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 a one-tailed hypothesis
  2. BUsing only volunteers
  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

Why do many researchers argue that p-values should be reported together with effect sizes and confidence intervals?

  1. AConfidence intervals are only needed in qualitative research
  2. BP-values show the probability the hypothesis is true
  3. CEffect sizes make sample size irrelevant to significance
  4. DA p-value does not show how large or precise the effect is
3.Descriptive statisticsMedium
2530354045

Find the sample standard deviation s of the data: 49, 26, 31, 33, 27, 25.

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

A quantity is normally distributed with mean 100 and standard deviation 5. What proportion P of values is above 110?

μ = 100σ = 5x = 110
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
n = 12, p = 0.25: P(X ≤ 1) = ?k0123456789101112

12 components are tested, and each one fails with probability 0.25, independently. What is the probability P that at most one fails?

n = 12p = 0.25
P =
7.Normal distributionMedium
6085P = ?

The exam scores are normally distributed with mean 65 points and standard deviation 10 points. What fraction lies between 60 and 85 points? Give P as a decimal.

μ = 65σ = 10x1 = 60x2 = 85
P =
8.Normal distributionHard
35455565758595pointsbottom 85%

The exam scores are normally distributed with mean 65 points and standard deviation 10 points. Drag the cut-off so the shaded bottom 85% is to the left of it. Find the cut-off x.

μ = 65 pointsσ = 10 points85%
x = points
9.Confidence intervalsMedium
n = 15, s = 13.5LUx̄ = 42.5E = ?95%

A sample of 15 items has mean 42.5 and standard deviation s = 13.5. The population is roughly normal. Find the margin of error E of a 95% confidence interval for the mean.

n = 15x̄ = 42.5s = 13.595%t(14) = 2.145
E =
10.Confidence intervalsHard
p ≈ 0.8, n = ?p̂ − Ep̂ + Ep̂E = 0.0290%

A pilot study suggests that about 0.8 of products are defective. How many products n must be inspected so that a 90% confidence interval for the proportion has a margin of error of at most 0.02?

p ≈ 0.8E = 0.0290%
n =
11.Hypothesis tests & confidence intervalsMedium
t −2.06402.064

A sample of 25 students has a mean score of 77.1 and a sample standard deviation of 5. Find the test statistic t for H₀: μ = 75.

μ₀ = 75x̄ = 77.1s = 5n = 25
t =
12.Hypothesis tests & confidence intervalsHard
−1.961.96z = ?

In a trial, 75 of 300 patients on treatment A recover, and 37 of 120 patients on treatment B. Using the pooled proportion, find the test statistic z for H₀: p₁ = p₂.

x₁ = 75n₁ = 300x₂ = 37n₂ = 120
z =
13.Linear regression & correlationMedium
xy244106148171022xy5101020

The table gives 5 paired observations of x = load in kN and y = extension in mm. Their sums are given. Find Pearson's correlation coefficient r from these sums.

n = 5Σx = 30Σy = 67Σx² = 220Σy² = 1085Σxy = 488
r =
14.Linear regression & correlationHard
xy2394546628741091xy510406080x = 7

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. Predict ŷ for x = 7.

n = 5x = 7
ŷ =
15.Chi-square testsMedium
yesnoTotalmen8337120women374380Total12080200

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 = 2002 × 2
χ² =
16.Chi-square testsHard

A chi-square test of independence is carried out on a contingency table with 2 rows and 6 columns (not counting the totals). How many degrees of freedom df does the test have?

2 × 6
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: D · A p-value does not show how large or precise the effect is

Why this is the answer

A p-value depends heavily on sample size and only indicates how surprising the data are under the null. Effect sizes show magnitude and intervals show precision.

Answer

D · A p-value does not show how large or precise the effect is
3.Descriptive statistics

Answer: s = 8.954

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 = 191/6 = 31.83

Squared deviations

Σ(x − x̄)² = 17.17² + (−5.833)² + (−0.8333)² + 1.167² + (−4.833)² + (−6.833)² = 400.8

Standard deviation

s = √(400.8/5) = 8.954

Answer

s = 8.954
4.Descriptive statistics

Answer: P = 0.02275

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) = 0.9772
P = 1 − 0.9772 = 0.02275

Answer

P = 0.02275
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: P = 0.1584

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.

Zero and one failure

P(X = 0) = 0.7512 = 0.03168
P(X = 1) = 12·0.25·0.7511 = 0.1267

Add them

P = 0.03168 + 0.1267 = 0.1584

Answer

P = 0.1584
7.Normal distribution

Answer: P = 0.6687

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 = (60 − 65)/10 = −0.5
z₂ = (85 − 65)/10 = 2

Area between

P = Φ(2) − Φ(−0.5) = 0.9772 − 0.3085 = 0.6687

Answer

P = 0.6687
8.Normal distribution

Answer: x = 75.4 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) = 0.85

z from the table

z ≈ 1.04

Back to the original units

x = 65 + 1.04·10 = 75.4 points

Answer

x = 75.4 points
9.Confidence intervals

Answer: E = 7.477

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 = 14, t = 2.145

Margin of error

E = 2.145·13.5/√15 = 7.477

Answer

E = 7.477
10.Confidence intervals

Answer: n = 1083

Confidence interval for a proportion: p̂ ± z·√(p̂(1 − p̂)/n), where p̂ = x/n. To reach a margin E you need n ≥ z²·p(1 − p)/E², rounded up; without a prior estimate use p = 0.5, which gives the largest n.

Solve for n

n = 1.645²·0.8·0.2/0.02² = 1082

Round up

n = 1083

Answer

n = 1083
11.Hypothesis tests & confidence intervals

Answer: t = 2.1

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 = 5/√25 = 5/5 = 1

Test statistic

t = (77.1 − 75)/1 = 2.1

Answer

t = 2.1
t0.025; df = 24 = 2.064

|t| > 2.064, so H₀ is rejected at the 5% level (two-sided).

12.Hypothesis tests & confidence intervals

Answer: z = −1.221

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.

Sample proportions

p̂₁ = 75/300 = 0.25
p̂₂ = 37/120 = 0.3083
p̂ = 112/420 = 0.2667

Standard error under H₀

SE = √(0.2667·0.7333·(1/300 + 1/120)) = 0.04776

Test statistic

z = (0.25 − 0.3083)/0.04776 = −1.221

Answer

z = −1.221

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

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 = 488 − 30·67/5 = 86
Sxx = 220 − 30²/5 = 40
Syy = 1085 − 67²/5 = 187.2

Correlation coefficient

r = 86/√(40·187.2) = 0.9938

Answer

r = 0.9938
14.Linear regression & correlation

Answer: ŷ = 70.2

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 = 30/5 = 6
ȳ = 320/5 = 64

Sums of squares and products

Sxx = Σ(x − x̄)² = 40
Sxy = Σ(x − x̄)(y − ȳ) = 248

Regression line

b = 248/40 = 6.2
a = 64 − 6.2·6 = 26.8
ŷ = 26.8 + 6.2·x

Prediction

ŷ = 26.8 + 6.2·7 = 70.2

Answer

ŷ = 70.2
15.Chi-square tests

Answer: χ² = 10.5

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

men: 120·120/200 = 72, 120·80/200 = 48
women: 80·120/200 = 48, 80·80/200 = 32

Chi-square statistic

χ² = (83 − 72)²/72 + (37 − 48)²/48 + (37 − 48)²/48 + (43 − 32)²/32
χ² = 10.5

Answer

χ² = 10.5

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

16.Chi-square tests

Answer: df = 5

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)·(6 − 1) = 5

Answer

df = 5

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