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 repeated measures design, order effects can be controlled by:

  1. AUsing a larger sample only
  2. BSingle-blind procedures
  3. CCounterbalancing
  4. DRandom sampling
2.Research methods in psychologyHard

Effect size measures such as Cohen's d are reported because they:

  1. AShow how large an effect is, regardless of sample size
  2. BTell how many participants dropped out
  3. CShow whether the result is statistically significant at the chosen alpha level
  4. DReplace the need for any hypothesis
3.Descriptive statisticsMedium
152025303540

Find the sample standard deviation s of the data: 25, 42, 12, 19, 21, 43.

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

Exam scores are normally distributed with mean 60 and standard deviation 20. Find the z-score z of a score of 36.

μ = 60σ = 20x = 36
z =
5.ProbabilityMedium
n = 11k = 5N = ?

In how many ways N can a committee of 5 people be chosen from a group of 11? The order does not matter.

n = 11k = 5
N =
6.ProbabilityHard
xp(x)100.05200.3300.35400.3μ

X has the distribution shown, and Y = 2X − 3. Find the standard deviation σY of Y.

x: 10, 20, 30, 40p(x): 0.05, 0.3, 0.35, 0.3Y = 2X − 3
σY =
7.Normal distributionMedium
680 kg730.9 kgT = X₁ + … + X₈?

The masses of passengers with luggage are independent and normal with mean 85 kg and standard deviation 12 kg. Find the probability P that the total mass of 8 of them exceeds 730.9 kg. Round z to two decimals.

μ = 85 kgσ = 12 kgn = 8> 730.9 kg
P =
8.Normal distributionHard
380420460500540580620hbottom 20%

The battery lifetimes are normally distributed with mean 500 h and standard deviation 40 h. Drag the cut-off so the shaded bottom 20% is to the left of it. Find the cut-off x.

μ = 500 hσ = 40 h20%
x = h
9.Confidence intervalsMedium
n = 30, s = 9.5LU = ?x̄ = 103.5E99%

A sample of 30 items has mean 103.5 and standard deviation s = 9.5. The population is roughly normal. Find the upper limit of the 99% confidence interval for the mean.

n = 30x̄ = 103.5s = 9.599%t(29) = 2.756
upper limit =
10.Confidence intervalsHard
n = 11σ = ?Us = 10.395%

The diameters of 11 machined shafts have a sample standard deviation of 10.3 µm. Assuming the diameters are normally distributed, find the lower 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 = 10.3 µmχ² = 3.247, 20.483
σ = µm
11.Hypothesis tests & confidence intervalsMedium
−1.961.96z = ?

Two production lines make the same part. Line 1: a sample of 100 parts has mean length 142.5 mm, and the line's known standard deviation is 6 mm. Line 2: a sample of 50 parts has mean 139 mm, known standard deviation 9 mm. Find the test statistic z for H₀: μ₁ = μ₂, taking the difference as x̄₁ − x̄₂ (line 1 minus line 2).

n₁ = 100x̄₁ = 142.5 mmσ₁ = 6 mmn₂ = 50x̄₂ = 139 mmσ₂ = 9 mm
z =
12.Hypothesis tests & confidence intervalsHard
z −1.9601.96

A sample of 25 measurements has mean 116.6. The population standard deviation is 12. Find the upper limit U of the 95% confidence interval for the mean.

x̄ = 116.6σ = 12n = 2595%
U =
13.Linear regression & correlationMedium
xy3505637649761187xy5106080

The table gives 5 paired observations of x = advertising spend in k€ and y = sales in k€. Find the coefficient of determination r² of the least-squares line.

n = 5
r² =
14.Linear regression & correlationHard
xy150252356461558xy24505560x = 1.5

The table gives 5 paired observations of x = hours studied and y = exam score. Use the least-squares line ŷ = a + b·x. A new observation (1.5, 50) is made. Find its residual e = y − ŷ.

n = 5(1.5, 50)
e =
15.Chi-square testsMedium
improvednot improvedTotalnew drug3189120placebo493180Total80120200

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
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 test of independence is carried out on a contingency table with 2 rows and 2 columns (not counting the totals). 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: C · Counterbalancing

Why this is the answer

Counterbalancing distributes practice and fatigue effects evenly across conditions.

Answer

C · Counterbalancing
2.Research methods in psychology

Answer: A · Show how large an effect is, regardless of sample size

Why this is the answer

With large samples tiny effects can be significant. Effect sizes express the magnitude of a difference, e.g. d = 0.2 small, 0.5 medium, 0.8 large.

Answer

A · Show how large an effect is, regardless of sample size
3.Descriptive statistics

Answer: s = 12.73

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 = 162/6 = 27

Squared deviations

Σ(x − x̄)² = (−2)² + 15² + (−15)² + (−8)² + (−6)² + 16² = 810

Standard deviation

s = √(810/5) = 12.73

Answer

s = 12.73
4.Descriptive statistics

Answer: z = −1.2

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 = (36 − 60)/20 = −1.2

Answer

z = −1.2
5.Probability

Answer: N = 462

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.

Combinations

C(11, 5) = (11·10·9·8·7)/(5·4·3·2·1) = 462

Answer

N = 462
6.Probability

Answer: σY = 17.78

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 = 29, Var(X) = 79, σX = 8.888

Linear change

σY = |2|·8.888 = 17.78

Answer

σY = 17.78

The mean does move: E(Y) = 2·29 − 3 = 55.

7.Normal distribution

Answer: P = 0.06681

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 = 680 kg, σT = 12·√8 = 33.94 kg

Standardise

z = (730.9 − 680)/33.94 = 1.5

Probability

P = 1 − Φ(1.5) = 0.06681

Answer

P = 0.06681
8.Normal distribution

Answer: x = 466 h

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) = 0.2

z from the table

z ≈ −0.84

Back to the original units

x = 500 − 0.84·40 = 466 h

Answer

x = 466 h
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 = 29, t = 2.756

Margin of error

E = 2.756·9.5/√30 = 4.78

Interval

103.5 ± 4.78 → [98.72, 108.3]

Answer

upper limit = 108.3
10.Confidence intervals

Answer: σ = 7.197 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 = 10·10.3² = 1061

Limits for σ²

1061/20.483 = 51.79
1061/3.247 = 326.7

Limit for σ

σ = √51.79 = 7.197 µm

Answer

σ = 7.197 µm

95% interval for σ: [7.197, 18.08] µm.

11.Hypothesis tests & confidence intervals

Answer: z = 2.487

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.

Standard error of the difference

SE = √(6²/100 + 9²/50) = √1.98 = 1.407

Test statistic

z = (142.5 − 139)/1.407 = 2.487

Answer

z = 2.487

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

12.Hypothesis tests & confidence intervals

Answer: U = 121.3

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).

Margin of error

E = 1.96·12/√25 = 4.704

Interval limit

U = 116.6 + 4.704 = 121.3

Answer

U = 121.3
13.Linear regression & correlation

Answer: r² = 0.9581

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 = 35/5 = 7
ȳ = 340/5 = 68

Sums of squares and products

Sxx = Σ(x − x̄)² = 40
Sxy = Σ(x − x̄)(y − ȳ) = 174
Syy = Σ(y − ȳ)² = 790

Coefficient of determination

r² = 174²/(40·790) = 0.9581

Answer

r² = 0.9581

About 95.81% of the variation in y is explained by the line.

14.Linear regression & correlation

Answer: e = −1.65

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 = 15/5 = 3
ȳ = 277/5 = 55.4

Sums of squares and products

Sxx = Σ(x − x̄)² = 10
Sxy = Σ(x − x̄)(y − ȳ) = 25

Regression line

b = 25/10 = 2.5
a = 55.4 − 2.5·3 = 47.9
ŷ = 47.9 + 2.5·x

Predicted value

ŷ = 47.9 + 2.5·1.5 = 51.65

Residual

e = 50 − 51.65 = −1.65

Answer

e = −1.65

The point lies below the line.

15.Chi-square tests

Answer: χ² = 25.09

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

Chi-square statistic

χ² = (31 − 48)²/48 + (89 − 72)²/72 + (49 − 32)²/32 + (31 − 48)²/48
χ² = 25.09

Answer

χ² = 25.09

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

16.Chi-square tests

Answer: χ²crit = 3.841

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)·(2 − 1) = 1

Table

χ²crit(df = 1, α = 0.05) = 3.841

Answer

χ²crit = 3.841

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

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