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. ACounterbalancing
  2. BRandom sampling
  3. CUsing a larger sample only
  4. DSingle-blind procedures
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 the dependent variable
  3. CThe study has too many conditions
  4. DTime of day is a confounding variable
3.Descriptive statisticsMedium
303540455055

Find the sample standard deviation s of the data: 30, 38, 46, 57, 31.

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

A quantity is normally distributed with mean 500 and standard deviation 50. What proportion P of values is below 562.5?

μ = 500σ = 50x = 562.5
P =
5.ProbabilityMedium
xp(x)10.3520.330.340.05μ = 2.05

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

x: 1, 2, 3, 4p(x): 0.35, 0.3, 0.3, 0.05
Var(X) =
6.ProbabilityHard
n = 10, p = 0.25: P(X = 2) = ?k012345678910

A test has 10 multiple-choice questions. A student who guesses gets each one right with probability 0.25, independently. What is the probability P of exactly 2 correct answers?

n = 10p = 0.25k = 2
P =
7.Normal distributionMedium
166.8192P = ?

The heights of adult men are normally distributed with mean 178 cm and standard deviation 7 cm. What fraction lies between 166.8 and 192 cm? Give P as a decimal.

μ = 178σ = 7x1 = 166.8x2 = 192
P =
8.Normal distributionHard
235240245250255260265gtop 40%

The bag weights are normally distributed with mean 250 g and standard deviation 5 g. Drag the cut-off so the shaded top 40% is to the right of it. Find the cut-off x.

μ = 250 gσ = 5 g40%
x = g
9.Confidence intervalsMedium
n = 12, s = 18LU = ?x̄ = 15.5E90%

A sample of 12 items has mean 15.5 and standard deviation s = 18. The population is roughly normal. Find the upper limit of the 90% confidence interval for the mean.

n = 12x̄ = 15.5s = 1890%t(11) = 1.796
upper limit =
10.Confidence intervalsHard
n = 11σ = ?Us = 11.195%

The diameters of 11 machined shafts have a sample standard deviation of 11.1 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 = 11.1 µmχ² = 3.247, 20.483
σ = µm
11.Hypothesis tests & confidence intervalsMedium
t −1.98401.984

A sample of 100 students has a mean score of 78.48 and a sample standard deviation of 12. Find the test statistic t for H₀: μ = 75.

μ₀ = 75x̄ = 78.48s = 12n = 100
t =
12.Hypothesis tests & confidence intervalsHard
−1.961.96z = ?

In a trial, 98 of 150 patients on treatment A recover, and 116 of 200 patients on treatment B. Using the pooled proportion, find the test statistic z for H₀: p₁ = p₂.

x₁ = 98n₁ = 150x₂ = 116n₂ = 200
z =
13.Linear regression & correlationMedium
xy155267367477579xy24607080

The table gives 5 paired observations of x = hours studied and y = exam score. Find the coefficient of determination r² of the least-squares line.

n = 5
r² =
14.Linear regression & correlationHard
xy1333455637729791191xy510406080x = 8

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. A new observation (8, 71) is made. Find its residual e = y − ŷ.

n = 6(8, 71)
e =
15.Chi-square testsMedium
busbikecarTotalstudents19322879staff17393793Total367165172

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

row total = 79column total = 36N = 172
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 test of independence is carried out on a contingency table with 2 rows and 3 columns (not counting the totals). 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 · Counterbalancing

Why this is the answer

Counterbalancing distributes practice and fatigue effects evenly across conditions.

Answer

A · Counterbalancing
2.Research methods in psychology

Answer: D · 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

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

Answer: s = 11.28

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 = 202/5 = 40.4

Squared deviations

Σ(x − x̄)² = (−10.4)² + (−2.4)² + 5.6² + 16.6² + (−9.4)² = 509.2

Standard deviation

s = √(509.2/4) = 11.28

Answer

s = 11.28
4.Descriptive statistics

Answer: P = 0.8944

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 = 1.25

Standard normal table

Φ(1.25) = 0.8944
P = 0.8944

Answer

P = 0.8944
5.Probability

Answer: Var(X) = 0.8475

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 = 1·0.35 + 2·0.3 + 3·0.3 + 4·0.05 = 2.05

Mean of the squares

E(X²) = 1·0.35 + 4·0.3 + 9·0.3 + 16·0.05 = 5.05

Variance

Var(X) = 5.05 − 2.05² = 0.8475

Answer

Var(X) = 0.8475
6.Probability

Answer: P = 0.2816

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.

Binomial formula

P(X = 2) = 45·0.252·0.758 = 45·0.0625·0.1001 = 0.2816

Answer

P = 0.2816
7.Normal distribution

Answer: P = 0.9225

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 = (166.8 − 178)/7 = −1.6
z₂ = (192 − 178)/7 = 2

Area between

P = Φ(2) − Φ(−1.6) = 0.9772 − 0.0548 = 0.9225

Answer

P = 0.9225
8.Normal distribution

Answer: x = 251.3 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) = 0.6

z from the table

z ≈ 0.25

Back to the original units

x = 250 + 0.25·5 = 251.3 g

Answer

x = 251.3 g
9.Confidence intervals

Answer: upper limit = 24.83

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 = 11, t = 1.796

Margin of error

E = 1.796·18/√12 = 9.332

Interval

15.5 ± 9.332 → [6.168, 24.83]

Answer

upper limit = 24.83
10.Confidence intervals

Answer: σ = 7.756 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 = 10·11.1² = 1232

Limits for σ²

1232/20.483 = 60.15
1232/3.247 = 379.5

Limit for σ

σ = √60.15 = 7.756 µm

Answer

σ = 7.756 µm

95% interval for σ: [7.756, 19.48] µm.

11.Hypothesis tests & confidence intervals

Answer: t = 2.9

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 = 12/√100 = 12/10 = 1.2

Test statistic

t = (78.48 − 75)/1.2 = 2.9

Answer

t = 2.9
t0.025; df = 99 = 1.984

|t| > 1.984, so H₀ is rejected at the 5% level (two-sided).

12.Hypothesis tests & confidence intervals

Answer: z = 1.393

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 = 98/150 = 0.6533
p̂₂ = 116/200 = 0.58
p̂ = 214/350 = 0.6114

Standard error under H₀

SE = √(0.6114·0.3886·(1/150 + 1/200)) = 0.05265

Test statistic

z = (0.6533 − 0.58)/0.05265 = 1.393

Answer

z = 1.393

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

13.Linear regression & correlation

Answer: r² = 0.9141

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 = 345/5 = 69

Sums of squares and products

Sxx = Σ(x − x̄)² = 10
Sxy = Σ(x − x̄)(y − ȳ) = 58
Syy = Σ(y − ȳ)² = 368

Coefficient of determination

r² = 58²/(10·368) = 0.9141

Answer

r² = 0.9141

About 91.41% of the variation in y is explained by the line.

14.Linear regression & correlation

Answer: e = −4.29

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 = 36/6 = 6
ȳ = 383/6 = 63.83

Sums of squares and products

Sxx = Σ(x − x̄)² = 70
Sxy = Σ(x − x̄)(y − ȳ) = 401

Regression line

b = 401/70 = 5.729
a = 63.83 − 5.729·6 = 29.46
ŷ = 29.46 + 5.729·x

Predicted value

ŷ = 29.46 + 5.729·8 = 75.29

Residual

e = 71 − 75.29 = −4.29

Answer

e = −4.29

The point lies below the line.

15.Chi-square tests

Answer: E = 16.53

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 = 79·36/172 = 16.53

Answer

E = 16.53
16.Chi-square tests

Answer: χ²crit = 9.21

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)·(3 − 1) = 2

Table

χ²crit(df = 2, α = 0.01) = 9.210

Answer

χ²crit = 9.210

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

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