whose mean is estimated from the same data. How many degrees of freedom df does the test have?

k = 7

df =

3Solutions

1.Research methods in psychology

Answer: D · The mean, median and mode are equal

Why this is the answer

A normal distribution is symmetrical and bell-shaped, so all three measures of central tendency fall at the centre.

Answer

D · The mean, median and mode are equal

2.Research methods in psychology

Answer: B · Acquiescence bias; include reverse-worded items

Why this is the answer

Acquiescence is a tendency to agree with items. Mixing positively and negatively worded items means that agreeing with everything no longer produces an extreme score.

Answer

B · Acquiescence bias; include reverse-worded items

3.Descriptive statistics

Answer: s = 15

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 = 202/7 = 28.86

Squared deviations

Σ(x − x̄)² = 19.14² + 18.14² + (−13.86)² + 6.143² + (−0.8571)² + (−11.86)² + (−16.86)² = 1351

Standard deviation

s = √(1351/6) = 15

Answer

s = 15

4.Descriptive statistics

Answer: P = 0.3085

Standardise

z = −0.5

Standard normal table

Φ(−0.5) = 0.3085

P = 0.3085

Answer

P = 0.3085

5.Probability

Answer: P = 0.3571

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.

First and second draw

P = 5/8 · 4/7 = 20/56 = 0.3571

Answer

P = 0.3571

6.Probability

Answer: P = 0.01578

Binomial formula

P(X = 4) = 330·0.1⁴·0.9⁷ = 330·0.0001·0.4783 = 0.01578

Answer

P = 0.01578

7.Normal distribution

Answer: z = −0.45

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 − 0.6736 = 0.3264

Read the z-table backwards

Find 0.3264 inside the table and read off z.

z = −0.45

Answer

z = −0.45

8.Normal distribution

Answer: x = 22.13 min

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) = 0.05 → z = −1.645

Back to x

x = 32 + −1.645·6 = 22.13 min

Answer

x = 22.13 min

9.Confidence intervals

Answer: upper limit = 9.285 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 = 11, t = 1.796

Margin of error

E = 1.796·1.9/√12 = 0.9851

Interval

8.3 ± 0.9851 → [7.315, 9.285]

Answer

upper limit = 9.285 mmHg

10.Confidence intervals

Answer: upper limit = 7.688 kg

Pooled standard deviation

s_p² = (4·10² + 6·4²)/10 = 49.6

s_p = 7.043

Margin of error

E = 2.228·7.043·√(1/5 + 1/7) = 9.188

Interval

−1.5 ± 9.188 → [−10.69, 7.688]

Answer

upper limit = 7.688 kg

11.Hypothesis tests & confidence intervals

Answer: t = 1.2

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/√16 = 20/4 = 5

Test statistic

t = (106 − 100)/5 = 1.2

Answer

t = 1.2

t_{0.025; df = 15} = 2.131

|t| < 2.131, so H₀ is not rejected at the 5% level (two-sided).

12.Hypothesis tests & confidence intervals

Answer: U = 119.5

Margin of error

E = 1.645·5/√49 = 1.175

Interval limit

U = 118.3 + 1.175 = 119.5

Answer

U = 119.5

13.Linear regression & correlation

Answer: r = 0.9935

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

S_xy = 4153 − 48·473/6 = 369

S_xx = 454 − 48²/6 = 70

S_yy = 39259 − 473²/6 = 1971

Correlation coefficient

r = 369/√(70·1971) = 0.9935

Answer

r = 0.9935

14.Linear regression & correlation

Answer: x = 24.91

Means

x̄ = 138/6 = 23

ȳ = 524/6 = 87.33

Sums of squares and products

S_xx = Σ(x − x̄)² = 70

S_xy = Σ(x − x̄)(y − ȳ) = 354

Regression line

b = 354/70 = 5.057

a = 87.33 − 5.057·23 = −28.98

ŷ = −28.98 + 5.057·x

Solve for x

x = (97 − (−28.98))/5.057 = 24.91

Answer

x = 24.91

15.Chi-square tests

Answer: χ² = 2.344

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

smokers: 40·104/160 = 26, 40·56/160 = 14

non-smokers: 120·104/160 = 78, 120·56/160 = 42

Chi-square statistic

χ² = (30 − 26)²/26 + (10 − 14)²/14 + (74 − 78)²/78 + (46 − 42)²/42

χ² = 2.344

Answer

χ² = 2.344

df = 1. At α = 0.05 the critical value is 3.841, so H₀ (independence) is not rejected.

16.Chi-square tests

Answer: df = 5

Degrees of freedom

df = 7 − 1 − 1 = 5

Answer

df = 5

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