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

Which test compares two independent groups when the data are ordinal or clearly not normal?

  1. AWilcoxon signed-rank test
  2. BIndependent samples t-test
  3. CMann–Whitney U test
  4. DChi-square test
2.Research methods in psychologyHard

A study reports t(28) = 2.50, p = .018 for two independent groups of equal size. How many participants took part?

  1. A56
  2. B29
  3. C30
  4. D28
3.Descriptive statisticsMedium
1520253035404550

Find the sample standard deviation s of the data: 21, 27, 54, 14, 21, 40.

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

Exam scores are normally distributed with mean 70 and standard deviation 10. Find the z-score z of a score of 50.

μ = 70σ = 10x = 50
z =
5.ProbabilityMedium
P(at least one six) = ??1?2?3?4?5?6P(six) = 1/6 on each roll

A fair die is rolled 6 times. What is the probability P of getting at least one six?

n = 6
P =
6.ProbabilityHard
n = 7, p = 0.6: P(X = 5) = ?k01234567

A test has 7 multiple-choice questions. A student who guesses gets each one right with probability 0.6, independently. What is the probability P of exactly 5 correct answers?

n = 7p = 0.6k = 5
P =
7.Normal distributionMedium
−3−2−10123zshaded area to the right of z = 0.9812

Drag z so that the shaded area under the standard normal curve, to the right of z, is 0.9812. Find z to two decimals.

A = 0.9812
z =
8.Normal distributionHard
μ = 250x = ?5%

The bag weights are normally distributed with mean 250 g and standard deviation 5 g. Find the value x below which 5% of them lie (the 5th percentile).

μ = 250σ = 5P = 5%
x = g
9.Confidence intervalsMedium
x̄E₁ = 2.290%, n = 25x̄E₂ = ?95%, n = 25

A 90% confidence interval for a mean, based on 25 observations and a known σ, has a margin of error of 2.2. What is the margin of error E₂ of a 95% interval from the same data?

90%n₁ = 25E₁ = 2.295%
E₂ =
10.Confidence intervalsHard
p = 0.5, n = ?p̂ − Ep̂ + Ep̂E = 0.0499%

Nothing is known in advance about the proportion of defective products. How many products n must be inspected so that a 99% confidence interval for the proportion has a margin of error of at most 0.04?

p = 0.5E = 0.0499%
n =
11.Hypothesis tests & confidence intervalsMedium
1.22p = ?

The mean of a population is claimed to be 100; the population standard deviation is known to be 10. A random sample of 100 observations has a mean of 101.22. Test H₀: μ = 100 against H₁: μ > 100. Find the p-value p.

μ₀ = 100σ = 10n = 100x̄ = 101.22H₁: μ > μ₀
p =
12.Hypothesis tests & confidence intervalsHard
μ₀x̄c

A one-sided z test of H₀: μ = 50 against H₁: μ < 50 is done at α = 0.01 with a sample of 36; the standard deviation is known to be 9. In reality the mean is 46.4. Find the power 1 − β of the test in percent.

μ₀ = 50μ = 46.4σ = 9n = 36α = 0.01
1 − β = %
13.Linear regression & correlationMedium
xy2841061682010251226xy5101020

The table gives 6 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 = 6Σx = 42Σy = 105Σx² = 364Σy² = 2121Σxy = 874
r =
14.Linear regression & correlationHard
xy352452559661761867xy468556065ŷ = 60

The table gives 6 paired observations of x = hours studied and y = exam score. Use the least-squares line ŷ = a + b·x. For which value x does the line predict ŷ = 60?

n = 6ŷ = 60
x =
15.Chi-square testsMedium
improvednot improvedTotalnew drug344680placebo4674120Total80120200

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 goodness-of-fit test checks whether counts in 6 categories follow a Poisson distribution whose mean is estimated from the same data. How many degrees of freedom df does the test have?

k = 6
df =

3Solutions

1.Research methods in psychology

Answer: C · Mann–Whitney U test

Why this is the answer

The Mann–Whitney U test compares ranks between two unrelated groups. The Wilcoxon signed-rank test is for related samples.

Answer

C · Mann–Whitney U test
2.Research methods in psychology

Answer: C · 30

Why this is the answer

Degrees of freedom of 28 mean 28 + 2 = 30 participants, 15 in each group.

Answer

C · 30
3.Descriptive statistics

Answer: s = 14.84

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 = 177/6 = 29.5

Squared deviations

Σ(x − x̄)² = (−8.5)² + (−2.5)² + 24.5² + (−15.5)² + (−8.5)² + 10.5² = 1102

Standard deviation

s = √(1102/5) = 14.84

Answer

s = 14.84
4.Descriptive statistics

Answer: z = −2

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 = (50 − 70)/10 = −2

Answer

z = −2
5.Probability

Answer: P = 0.6651

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)6 = 0.3349
P = 1 − 0.3349 = 0.6651

Answer

P = 0.6651
6.Probability

Answer: P = 0.2613

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 = 5) = 21·0.65·0.42 = 21·0.07776·0.16 = 0.2613

Answer

P = 0.2613
7.Normal distribution

Answer: z = −2.08

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 − 0.9812 = 0.0188

Read the z-table backwards

Find 0.0188 inside the table and read off z.

z = −2.08

Answer

z = −2.08
8.Normal distribution

Answer: x = 241.8 g

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 = 250 + −1.645·5 = 241.78 g

Answer

x = 241.78 g
9.Confidence intervals

Answer: E₂ = 2.621

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.

Ratio of margins

E₂/E₁ = (1.96/1.645)·√(25/25) = 1.191

New margin

E₂ = 2.2·1.191 = 2.621

Answer

E₂ = 2.621
10.Confidence intervals

Answer: n = 1037

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 = 2.576²·0.5·0.5/0.04² = 1037

Round up

n = 1037

Answer

n = 1037
11.Hypothesis tests & confidence intervals

Answer: p = 0.1112

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.

Test statistic

SE = σ/√n = 10/√100 = 1
z = (101.22 − 100)/1 = 1.22

Tail area

Φ(1.22) = 0.8888
1 − Φ(1.22) = 0.1112

p-value

p = 0.1112

Answer

p = 0.1112

p > 0.05, so H₀ is not rejected at the 5% level.

12.Hypothesis tests & confidence intervals

Answer: 1 − β = 52.95 %

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 = 9/√36 = 1.5
x̄c = 50 − 2.326·1.5 = 46.51

z under the true mean

z = (46.51 − 46.4)/1.5 = 0.074

Power

power = P(Z < 0.074) = 52.95%
1 − β = 52.95%

Answer

1 − β = 52.95%
13.Linear regression & correlation

Answer: r = 0.9867

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 = 874 − 42·105/6 = 139
Sxx = 364 − 42²/6 = 70
Syy = 2121 − 105²/6 = 283.5

Correlation coefficient

r = 139/√(70·283.5) = 0.9867

Answer

r = 0.9867
14.Linear regression & correlation

Answer: x = 5.949

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 = 33/6 = 5.5
ȳ = 352/6 = 58.67

Sums of squares and products

Sxx = Σ(x − x̄)² = 17.5
Sxy = Σ(x − x̄)(y − ȳ) = 52

Regression line

b = 52/17.5 = 2.971
a = 58.67 − 2.971·5.5 = 42.32
ŷ = 42.32 + 2.971·x

Solve for x

x = (60 − 42.32)/2.971 = 5.949

Answer

x = 5.949
15.Chi-square tests

Answer: χ² = 0.3472

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: 80·80/200 = 32, 80·120/200 = 48
placebo: 120·80/200 = 48, 120·120/200 = 72

Chi-square statistic

χ² = (34 − 32)²/32 + (46 − 48)²/48 + (46 − 48)²/48 + (74 − 72)²/72
χ² = 0.3472

Answer

χ² = 0.3472

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

16.Chi-square tests

Answer: df = 4

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 = 6 − 1 − 1 = 4

Answer

df = 4

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