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

A researcher sends a questionnaire to all students and then asks respondents to recruit friends who also use a rare sleep app. Which sampling method is used for the second step?

  1. ASnowball sampling
  2. BSystematic sampling
  3. CStratified sampling
  4. DRandom sampling
2.Research methods in psychologyHard

What is the main purpose of random allocation to conditions in an experiment?

  1. ATo ensure participants do not know the aim
  2. BTo spread participant variables evenly so they do not confound the results
  3. CTo make the sample representative of the target population so results generalise
  4. DTo increase the response rate
3.Descriptive statisticsMedium
1520253035404550

Find the sample standard deviation s of the data: 12, 31, 21, 50, 13.

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

Exam scores are normally distributed with mean 100 and standard deviation 8. Find the z-score z of a score of 116.

μ = 100σ = 8x = 116
z =
5.ProbabilityMedium
P(at least one six) = ??1?2?3?4P(six) = 1/6 on each roll

A fair die is rolled 4 times. What is the probability P of getting at least one six?

n = 4
P =
6.ProbabilityHard
n = 9, p = 0.2: P(X = 1) = ?k0123456789

A test has 9 multiple-choice questions. A student who guesses gets each one right with probability 0.2, independently. What is the probability P of exactly 1 correct answers?

n = 9p = 0.2k = 1
P =
7.Normal distributionMedium
−3−2−10123zshaded area between −z and z = 0.7063

Drag z so that the shaded area between −z and z under the standard normal curve is 0.7063. Find z to two decimals.

A = 0.7063
z =
8.Normal distributionHard
557085100115130145pointstop 40%

The IQ scores are normally distributed with mean 100 points and standard deviation 15 points. Drag the cut-off so the shaded top 40% is to the right of it. Find the cut-off x.

μ = 100 pointsσ = 15 points40%
x = points
9.Confidence intervalsMedium
n = 21, s = 20L = ?Ux̄ = 60.5E99%

A sample of 21 items has mean 60.5 and standard deviation s = 20. The population is roughly normal. Find the lower limit of the 99% confidence interval for the mean.

n = 21x̄ = 60.5s = 2099%t(20) = 2.845
lower limit =
10.Confidence intervalsHard
n = 15σ = ?Us = 6.695%

The diameters of 15 machined shafts have a sample standard deviation of 6.6 µm. Assuming the diameters are normally distributed, find the lower limit σ of the 95% confidence interval for the population standard deviation. With df = 14, 2.5% of the χ² distribution lies below 5.629 and 2.5% lies above 26.119.

n = 15s = 6.6 µmχ² = 5.629, 26.119
σ = µm
11.Hypothesis tests & confidence intervalsMedium
t −2.0302.03

A sample of 36 students has a mean score of 53.5 and a sample standard deviation of 10. Find the test statistic t for H₀: μ = 50.

μ₀ = 50x̄ = 53.5s = 10n = 36
t =
12.Hypothesis tests & confidence intervalsHard
μ₀x̄c

A one-sided z test of H₀: μ = 80 against H₁: μ > 80 is done at α = 0.10 with a sample of 36; the standard deviation is known to be 6. In reality the mean is 83. Find the probability of a Type II error β in percent.

μ₀ = 80μ = 83σ = 6n = 36α = 0.10
β = %
13.Linear regression & correlationMedium
xy125226324423522619xy2462025

The table gives 6 paired observations of x = car age in years and y = price in k€. Find the coefficient of determination r² of the least-squares line.

n = 6
r² =
14.Linear regression & correlationHard
xy234961381510161221xy5101020x = 9

The table gives 6 paired observations of x = load in kN and y = extension in mm. Use the least-squares line ŷ = a + b·x. Predict ŷ for x = 9.

n = 6x = 9
ŷ =
15.Chi-square testsMedium
lowmediumhighTotalurban22102254rural26363597Total484657151

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

row total = 54column total = 46N = 151
E =
16.Chi-square testsHard

A chi-square goodness-of-fit test checks whether counts in 7 categories follow a Poisson distribution whose mean is estimated from the same data. How many degrees of freedom df does the test have?

k = 7
df =

3Solutions

1.Research methods in psychology

Answer: A · Snowball sampling

Why this is the answer

Snowball sampling is useful for hard-to-reach groups, but the sample is likely to share characteristics and is not representative.

Answer

A · Snowball sampling
2.Research methods in psychology

Answer: B · To spread participant variables evenly so they do not confound the results

Why this is the answer

Random allocation balances individual differences across conditions, supporting causal conclusions. Random sampling concerns generalisation to a population.

Answer

B · To spread participant variables evenly so they do not confound the results
3.Descriptive statistics

Answer: s = 15.73

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 = 127/5 = 25.4

Squared deviations

Σ(x − x̄)² = (−13.4)² + 5.6² + (−4.4)² + 24.6² + (−12.4)² = 989.2

Standard deviation

s = √(989.2/4) = 15.73

Answer

s = 15.73
4.Descriptive statistics

Answer: z = 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 = (116 − 100)/8 = 2

Answer

z = 2
5.Probability

Answer: P = 0.5177

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.

Complement

P(no six) = (5/6)4 = 0.4823
P = 1 − 0.4823 = 0.5177

Answer

P = 0.5177
6.Probability

Answer: P = 0.302

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 = 1) = 9·0.21·0.88 = 9·0.2·0.1678 = 0.302

Answer

P = 0.302
7.Normal distribution

Answer: z = 1.05

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 + 0.7063)/2 = 0.8531

Read the z-table backwards

Find 0.8531 inside the table and read off z.

z = 1.05

Answer

z = 1.05
8.Normal distribution

Answer: x = 103.8 points

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 = 100 + 0.25·15 = 103.8 points

Answer

x = 103.8 points
9.Confidence intervals

Answer: lower limit = 48.08

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 = 20, t = 2.845

Margin of error

E = 2.845·20/√21 = 12.42

Interval

60.5 ± 12.42 → [48.08, 72.92]

Answer

lower limit = 48.08
10.Confidence intervals

Answer: σ = 4.832 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 = 14·6.6² = 609.8

Limits for σ²

609.8/26.119 = 23.35
609.8/5.629 = 108.3

Limit for σ

σ = √23.35 = 4.832 µm

Answer

σ = 4.832 µm

95% interval for σ: [4.832, 10.41] µm.

11.Hypothesis tests & confidence intervals

Answer: t = 2.1

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/√36 = 10/6 = 1.667

Test statistic

t = (53.5 − 50)/1.667 = 2.1

Answer

t = 2.1
t0.025; df = 35 = 2.03

|t| > 2.03, so H₀ is rejected at the 5% level (two-sided).

12.Hypothesis tests & confidence intervals

Answer: β = 4.29 %

Power of a test: power = 1 − β is the probability of rejecting H₀ when it is false. Find the cutoff x̄c from μ₀ and z_α, then compute the probability of landing beyond x̄c using the true mean: z = (x̄c − μ)/(σ/√n).

Critical sample mean

SE = 6/√36 = 1
x̄c = 80 + 1.282·1 = 81.28

z under the true mean

z = (81.28 − 83)/1 = −1.718

Type II error

power = P(Z > −1.718) = 95.71%
β = 100% − 95.71% = 4.29%

Answer

β = 4.29%
13.Linear regression & correlation

Answer: r² = 0.8567

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 = 139/6 = 23.17

Sums of squares and products

Sxx = Σ(x − x̄)² = 17.5
Sxy = Σ(x − x̄)(y − ȳ) = −21.5
Syy = Σ(y − ȳ)² = 30.83

Coefficient of determination

r² = −21.5²/(17.5·30.83) = 0.8567

Answer

r² = 0.8567

About 85.67% of the variation in y is explained by the line.

14.Linear regression & correlation

Answer: ŷ = 16.06

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 = 42/6 = 7
ȳ = 77/6 = 12.83

Sums of squares and products

Sxx = Σ(x − x̄)² = 70
Sxy = Σ(x − x̄)(y − ȳ) = 113

Regression line

b = 113/70 = 1.614
a = 12.83 − 1.614·7 = 1.533
ŷ = 1.533 + 1.614·x

Prediction

ŷ = 1.533 + 1.614·9 = 16.06

Answer

ŷ = 16.06
15.Chi-square tests

Answer: E = 16.45

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 = 54·46/151 = 16.45

Answer

E = 16.45
16.Chi-square tests

Answer: df = 5

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 = 7 − 1 − 1 = 5

Answer

df = 5

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