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 a perfectly normal distribution:

  1. AThe mode is the lowest score
  2. BMost scores lie in the tails
  3. CThe mean, median and mode are equal
  4. DThe mean is always larger than the median
2.Research methods in psychologyHard

A study finds that the effect of stress on illness is weaker in people with strong social support. Social support is acting as a:

  1. AMediator
  2. BConfound
  3. CModerator
  4. DDependent variable
3.Descriptive statisticsMedium
25303540455055

Find the sample standard deviation s of the data: 52, 42, 36, 25, 48, 59.

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 above 62.5?

μ = 70σ = 5x = 62.5
P =
5.ProbabilityMedium
xp(x)10.2520.1530.2540.1550.2μ = 2.9

For the distribution shown, find the standard deviation σ of X.

x: 1, 2, 3, 4, 5p(x): 0.25, 0.15, 0.25, 0.15, 0.2
σ =
6.ProbabilityHard
n = 6, p = 0.25: P(X ≤ 1) = ?k0123456

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

n = 6p = 0.25
P =
7.Normal distributionMedium
85.75112P = ?

The IQ scores are normally distributed with mean 100 points and standard deviation 15 points. What fraction lies between 85.75 and 112 points? Give P as a decimal.

μ = 100σ = 15x1 = 85.75x2 = 112
P =
8.Normal distributionHard
0.02 mm0D − d?

Hole diameters D are normal with mean 25.02 mm and standard deviation 0.02 mm. Shaft diameters d are independent and normal with mean 25 mm and standard deviation 0.01 mm. A randomly chosen shaft fits when D − d > 0. Find the probability P that it fits. Round z to two decimals.

hole: μ = 25.02 mm, σ = 0.02 mmshaft: μ = 25 mm, σ = 0.01 mm
P =
9.Confidence intervalsMedium
x̄E₁ = 1.195%, n = 64x̄E₂ = ?90%, n = 128

A 95% confidence interval for a mean, based on 64 observations and a known σ, has a margin of error of 1.1. What is the margin of error E₂ of a 90% interval based on 128 observations?

95%n₁ = 64E₁ = 1.190%n₂ = 128
E₂ =
10.Confidence intervalsHard
p ≈ 0.8, n = ?p̂ − Ep̂ + Ep̂E = 0.0195%

A pilot study suggests that about 0.8 of products are defective. How many products n must be inspected so that a 95% confidence interval for the proportion has a margin of error of at most 0.01?

p ≈ 0.8E = 0.0195%
n =
11.Hypothesis tests & confidence intervalsMedium
−2.3062.306t = ?df = 8

9 employees take a typing test before and after a training course. For each person the difference d = after − before (words per minute) is recorded. The mean difference is 3.15 and the standard deviation of the differences is 2.7. Find the paired test statistic t for H₀: μd = 0.

n = 9d̄ = 3.15sd = 2.7
t =
12.Hypothesis tests & confidence intervalsHard
−1.961.96z = ?

In a trial, 60 of 100 patients on treatment A recover, and 67 of 100 patients on treatment B. Using the pooled proportion, find the test statistic z for H₀: p₁ = p₂.

x₁ = 60n₁ = 100x₂ = 67n₂ = 100
z =
13.Linear regression & correlationMedium
xy175920762381269429114xy2025306080100120

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

n = 5
r =
14.Linear regression & correlationHard
xy2104176258321043xy5102040x = 5

The table gives 5 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 = 5x = 5
ŷ =
15.Chi-square testsMedium
lowmediumhighTotalurban14221652rural31154187Total453757139

The contingency table shows observed counts. For a chi-square test of independence, find the expected count E for the cell “urban, medium”.

row total = 52column total = 37N = 139
E =
16.Chi-square testsHard
dfα = 0.05α = 0.0113.8416.63525.9919.21037.81511.34549.48813.277511.07015.086612.59216.812714.06718.475815.50720.090916.91921.6661018.30723.209

A chi-square goodness-of-fit test checks whether counts in 9 categories follow a Poisson distribution whose mean is estimated from the same data. Using the table, find the critical value χ²crit at significance level α = 0.01.

α = 0.01
χ²crit =

3Solutions

1.Research methods in psychology

Answer: C · The mean, median and mode are equal

Why this is the answer

A normal distribution is symmetrical and bell-shaped, so all three measures of central tendency fall at the centre.

Answer

C · The mean, median and mode are equal
2.Research methods in psychology

Answer: C · Moderator

Why this is the answer

A moderator alters the strength or direction of the relationship between two variables. A mediator explains the mechanism through which one variable affects another.

Answer

C · Moderator
3.Descriptive statistics

Answer: s = 12.11

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 = 262/6 = 43.67

Squared deviations

Σ(x − x̄)² = 8.333² + (−1.667)² + (−7.667)² + (−18.67)² + 4.333² + 15.33² = 733.3

Standard deviation

s = √(733.3/5) = 12.11

Answer

s = 12.11
4.Descriptive statistics

Answer: P = 0.9332

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 = 1 − 0.06681 = 0.9332

Answer

P = 0.9332
5.Probability

Answer: σ = 1.446

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 = 1·0.25 + 2·0.15 + 3·0.25 + 4·0.15 + 5·0.2 = 2.9

Mean of the squares

E(X²) = 1·0.25 + 4·0.15 + 9·0.25 + 16·0.15 + 25·0.2 = 10.5

Variance

Var(X) = 10.5 − 2.9² = 2.09
σ = √2.09 = 1.446

Answer

σ = 1.446
6.Probability

Answer: P = 0.5339

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) = 0.756 = 0.178
P(X = 1) = 6·0.25·0.755 = 0.356

Add them

P = 0.178 + 0.356 = 0.5339

Answer

P = 0.5339
7.Normal distribution

Answer: P = 0.6171

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 = (85.75 − 100)/15 = −0.95
z₂ = (112 − 100)/15 = 0.8

Area between

P = Φ(0.8) − Φ(−0.95) = 0.7881 − 0.1711 = 0.6171

Answer

P = 0.6171
8.Normal distribution

Answer: P = 0.8133

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 = 25.02 − 25 = 0.02 mm, σ = 0.02236 mm

Standardise

z = (0 − 0.02)/0.02236 = −0.89

Probability

P = 1 − Φ(−0.89) = 0.8133

Answer

P = 0.8133
9.Confidence intervals

Answer: E₂ = 0.6528

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 = (1.645/1.96)·√(64/128) = 0.5935

New margin

E₂ = 1.1·0.5935 = 0.6528

Answer

E₂ = 0.6528
10.Confidence intervals

Answer: n = 6147

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 = 1.96²·0.8·0.2/0.01² = 6147

Round up

n = 6147

Answer

n = 6147
11.Hypothesis tests & confidence intervals

Answer: t = 3.5

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 of the mean difference

SE = 2.7/√9 = 0.9

Test statistic

t = 3.15/0.9 = 3.5

Answer

t = 3.5

With df = 8, the two-sided 5% critical value is 2.306; |t| is larger, so H₀ is rejected.

12.Hypothesis tests & confidence intervals

Answer: z = −1.028

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 = 60/100 = 0.6
p̂₂ = 67/100 = 0.67
p̂ = 127/200 = 0.635

Standard error under H₀

SE = √(0.635·0.365·(1/100 + 1/100)) = 0.06808

Test statistic

z = (0.6 − 0.67)/0.06808 = −1.028

Answer

z = −1.028

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

13.Linear regression & correlation

Answer: r = 0.9832

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 = 115/5 = 23
ȳ = 424/5 = 84.8

Sums of squares and products

Sxx = Σ(x − x̄)² = 90
Sxy = Σ(x − x̄)(y − ȳ) = 384
Syy = Σ(y − ȳ)² = 1695

Correlation coefficient

r = 384/√(90·1695) = 0.9832

Answer

r = 0.9832
14.Linear regression & correlation

Answer: ŷ = 21.35

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 = 127/5 = 25.4

Sums of squares and products

Sxx = Σ(x − x̄)² = 40
Sxy = Σ(x − x̄)(y − ȳ) = 162

Regression line

b = 162/40 = 4.05
a = 25.4 − 4.05·6 = 1.1
ŷ = 1.1 + 4.05·x

Prediction

ŷ = 1.1 + 4.05·5 = 21.35

Answer

ŷ = 21.35
15.Chi-square tests

Answer: E = 13.84

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 count

E = 52·37/139 = 13.84

Answer

E = 13.84
16.Chi-square tests

Answer: χ²crit = 18.48

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.

Degrees of freedom

df = 9 − 1 − 1 = 7

Table

χ²crit(df = 7, α = 0.01) = 18.475

Answer

χ²crit = 18.475

Reject H₀ if the computed χ² is larger than this value.

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