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

The standard error of the mean becomes smaller when:

  1. AThe mean increases
  2. BThe sample size decreases
  3. CThe variable is measured at ordinal level
  4. DThe sample size increases
2.Research methods in psychologyHard

A study shows that a questionnaire correlates strongly with an established test of the same construct. This supports:

  1. ATest–retest reliability
  2. BFace validity
  3. CEcological validity
  4. DConcurrent validity
3.Descriptive statisticsMedium
1520253035404550

Find the sample standard deviation s of the data: 46, 11, 29, 41, 54, 51.

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

A quantity is normally distributed with mean 70 and standard deviation 5. What proportion P of values is below 62.5?

μ = 70σ = 5x = 62.5
P =
5.ProbabilityMedium
red: 7blue: 4

A bag holds 7 red and 4 blue balls. You draw two balls without putting the first one back. What is the probability P that both are red?

7 / 4
P =
6.ProbabilityHard
€p(x)−10.7530.2190.05

A game costs €1 to play. You win €4 with probability 0.2 and €20 with probability 0.05; otherwise you win nothing. The stake is not returned. Find the expected net gain E(G) per game.

stake = €1€4: p = 0.2€20: p = 0.05
E(G) = €
7.Normal distributionMedium
490520P = ?

The battery lifetimes are normally distributed with mean 500 h and standard deviation 40 h. What fraction lies between 490 and 520 h? Give P as a decimal.

μ = 500σ = 40x1 = 490x2 = 520
P =
8.Normal distributionHard
10 g0X − Y?

The mass X of a product from line 1 and Y from line 2 are independent and normal: X has mean 500 g and standard deviation 8 g, Y has mean 490 g and standard deviation 6 g. Find the probability P that X is greater than Y. Round z to two decimals.

μX = 500 g, σX = 8 gμY = 490 g, σY = 6 g
P =
9.Confidence intervalsMedium
n = 12, s = 16.5LUx̄ = 37E = ?90%

A sample of 12 items has mean 37 and standard deviation s = 16.5. The population is roughly normal. Find the margin of error E of a 90% confidence interval for the mean.

n = 12x̄ = 37s = 16.590%t(11) = 1.796
E =
10.Confidence intervalsHard
σ = 13, n = ?x̄ − Ex̄ + Ex̄E = 1.595%

The population standard deviation is 13. How large a sample n do you need so that a 95% confidence interval for the mean has a margin of error of at most 1.5?

σ = 13E = 1.595%
n =
11.Hypothesis tests & confidence intervalsMedium
t −2.0302.03

A sample of 36 students has a mean score of 97.75 and a sample standard deviation of 15. Find the test statistic t for H₀: μ = 100.

μ₀ = 100x̄ = 97.75s = 15n = 36
t =
12.Hypothesis tests & confidence intervalsHard
σ = 12, n = ?x̄ − Ex̄ + Ex̄E = 299%

How large must a sample be so that the 99% confidence interval for a mean has a margin of error of at most 2? The standard deviation is 12. Find the smallest sample size n.

σ = 12E = 299%
n =
13.Linear regression & correlationMedium
xy1669188520942211224119xy2080100120

The table gives 5 paired observations of x = temperature in °C and y = ice creams sold. Their sums are given. Find Pearson's correlation coefficient r from these sums.

n = 5Σx = 100Σy = 479Σx² = 2040Σy² = 47527Σxy = 9834
r =
14.Linear regression & correlationHard
xy2841061381710211228xy510102030x = 5

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 = 5.

n = 6x = 5
ŷ =
15.Chi-square testsMedium
improvednot improvedTotalnew drug421860placebo5763120Total9981180

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 × 2
χ² =
16.Chi-square testsHard
passfailTotalline 1202040line 2312960line 333740Total8456140dfα = 0.05α = 0.0113.8416.63525.9919.21037.81511.34549.48813.277511.07015.086612.59216.812714.06718.475815.50720.090916.91921.6661018.30723.209

The table shows observed counts. Test at α = 0.01 whether the two variables are independent (no continuity correction). Compute the test statistic χ², then compare it with the critical value from the table.

N = 1403 × 2α = 0.01
χ² =

3Solutions

1.Research methods in psychology

Answer: D · The sample size increases

Why this is the answer

The standard error estimates how much sample means vary around the population mean. Larger samples give more precise estimates.

Answer

D · The sample size increases
2.Research methods in psychology

Answer: D · Concurrent validity

Why this is the answer

Concurrent validity is shown when a new measure agrees with an established measure taken at about the same time.

Answer

D · Concurrent validity
3.Descriptive statistics

Answer: s = 16.16

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 = 232/6 = 38.67

Squared deviations

Σ(x − x̄)² = 7.333² + (−27.67)² + (−9.667)² + 2.333² + 15.33² + 12.33² = 1305

Standard deviation

s = √(1305/5) = 16.16

Answer

s = 16.16
4.Descriptive statistics

Answer: P = 0.06681

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 = −1.5

Standard normal table

Φ(−1.5) = 0.06681
P = 0.06681

Answer

P = 0.06681
5.Probability

Answer: P = 0.3818

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 = 7/11 · 6/10 = 42/110 = 0.3818

Answer

P = 0.3818
6.Probability

Answer: E(G) = 0.8 €

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.

Net outcomes

G: 3 (p = 0.2), 19 (p = 0.05), −1 (p = 0.75)

Expected value

E(G) = 3·0.2 + 19·0.05 − 1·0.75 = 0.8 €

Answer

E(G) = 0.8 €

A positive value means you win money on average.

7.Normal distribution

Answer: P = 0.2902

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 = (490 − 500)/40 = −0.25
z₂ = (520 − 500)/40 = 0.5

Area between

P = Φ(0.5) − Φ(−0.25) = 0.6915 − 0.4013 = 0.2902

Answer

P = 0.2902
8.Normal distribution

Answer: P = 0.8413

Sums and differences of normal variables: if X and Y are independent and normal, X + Y and X − Y are normal too. Means add or subtract: E(X ± Y) = μX ± μY. Variances always add, also for a difference: Var(X ± Y) = σX² + σY², so σ = √(σX² + σY²). Standard deviations never add directly. The total of n independent items with the same distribution has mean n·μ and standard deviation σ·√n. Then standardise as usual: z = (x − mean)/σ and read Φ(z).

Distribution of the difference

μ = 500 − 490 = 10 g, σ = 10 g

Standardise

z = (0 − 10)/10 = −1

Probability

P = 1 − Φ(−1) = 0.8413

Answer

P = 0.8413
9.Confidence intervals

Answer: E = 8.555

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 = 11, t = 1.796

Margin of error

E = 1.796·16.5/√12 = 8.555

Answer

E = 8.555
10.Confidence intervals

Answer: n = 289

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 = (1.96·13/1.5)² = 288.5

Round up

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

n = 289

Answer

n = 289
11.Hypothesis tests & confidence intervals

Answer: t = −0.9

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 = 15/√36 = 15/6 = 2.5

Test statistic

t = (97.75 − 100)/2.5 = −0.9

Answer

t = −0.9
t0.025; df = 35 = 2.03

|t| < 2.03, so H₀ is not rejected at the 5% level (two-sided).

12.Hypothesis tests & confidence intervals

Answer: n = 239

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 = (2.576·12/2)² = 238.9

Round up

n = 239

Answer

n = 239
13.Linear regression & correlation

Answer: r = 0.9921

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 = 9834 − 100·479/5 = 254
Sxx = 2040 − 100²/5 = 40
Syy = 47527 − 479²/5 = 1639

Correlation coefficient

r = 254/√(40·1639) = 0.9921

Answer

r = 0.9921
14.Linear regression & correlation

Answer: ŷ = 12.25

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 = 97/6 = 16.17

Sums of squares and products

Sxx = Σ(x − x̄)² = 70
Sxy = Σ(x − x̄)(y − ȳ) = 137

Regression line

b = 137/70 = 1.957
a = 16.17 − 1.957·7 = 2.467
ŷ = 2.467 + 1.957·x

Prediction

ŷ = 2.467 + 1.957·5 = 12.25

Answer

ŷ = 12.25
15.Chi-square tests

Answer: χ² = 8.182

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

new drug: 60·99/180 = 33, 60·81/180 = 27
placebo: 120·99/180 = 66, 120·81/180 = 54

Chi-square statistic

χ² = (42 − 33)²/33 + (18 − 27)²/27 + (57 − 66)²/66 + (63 − 54)²/54
χ² = 8.182

Answer

χ² = 8.182

df = 1. At α = 0.05 the critical value is 3.841, so H₀ (independence) is rejected.

16.Chi-square tests

Answer: χ² = 11.84

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

line 1: 24, 16
line 2: 36, 24
line 3: 24, 16

Chi-square statistic

χ² = (20 − 24)²/24 + (20 − 16)²/16 + (31 − 36)²/36 + (29 − 24)²/24 + (33 − 24)²/24 + (7 − 16)²/16
χ² = 11.84

Decision

df = 2, χ²crit = 9.210
11.84 > 9.210

Answer

χ² = 11.84

χ² is above the critical value, so H₀ is rejected: the variables appear to be related.

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