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

Experimenter bias is reduced most directly by:

  1. AUsing only volunteers
  2. BUsing a one-tailed hypothesis
  3. CUsing a researcher who does not know the hypothesis or conditions to collect data
  4. DUsing a larger sample
2.Research methods in psychologyHard

A 2 × 3 factorial ANOVA shows that a drug improves memory in young adults but not in older adults. What has been found?

  1. AOnly a main effect of age
  2. BAn interaction between drug and age
  3. CA floor effect
  4. DA Type I error
3.Descriptive statisticsMedium
2025303540455055

Find the sample standard deviation s of the data: 17, 44, 59, 32, 17.

n = 5
s =
4.Descriptive statisticsHard
z −2−1012

A quantity is normally distributed with mean 70 and standard deviation 12. What proportion P of values is above 88?

μ = 70σ = 12x = 88
P =
5.ProbabilityMedium
red: 7blue: 3

A bag holds 7 red and 3 blue balls. You draw two balls without putting the first one back. What is the probability P that both are red?

7 / 3
P =
6.ProbabilityHard
xp(x)10.120.0530.340.250.35μ

X has the distribution shown, and Y = 10X + 1. Find the standard deviation σY of Y.

x: 1, 2, 3, 4, 5p(x): 0.1, 0.05, 0.3, 0.2, 0.35Y = 10X + 1
σY =
7.Normal distributionMedium
178181.85P = ?

The heights of adult men are normally distributed with mean 178 cm and standard deviation 7 cm. What fraction lies between 178 and 181.85 cm? Give P as a decimal.

μ = 178σ = 7x1 = 178x2 = 181.85
P =
8.Normal distributionHard
10 cm0X − Y?

The heights X of men and Y of women in a group are independent and normal: X has mean 180 cm and standard deviation 10 cm, Y has mean 170 cm and standard deviation 12 cm. Find the probability P that X is greater than Y. Round z to two decimals.

μX = 180 cm, σX = 10 cmμY = 170 cm, σY = 12 cm
P =
9.Confidence intervalsMedium
n = 21, s = 4.5LUx̄ = 68.5E = ?95%

A sample of 21 items has mean 68.5 and standard deviation s = 4.5. The population is roughly normal. Find the margin of error E of a 95% confidence interval for the mean.

n = 21x̄ = 68.5s = 4.595%t(20) = 2.086
E =
10.Confidence intervalsHard
σ = 39, n = ?x̄ − Ex̄ + Ex̄E = 0.590%

The population standard deviation is 39. 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 0.5?

σ = 39E = 0.590%
n =
11.Hypothesis tests & confidence intervalsMedium
−1.961.96z = ?

Two production lines make the same part. Line 1: a sample of 60 parts has mean length 53 mm, and the line's known standard deviation is 10 mm. Line 2: a sample of 30 parts has mean 54.8 mm, known standard deviation 7 mm. Find the test statistic z for H₀: μ₁ = μ₂, taking the difference as x̄₁ − x̄₂ (line 1 minus line 2).

n₁ = 60x̄₁ = 53 mmσ₁ = 10 mmn₂ = 30x̄₂ = 54.8 mmσ₂ = 7 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 = 0.10 with a sample of 25; the standard deviation is known to be 2.5. In reality the mean is 50.85. Find the power 1 − β of the test in percent.

μ₀ = 50μ = 50.85σ = 2.5n = 25α = 0.10
1 − β = %
13.Linear regression & correlationMedium
xy166319722287251092811931139xy15202530100

The table gives 6 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 = 6Σx = 141Σy = 589Σx² = 3471Σy² = 62085Σxy = 14656
r =
14.Linear regression & correlationHard
xy146017712088239126117xy1520256080100120x = 15.5

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 = 15.5.

n = 5x = 15.5
ŷ =
15.Chi-square testsMedium
lowmediumhighTotalurban651619100rural15242160Total804040160

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 4 rows and 4 columns (not counting the totals). How many degrees of freedom df does the test have?

4 × 4
df =

3Solutions

1.Research methods in psychology

Answer: C · Using a researcher who does not know the hypothesis or conditions to collect data

Why this is the answer

Experimenters who know the hypothesis may unintentionally influence participants or recordings. Single- or double-blind procedures and standardised instructions reduce this.

Answer

C · Using a researcher who does not know the hypothesis or conditions to collect data
2.Research methods in psychology

Answer: B · An interaction between drug and age

Why this is the answer

An interaction occurs when the effect of one factor differs across levels of another. Here the drug effect depends on age group.

Answer

B · An interaction between drug and age
3.Descriptive statistics

Answer: s = 18.07

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 = 169/5 = 33.8

Squared deviations

Σ(x − x̄)² = (−16.8)² + 10.2² + 25.2² + (−1.8)² + (−16.8)² = 1307

Standard deviation

s = √(1307/4) = 18.07

Answer

s = 18.07
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) = 0.9332
P = 1 − 0.9332 = 0.06681

Answer

P = 0.06681
5.Probability

Answer: P = 0.4667

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 = 7/10 · 6/9 = 42/90 = 0.4667

Answer

P = 0.4667
6.Probability

Answer: σY = 12.76

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 = 3.65, Var(X) = 1.627, σX = 1.276

Linear change

σY = |10|·1.276 = 12.76

Answer

σY = 12.76

The mean does move: E(Y) = 10·3.65 + 1 = 37.5.

7.Normal distribution

Answer: P = 0.2088

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 = (178 − 178)/7 = 0
z₂ = (181.85 − 178)/7 = 0.55

Area between

P = Φ(0.55) − Φ(0) = 0.7088 − 0.5 = 0.2088

Answer

P = 0.2088
8.Normal distribution

Answer: P = 0.7389

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 = 180 − 170 = 10 cm, σ = 15.62 cm

Standardise

z = (0 − 10)/15.62 = −0.64

Probability

P = 1 − Φ(−0.64) = 0.7389

Answer

P = 0.7389
9.Confidence intervals

Answer: E = 2.048

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 = 20, t = 2.086

Margin of error

E = 2.086·4.5/√21 = 2.048

Answer

E = 2.048
10.Confidence intervals

Answer: n = 16464

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 = (1.645·39/0.5)² = 16463

Round up

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

n = 16464

Answer

n = 16464
11.Hypothesis tests & confidence intervals

Answer: z = −0.9909

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 = √(10²/60 + 7²/30) = √3.3 = 1.817

Test statistic

z = (53 − 54.8)/1.817 = −0.9909

Answer

z = −0.9909

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

12.Hypothesis tests & confidence intervals

Answer: 1 − β = 66.2 %

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 = 2.5/√25 = 0.5
x̄c = 50 + 1.282·0.5 = 50.64

z under the true mean

z = (50.64 − 50.85)/0.5 = −0.418

Power

power = P(Z > −0.418) = 66.2%
1 − β = 66.2%

Answer

1 − β = 66.2%
13.Linear regression & correlation

Answer: r = 0.9938

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 = 14656 − 141·589/6 = 814.5
Sxx = 3471 − 141²/6 = 157.5
Syy = 62085 − 589²/6 = 4265

Correlation coefficient

r = 814.5/√(157.5·4265) = 0.9938

Answer

r = 0.9938
14.Linear regression & correlation

Answer: ŷ = 65.3

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 = 100/5 = 20
ȳ = 427/5 = 85.4

Sums of squares and products

Sxx = Σ(x − x̄)² = 90
Sxy = Σ(x − x̄)(y − ȳ) = 402

Regression line

b = 402/90 = 4.467
a = 85.4 − 4.467·20 = −3.933
ŷ = −3.933 + 4.467·x

Prediction

ŷ = −3.933 + 4.467·15.5 = 65.3

Answer

ŷ = 65.3
15.Chi-square tests

Answer: χ² = 24.48

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

Chi-square statistic

χ² = (65 − 50)²/50 + (16 − 25)²/25 + (19 − 25)²/25 + (15 − 30)²/30 + (24 − 15)²/15 + (21 − 15)²/15
χ² = 24.48

Answer

χ² = 24.48

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

16.Chi-square tests

Answer: df = 9

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 = (4 − 1)·(4 − 1) = 9

Answer

df = 9

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