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. AMost scores lie in the tails
  2. BThe mode is the lowest score
  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. BDependent variable
  3. CConfound
  4. DModerator
3.Descriptive statisticsMedium
20253035404550

Find the sample standard deviation s of the data: 32, 23, 40, 52, 17, 16, 30.

n = 7
s =
4.Descriptive statisticsHard
z −2−1012

A quantity is normally distributed with mean 100 and standard deviation 8. What proportion P of values is above 93.6?

μ = 100σ = 8x = 93.6
P =
5.ProbabilityMedium
xp(x)00.3510.3520.3μ = 0.95

For the distribution shown, find the variance Var(X) of X.

x: 0, 1, 2p(x): 0.35, 0.35, 0.3
Var(X) =
6.ProbabilityHard
n = 12, p = 0.1: P(X ≤ 1) = ?k0123456789101112

12 components are tested, and each one fails with probability 0.1, independently. What is the probability P that at most one fails?

n = 12p = 0.1
P =
7.Normal distributionMedium
72 kg69.1 kgT = X₁ + … + X₆?

The masses of boxes are independent and normal with mean 12 kg and standard deviation 0.8 kg. Find the probability P that the total mass of 6 of them is below 69.1 kg. Round z to two decimals.

μ = 12 kgσ = 0.8 kgn = 6< 69.1 kg
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 = 16, s = 4LU = ?x̄ = 106.5E90%

A sample of 16 items has mean 106.5 and standard deviation s = 4. The population is roughly normal. Find the upper limit of the 90% confidence interval for the mean.

n = 16x̄ = 106.5s = 490%t(15) = 1.753
upper limit =
10.Confidence intervalsHard
n = 21σ = ?Us = 5.895%

The diameters of 21 machined shafts have a sample standard deviation of 5.8 µm. Assuming the diameters are normally distributed, find the lower limit σ of the 95% confidence interval for the population standard deviation. With df = 20, 2.5% of the χ² distribution lies below 9.591 and 2.5% lies above 34.17.

n = 21s = 5.8 µmχ² = 9.591, 34.17
σ = µm
11.Hypothesis tests & confidence intervalsMedium
−2.3652.365t = ?df = 7

8 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 3.62 and the standard deviation of the differences is 3.1. Find the paired test statistic t for H₀: μd = 0.

n = 8d̄ = 3.62sd = 3.1
t =
12.Hypothesis tests & confidence intervalsHard
−1.961.96z = ?

In a trial, 75 of 150 patients on treatment A recover, and 195 of 300 patients on treatment B. Using the pooled proportion, find the test statistic z for H₀: p₁ = p₂.

x₁ = 75n₁ = 150x₂ = 195n₂ = 300
z =
13.Linear regression & correlationMedium
xy17491949216123602569xy2025506070

The table gives 5 paired observations of x = temperature in °C and y = ice creams sold. Find the coefficient of determination r² of the least-squares line.

n = 5
r² =
14.Linear regression & correlationHard
xy23466128161018xy51051015ŷ = 9

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. For which value x does the line predict ŷ = 9?

n = 5ŷ = 9
x =
15.Chi-square testsMedium
no defectminormajorTotalday shift21402788night shift30251469Total516541157

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

row total = 88column total = 41N = 157
E =
16.Chi-square testsHard

A chi-square goodness-of-fit test compares observed counts in 4 categories with fully specified expected proportions (no parameters are estimated from the data). How many degrees of freedom df does the test have?

k = 4
df =

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: D · 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

D · Moderator
3.Descriptive statistics

Answer: s = 12.92

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 = 210/7 = 30

Squared deviations

Σ(x − x̄)² = 2² + (−7)² + 10² + 22² + (−13)² + (−14)² + 0² = 1002

Standard deviation

s = √(1002/6) = 12.92

Answer

s = 12.92
4.Descriptive statistics

Answer: P = 0.7881

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 = −0.8

Standard normal table

Φ(−0.8) = 0.2119
P = 1 − 0.2119 = 0.7881

Answer

P = 0.7881
5.Probability

Answer: Var(X) = 0.6475

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 = 0·0.35 + 1·0.35 + 2·0.3 = 0.95

Mean of the squares

E(X²) = 0·0.35 + 1·0.35 + 4·0.3 = 1.55

Variance

Var(X) = 1.55 − 0.95² = 0.6475

Answer

Var(X) = 0.6475
6.Probability

Answer: P = 0.659

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) = 0.912 = 0.2824
P(X = 1) = 12·0.1·0.911 = 0.3766

Add them

P = 0.2824 + 0.3766 = 0.659

Answer

P = 0.659
7.Normal distribution

Answer: P = 0.06944

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 total

μT = 72 kg, σT = 0.8·√6 = 1.96 kg

Standardise

z = (69.1 − 72)/1.96 = −1.48

Probability

P = Φ(−1.48) = 0.06944

Answer

P = 0.06944
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: upper limit = 108.3

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 = 15, t = 1.753

Margin of error

E = 1.753·4/√16 = 1.753

Interval

106.5 ± 1.753 → [104.7, 108.3]

Answer

upper limit = 108.3
10.Confidence intervals

Answer: σ = 4.437 µm

Interval for a variance: (n − 1)s²/σ² has a χ² distribution with n − 1 degrees of freedom. The 95% interval for σ² runs from (n − 1)s² divided by the upper χ² value (2.5% above it) to (n − 1)s² divided by the lower χ² value (2.5% below it); the interval for σ is the square root of both limits. It is not symmetric around s.

Sum of squares

(n − 1)s² = 20·5.8² = 672.8

Limits for σ²

672.8/34.17 = 19.69
672.8/9.591 = 70.15

Limit for σ

σ = √19.69 = 4.437 µm

Answer

σ = 4.437 µm

95% interval for σ: [4.437, 8.376] µm.

11.Hypothesis tests & confidence intervals

Answer: t = 3.303

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 = 3.1/√8 = 1.096

Test statistic

t = 3.62/1.096 = 3.303

Answer

t = 3.303

With df = 7, the two-sided 5% critical value is 2.365; |t| is larger, so H₀ is rejected.

12.Hypothesis tests & confidence intervals

Answer: z = −3.062

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 = 75/150 = 0.5
p̂₂ = 195/300 = 0.65
p̂ = 270/450 = 0.6

Standard error under H₀

SE = √(0.6·0.4·(1/150 + 1/300)) = 0.04899

Test statistic

z = (0.5 − 0.65)/0.04899 = −3.062

Answer

z = −3.062

|z| > 1.96, so H₀ is rejected at the 5% level (two-sided).

13.Linear regression & correlation

Answer: r² = 0.8811

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 = 105/5 = 21
ȳ = 288/5 = 57.6

Sums of squares and products

Sxx = Σ(x − x̄)² = 40
Sxy = Σ(x − x̄)(y − ȳ) = 102
Syy = Σ(y − ȳ)² = 295.2

Coefficient of determination

r² = 102²/(40·295.2) = 0.8811

Answer

r² = 0.8811

About 88.11% of the variation in y is explained by the line.

14.Linear regression & correlation

Answer: x = 5

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 = 55/5 = 11

Sums of squares and products

Sxx = Σ(x − x̄)² = 40
Sxy = Σ(x − x̄)(y − ȳ) = 80

Regression line

b = 80/40 = 2
a = 11 − 2·6 = −1
ŷ = −1 + 2·x

Solve for x

x = (9 − (−1))/2 = 5

Answer

x = 5
15.Chi-square tests

Answer: E = 22.98

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 = 88·41/157 = 22.98

Answer

E = 22.98
16.Chi-square tests

Answer: df = 3

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 = 3

Answer

df = 3

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