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 field experiment, the researcher:

  1. AManipulates the IV in a real-world setting
  2. BAnalyses data collected by others
  3. CManipulates the IV in a laboratory
  4. DOnly observes without manipulating anything
2.Research methods in psychologyHard

A researcher measures the effect of noise on concentration. Participants in the noise condition happened to be tested in the morning and the quiet condition in the afternoon. What is the problem?

  1. AThe data are at nominal level
  2. BTime of day is a confounding variable
  3. CTime of day is the dependent variable
  4. DThe study has too many conditions
3.Descriptive statisticsMedium
25303540455055

Find the sample standard deviation s of the data: 49, 24, 39, 52, 57.

n = 5
s =
4.Descriptive statisticsHard
z −2−1012

A quantity is normally distributed with mean 60 and standard deviation 8. What proportion P of values is above 64.8?

μ = 60σ = 8x = 64.8
P =
5.ProbabilityMedium
xp(x)00.310.2520.2530.2μ = 1.35

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

x: 0, 1, 2, 3p(x): 0.3, 0.25, 0.25, 0.2
σ =
6.ProbabilityHard
€p(x)−50.73150.25950.02

A game costs €5 to play. You win €20 with probability 0.25 and €100 with probability 0.02; otherwise you win nothing. The stake is not returned. Find the expected net gain E(G) per game.

stake = €5€20: p = 0.25€100: p = 0.02
E(G) = €
7.Normal distributionMedium
−3−2−10123zshaded area between −z and z = 0.6265

Drag z so that the shaded area between −z and z under the standard normal curve is 0.6265. Find z to two decimals.

A = 0.6265
z =
8.Normal distributionHard
235240245250255260265gbottom 25%

The bag weights are normally distributed with mean 250 g and standard deviation 5 g. Drag the cut-off so the shaded bottom 25% is to the left of it. Find the cut-off x.

μ = 250 gσ = 5 g25%
x = g
9.Confidence intervalsMedium
n = 10, s = 5.5LUx̄ = 43E = ?95%

A sample of 10 items has mean 43 and standard deviation s = 5.5. The population is roughly normal. Find the margin of error E of a 95% confidence interval for the mean.

n = 10x̄ = 43s = 5.595%t(9) = 2.262
E =
10.Confidence intervalsHard
σ = 33, n = ?x̄ − Ex̄ + Ex̄E = 495%

The population standard deviation is 33. How large a sample n do you need so that a 95% confidence interval for the mean has a margin of error of at most 4?

σ = 33E = 495%
n =
11.Hypothesis tests & confidence intervalsMedium
−2.0452.045t = ?df = 29

Two teaching methods are compared. Group 1: 11 students, mean score 69.5, standard deviation 11. Group 2: 20 students, mean score 65.9, standard deviation 10.5. Assume equal population variances and use the pooled two-sample t test. Find the test statistic t for H₀: μ₁ = μ₂, taking the difference as x̄₁ − x̄₂ (group 1 minus group 2).

n₁ = 11x̄₁ = 69.5s₁ = 11n₂ = 20x̄₂ = 65.9s₂ = 10.5
t =
12.Hypothesis tests & confidence intervalsHard
−1.961.96z = ?

In a trial, 24 of 120 patients on treatment A recover, and 58 of 200 patients on treatment B. Using the pooled proportion, find the test statistic z for H₀: p₁ = p₂.

x₁ = 24n₁ = 120x₂ = 58n₂ = 200
z =
13.Linear regression & correlationMedium
xy36258078791041111413130xy510100

The table gives 6 paired observations of x = hours studied and y = exam score. Find Pearson's correlation coefficient r.

n = 6
r =
14.Linear regression & correlationHard
xy126225324423518620xy2462025x = 5.5

The table gives 6 paired observations of x = car age in years and y = price in k€. Use the least-squares line ŷ = a + b·x. A new observation (5.5, 23) is made. Find its residual e = y − ŷ.

n = 6(5.5, 23)
e =
15.Chi-square testsMedium
yesnoTotalmen243660women112940Total3565100

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 = 1002 × 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 goodness-of-fit test compares observed counts in 9 categories with fully specified expected proportions (no parameters are estimated from the 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: A · Manipulates the IV in a real-world setting

Why this is the answer

Field experiments, such as staged helping situations in the street, increase ecological validity but give less control over extraneous variables and raise consent issues.

Answer

A · Manipulates the IV in a real-world setting
2.Research methods in psychology

Answer: B · Time of day is a confounding variable

Why this is the answer

Because noise and testing time vary together, any difference could be due to either. Random allocation of testing times or holding time constant would remove the confound.

Answer

B · Time of day is a confounding variable
3.Descriptive statistics

Answer: s = 13.07

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 = 221/5 = 44.2

Squared deviations

Σ(x − x̄)² = 4.8² + (−20.2)² + (−5.2)² + 7.8² + 12.8² = 682.8

Standard deviation

s = √(682.8/4) = 13.07

Answer

s = 13.07
4.Descriptive statistics

Answer: P = 0.2743

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 = 0.6

Standard normal table

Φ(0.6) = 0.7257
P = 1 − 0.7257 = 0.2743

Answer

P = 0.2743
5.Probability

Answer: σ = 1.108

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 = 0·0.3 + 1·0.25 + 2·0.25 + 3·0.2 = 1.35

Mean of the squares

E(X²) = 0·0.3 + 1·0.25 + 4·0.25 + 9·0.2 = 3.05

Variance

Var(X) = 3.05 − 1.35² = 1.227
σ = √1.227 = 1.108

Answer

σ = 1.108
6.Probability

Answer: E(G) = 2 €

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.

Net outcomes

G: 15 (p = 0.25), 95 (p = 0.02), −5 (p = 0.73)

Expected value

E(G) = 15·0.25 + 95·0.02 − 5·0.73 = 2 €

Answer

E(G) = 2 €

A positive value means you win money on average.

7.Normal distribution

Answer: z = 0.89

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 + 0.6265)/2 = 0.8133

Read the z-table backwards

Find 0.8133 inside the table and read off z.

z = 0.89

Answer

z = 0.89
8.Normal distribution

Answer: x = 246.7 g

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) = 0.25

z from the table

z ≈ −0.67

Back to the original units

x = 250 − 0.67·5 = 246.7 g

Answer

x = 246.7 g
9.Confidence intervals

Answer: E = 3.934

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 = 9, t = 2.262

Margin of error

E = 2.262·5.5/√10 = 3.934

Answer

E = 3.934
10.Confidence intervals

Answer: n = 262

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.

Solve for n

n = (1.96·33/4)² = 261.5

Round up

A sample size is a whole number, and rounding down would make the margin too large.

n = 262

Answer

n = 262
11.Hypothesis tests & confidence intervals

Answer: t = 0.8984

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.

Pooled variance

sp² = (10·11² + 19·10.5²)/29 = 114
sp = 10.68

Standard error

SE = 10.68·√(1/11 + 1/20) = 4.007

Test statistic

t = (69.5 − 65.9)/4.007 = 0.8984

Answer

t = 0.8984

With df = 29, the two-sided 5% critical value is 2.045; |t| is smaller, so H₀ is not rejected.

12.Hypothesis tests & confidence intervals

Answer: z = −1.785

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 = 24/120 = 0.2
p̂₂ = 58/200 = 0.29
p̂ = 82/320 = 0.2562

Standard error under H₀

SE = √(0.2562·0.7438·(1/120 + 1/200)) = 0.05041

Test statistic

z = (0.2 − 0.29)/0.05041 = −1.785

Answer

z = −1.785

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

13.Linear regression & correlation

Answer: r = 0.9955

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 = 48/6 = 8
ȳ = 577/6 = 96.17

Sums of squares and products

Sxx = Σ(x − x̄)² = 70
Sxy = Σ(x − x̄)(y − ȳ) = 459
Syy = Σ(y − ȳ)² = 3037

Correlation coefficient

r = 459/√(70·3037) = 0.9955

Answer

r = 0.9955
14.Linear regression & correlation

Answer: e = 3.305

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 = 21/6 = 3.5
ȳ = 136/6 = 22.67

Sums of squares and products

Sxx = Σ(x − x̄)² = 17.5
Sxy = Σ(x − x̄)(y − ȳ) = −26

Regression line

b = −26/17.5 = −1.486
a = 22.67 − (−1.486)·3.5 = 27.87
ŷ = 27.87 − 1.486·x

Predicted value

ŷ = 27.87 + (−1.486)·5.5 = 19.7

Residual

e = 23 − 19.7 = 3.305

Answer

e = 3.305

The point lies above the line.

15.Chi-square tests

Answer: χ² = 1.648

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

men: 60·35/100 = 21, 60·65/100 = 39
women: 40·35/100 = 14, 40·65/100 = 26

Chi-square statistic

χ² = (24 − 21)²/21 + (36 − 39)²/39 + (11 − 14)²/14 + (29 − 26)²/26
χ² = 1.648

Answer

χ² = 1.648

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

16.Chi-square tests

Answer: χ²crit = 20.09

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.

Degrees of freedom

df = 9 − 1 = 8

Table

χ²crit(df = 8, α = 0.01) = 20.090

Answer

χ²crit = 20.090

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

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