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

In observational research, event sampling means:

  1. ARecording every time a specified behaviour occurs
  2. BRecording behaviour only at fixed time intervals
  3. CChoosing participants at random
  4. DRecording only events the participants report
2.Research methods in psychologyHard

A participant takes a drug or a sugar pill but does not know which, while the researcher does. What design is this and what bias remains possible?

  1. AOpen trial; placebo effects are fully controlled
  2. BSingle-blind; experimenter bias is still possible
  3. CSingle-blind; demand characteristics are impossible
  4. DDouble-blind; no bias is possible
3.Descriptive statisticsMedium
202530354045505560

Find the sample standard deviation s of the data: 25, 20, 60, 18, 55, 52.

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

A quantity is normally distributed with mean 70 and standard deviation 15. What proportion P of values is above 40?

μ = 70σ = 15x = 40
P =
5.ProbabilityMedium
xp(x)00.3510.3520.0530.1540.1μ = 1.3

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

x: 0, 1, 2, 3, 4p(x): 0.35, 0.35, 0.05, 0.15, 0.1
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 between −z and z = 0.6827

Drag z so that the shaded area between −z and z under the standard normal curve is 0.6827. Find z to two decimals.

A = 0.6827
z =
8.Normal distributionHard
380420460500540580620htop 70%

The battery lifetimes are normally distributed with mean 500 h and standard deviation 40 h. Drag the cut-off so the shaded top 70% is to the right of it. Find the cut-off x.

μ = 500 hσ = 40 h70%
x = h
9.Confidence intervalsMedium
n = 30, s = 12.5LU = ?x̄ = 64E99%

A sample of 30 items has mean 64 and standard deviation s = 12.5. The population is roughly normal. Find the upper limit of the 99% confidence interval for the mean.

n = 30x̄ = 64s = 12.599%t(29) = 2.756
upper limit =
10.Confidence intervalsHard
σ = 23, n = ?x̄ − Ex̄ + Ex̄E = 199%

The population standard deviation is 23. 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 1?

σ = 23E = 199%
n =
11.Hypothesis tests & confidence intervalsMedium
t −1.9901.99

A sample of 81 students has a mean score of 76.51 and a sample standard deviation of 8. Find the test statistic t for H₀: μ = 75.

μ₀ = 75x̄ = 76.51s = 8n = 81
t =
12.Hypothesis tests & confidence intervalsHard
σ = 10, 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 10. Find the smallest sample size n.

σ = 10E = 1.599%
n =
13.Linear regression & correlationMedium
xy138244349454557662xy246405060

The table gives 6 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 = 6
r² =
14.Linear regression & correlationHard
xy1657197722912510628123xy2025100x = 26.5

The table gives 5 paired observations of x = temperature in °C and y = ice creams sold. Use the least-squares line ŷ = a + b·x. Predict ŷ for x = 26.5.

n = 5x = 26.5
ŷ =
15.Chi-square testsMedium
busbikecarTotalstudents612910100staff19113060Total804040160

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 test of independence is carried out on a contingency table with 2 rows and 4 columns (not counting the totals). How many degrees of freedom df does the test have?

2 × 4
df =

3Solutions

1.Research methods in psychology

Answer: A · Recording every time a specified behaviour occurs

Why this is the answer

Event sampling counts each occurrence of target behaviours. Time sampling records what happens at set intervals, such as every 30 seconds.

Answer

A · Recording every time a specified behaviour occurs
2.Research methods in psychology

Answer: B · Single-blind; experimenter bias is still possible

Why this is the answer

In a single-blind design participants do not know their condition, which controls placebo expectations. The researcher's knowledge can still influence results, so double-blind designs are preferred.

Answer

B · Single-blind; experimenter bias is still possible
3.Descriptive statistics

Answer: s = 19.29

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 = 230/6 = 38.33

Squared deviations

Σ(x − x̄)² = (−13.33)² + (−18.33)² + 21.67² + (−20.33)² + 16.67² + 13.67² = 1861

Standard deviation

s = √(1861/5) = 19.29

Answer

s = 19.29
4.Descriptive statistics

Answer: P = 0.9772

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 = −2

Standard normal table

Φ(−2) = 0.02275
P = 1 − 0.02275 = 0.9772

Answer

P = 0.9772
5.Probability

Answer: Var(X) = 1.81

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 + 2·0.05 + 3·0.15 + 4·0.1 = 1.3

Mean of the squares

E(X²) = 0·0.35 + 1·0.35 + 4·0.05 + 9·0.15 + 16·0.1 = 3.5

Variance

Var(X) = 3.5 − 1.3² = 1.81

Answer

Var(X) = 1.81
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

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 + 0.6827)/2 = 0.8413

Read the z-table backwards

Find 0.8413 inside the table and read off z.

z = 1.00

Answer

z = 1.00
8.Normal distribution

Answer: x = 479 h

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) = 0.3

z from the table

z ≈ −0.52

Back to the original units

x = 500 − 0.52·40 = 479 h

Answer

x = 479 h
9.Confidence intervals

Answer: upper limit = 70.29

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 = 29, t = 2.756

Margin of error

E = 2.756·12.5/√30 = 6.29

Interval

64 ± 6.29 → [57.71, 70.29]

Answer

upper limit = 70.29
10.Confidence intervals

Answer: n = 3511

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 = (2.576·23/1)² = 3510

Round up

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

n = 3511

Answer

n = 3511
11.Hypothesis tests & confidence intervals

Answer: t = 1.699

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/√81 = 8/9 = 0.8889

Test statistic

t = (76.51 − 75)/0.8889 = 1.699

Answer

t = 1.699
t0.025; df = 80 = 1.99

|t| < 1.99, so H₀ is not rejected at the 5% level (two-sided).

12.Hypothesis tests & confidence intervals

Answer: n = 295

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 = (2.576·10/1.5)² = 294.9

Round up

n = 295

Answer

n = 295
13.Linear regression & correlation

Answer: r² = 0.992

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 = 304/6 = 50.67

Sums of squares and products

Sxx = Σ(x − x̄)² = 17.5
Sxy = Σ(x − x̄)(y − ȳ) = 82
Syy = Σ(y − ȳ)² = 387.3

Coefficient of determination

r² = 82²/(17.5·387.3) = 0.992

Answer

r² = 0.992

About 99.2% of the variation in y is explained by the line.

14.Linear regression & correlation

Answer: ŷ = 115

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 = 110/5 = 22
ȳ = 454/5 = 90.8

Sums of squares and products

Sxx = Σ(x − x̄)² = 90
Sxy = Σ(x − x̄)(y − ȳ) = 483

Regression line

b = 483/90 = 5.367
a = 90.8 − 5.367·22 = −27.27
ŷ = −27.27 + 5.367·x

Prediction

ŷ = −27.27 + 5.367·26.5 = 115

Answer

ŷ = 115
15.Chi-square tests

Answer: χ² = 32.16

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

students: 100·80/160 = 50, 100·40/160 = 25, 100·40/160 = 25
staff: 60·80/160 = 30, 60·40/160 = 15, 60·40/160 = 15

Chi-square statistic

χ² = (61 − 50)²/50 + (29 − 25)²/25 + (10 − 25)²/25 + (19 − 30)²/30 + (11 − 15)²/15 + (30 − 15)²/15
χ² = 32.16

Answer

χ² = 32.16

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

16.Chi-square tests

Answer: df = 3

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 = (2 − 1)·(4 − 1) = 3

Answer

df = 3

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