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 positively skewed distribution, such as income:

  1. AThere are no extreme scores
  2. BThe mean is less than the median
  3. CThe mean and median are always equal
  4. DThe mean is greater than the median
2.Research methods in psychologyHard

A researcher wants to test whether a new teaching method causes better grades, but cannot randomise students. Which design gives the strongest causal evidence in this situation?

  1. AComparing matched classes measured before and after, with a comparison group
  2. BMeasuring only the class using the new method after teaching
  3. CAsking students whether they think the method worked
  4. DCorrelating grades with how much students like the teacher
3.Descriptive statisticsMedium
1520253035404550

Find the sample standard deviation s of the data: 54, 26, 45, 15, 47, 44, 49.

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

Exam scores are normally distributed with mean 100 and standard deviation 20. Find the z-score z of a score of 125.

μ = 100σ = 20x = 125
z =
5.ProbabilityMedium
xp(x)00.3520.240.1560.3μ = 2.8

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

x: 0, 2, 4, 6p(x): 0.35, 0.2, 0.15, 0.3
σ =
6.ProbabilityHard
xp(x)00.3510.320.35μ

X has the distribution shown, and Y = 3X + 4. Find the standard deviation σY of Y.

x: 0, 1, 2p(x): 0.35, 0.3, 0.35Y = 3X + 4
σY =
7.Normal distributionMedium
−3−2−10123zshaded area between −z and z = 0.3616

Drag z so that the shaded area between −z and z under the standard normal curve is 0.3616. Find z to two decimals.

A = 0.3616
z =
8.Normal distributionHard
0.05 mm0D − d?

Hole diameters D are normal with mean 25.05 mm and standard deviation 0.01 mm. Shaft diameters d are independent and normal with mean 25 mm and standard deviation 0.02 mm. A randomly chosen shaft fits when D − d > 0. Find the probability P that it fits. Round z to two decimals.

hole: μ = 25.05 mm, σ = 0.01 mmshaft: μ = 25 mm, σ = 0.02 mm
P =
9.Confidence intervalsMedium
n = 20, s = 20L = ?Ux̄ = 74E99%

A sample of 20 items has mean 74 and standard deviation s = 20. The population is roughly normal. Find the lower limit of the 99% confidence interval for the mean.

n = 20x̄ = 74s = 2099%t(19) = 2.861
lower limit =
10.Confidence intervalsHard
n = 11Lσ = ?s = 0.695%

The diameters of 11 machined shafts have a sample standard deviation of 0.6 µm. Assuming the diameters are normally distributed, find the upper limit σ of the 95% confidence interval for the population standard deviation. With df = 10, 2.5% of the χ² distribution lies below 3.247 and 2.5% lies above 20.483.

n = 11s = 0.6 µmχ² = 3.247, 20.483
σ = µm
11.Hypothesis tests & confidence intervalsMedium
t −1.99801.998

A sample of 64 students has a mean score of 120.9 and a sample standard deviation of 10. Find the test statistic t for H₀: μ = 120.

μ₀ = 120x̄ = 120.9s = 10n = 64
t =
12.Hypothesis tests & confidence intervalsHard
−1.961.96z = ?

In a trial, 45 of 300 patients on treatment A recover, and 65 of 250 patients on treatment B. Using the pooled proportion, find the test statistic z for H₀: p₁ = p₂.

x₁ = 45n₁ = 300x₂ = 65n₂ = 250
z =
13.Linear regression & correlationMedium
xy294206298341044xy5102040

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 = 136Σx² = 220Σy² = 4414Σxy = 984
r =
14.Linear regression & correlationHard
xy244353450558654759xy24645505560x = 4.5

The table gives 6 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 = 4.5.

n = 6x = 4.5
ŷ =
15.Chi-square testsMedium
lowmediumhighTotalurban21201960rural21223780Total424256140

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 = 1402 × 3
χ² =
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 6 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 = 0.05.

α = 0.05
χ²crit =

3Solutions

1.Research methods in psychology

Answer: D · The mean is greater than the median

Why this is the answer

A few very high values in the right tail raise the mean above the median, which is why the median is often reported for skewed data.

Answer

D · The mean is greater than the median
2.Research methods in psychology

Answer: A · Comparing matched classes measured before and after, with a comparison group

Why this is the answer

A non-equivalent groups pretest-posttest design controls for initial differences and for change over time, though it is weaker than a randomised experiment.

Answer

A · Comparing matched classes measured before and after, with a comparison group
3.Descriptive statistics

Answer: s = 14.07

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 = 280/7 = 40

Squared deviations

Σ(x − x̄)² = 14² + (−14)² + 5² + (−25)² + 7² + 4² + 9² = 1188

Standard deviation

s = √(1188/6) = 14.07

Answer

s = 14.07
4.Descriptive statistics

Answer: z = 1.25

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 = (125 − 100)/20 = 1.25

Answer

z = 1.25
5.Probability

Answer: σ = 2.482

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 + 2·0.2 + 4·0.15 + 6·0.3 = 2.8

Mean of the squares

E(X²) = 0·0.35 + 4·0.2 + 16·0.15 + 36·0.3 = 14

Variance

Var(X) = 14 − 2.8² = 6.16
σ = √6.16 = 2.482

Answer

σ = 2.482
6.Probability

Answer: σY = 2.51

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 = 1, Var(X) = 0.7, σX = 0.8367

Linear change

σY = |3|·0.8367 = 2.51

Answer

σY = 2.51

The mean does move: E(Y) = 3·1 + 4 = 7.

7.Normal distribution

Answer: z = 0.47

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 + 0.3616)/2 = 0.6808

Read the z-table backwards

Find 0.6808 inside the table and read off z.

z = 0.47

Answer

z = 0.47
8.Normal distribution

Answer: P = 0.9875

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 = 25.05 − 25 = 0.05 mm, σ = 0.02236 mm

Standardise

z = (0 − 0.05)/0.02236 = −2.24

Probability

P = 1 − Φ(−2.24) = 0.9875

Answer

P = 0.9875
9.Confidence intervals

Answer: lower limit = 61.21

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 = 19, t = 2.861

Margin of error

E = 2.861·20/√20 = 12.79

Interval

74 ± 12.79 → [61.21, 86.79]

Answer

lower limit = 61.21
10.Confidence intervals

Answer: σ = 1.053 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 = 10·0.6² = 3.6

Limits for σ²

3.6/20.483 = 0.1758
3.6/3.247 = 1.109

Limit for σ

σ = √1.109 = 1.053 µm

Answer

σ = 1.053 µm

95% interval for σ: [0.4192, 1.053] µm.

11.Hypothesis tests & confidence intervals

Answer: t = 0.72

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 = 10/√64 = 10/8 = 1.25

Test statistic

t = (120.9 − 120)/1.25 = 0.72

Answer

t = 0.72
t0.025; df = 63 = 1.998

|t| < 1.998, so H₀ is not rejected at the 5% level (two-sided).

12.Hypothesis tests & confidence intervals

Answer: z = −3.211

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 = 45/300 = 0.15
p̂₂ = 65/250 = 0.26
p̂ = 110/550 = 0.2

Standard error under H₀

SE = √(0.2·0.8·(1/300 + 1/250)) = 0.03425

Test statistic

z = (0.15 − 0.26)/0.03425 = −3.211

Answer

z = −3.211

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

13.Linear regression & correlation

Answer: r = 0.9935

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 = 984 − 30·136/5 = 168
Sxx = 220 − 30²/5 = 40
Syy = 4414 − 136²/5 = 714.8

Correlation coefficient

r = 168/√(40·714.8) = 0.9935

Answer

r = 0.9935
14.Linear regression & correlation

Answer: ŷ = 53

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 = 27/6 = 4.5
ȳ = 318/6 = 53

Sums of squares and products

Sxx = Σ(x − x̄)² = 17.5
Sxy = Σ(x − x̄)(y − ȳ) = 43

Regression line

b = 43/17.5 = 2.457
a = 53 − 2.457·4.5 = 41.94
ŷ = 41.94 + 2.457·x

Prediction

ŷ = 41.94 + 2.457·4.5 = 53

Answer

ŷ = 53
15.Chi-square tests

Answer: χ² = 3.087

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: 60·42/140 = 18, 60·42/140 = 18, 60·56/140 = 24
rural: 80·42/140 = 24, 80·42/140 = 24, 80·56/140 = 32

Chi-square statistic

χ² = (21 − 18)²/18 + (20 − 18)²/18 + (19 − 24)²/24 + (21 − 24)²/24 + (22 − 24)²/24 + (37 − 32)²/32
χ² = 3.087

Answer

χ² = 3.087

df = 2. At α = 0.05 the critical value is 5.991, so H₀ (independence) is not rejected.

16.Chi-square tests

Answer: χ²crit = 9.488

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

Table

χ²crit(df = 4, α = 0.05) = 9.488

Answer

χ²crit = 9.488

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

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