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

Standardised instructions are used in an experiment so that:

  1. AParticipants can choose their own conditions
  2. BAll participants receive the same information in the same way
  3. CThe researcher can adapt the procedure to suit each participant's needs
  4. DThe results become statistically significant
2.Research methods in psychologyHard

A study of 10 participants finds a large but non-significant effect. What is the most appropriate conclusion?

  1. AThe study may have lacked power, so the result is inconclusive
  2. BThe effect size must be miscalculated
  3. CThe study proves the null hypothesis is true, since the effect was not significant
  4. DThe effect must be a Type I error
3.Descriptive statisticsMedium
15202530354045505560

Find the sample standard deviation s of the data: 50, 25, 60, 14, 20, 16.

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

A quantity is normally distributed with mean 500 and standard deviation 75. What proportion P of values is below 530?

μ = 500σ = 75x = 530
P =
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 = 10, p = 0.25: P(X = 1) = ?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 1 correct answers?

n = 10p = 0.25k = 1
P =
7.Normal distributionMedium
−3−2−10123zshaded area between −z and z = 0.5587

Drag z so that the shaded area between −z and z under the standard normal curve is 0.5587. Find z to two decimals.

A = 0.5587
z =
8.Normal distributionHard
10 cm0X − Y?

The heights X of men and Y of women in a group are independent and normal: X has mean 180 cm and standard deviation 10 cm, Y has mean 170 cm and standard deviation 12 cm. Find the probability P that X is greater than Y. Round z to two decimals.

μX = 180 cm, σX = 10 cmμY = 170 cm, σY = 12 cm
P =
9.Confidence intervalsMedium
n = 12, sd​ = 7.40L = ?Ud̄ = 7.3E99%

The blood pressure of 12 patients is measured before and after a treatment. The differences d = before − after (mmHg) have mean 7.3 and standard deviation 7.4. Find the lower limit of the 99% confidence interval for the mean difference.

n = 12d̄ = 7.3sd = 7.499%t(11) = 3.106
lower limit = mmHg
10.Confidence intervalsHard
σ = 22, n = ?x̄ − Ex̄ + Ex̄E = 590%

The population standard deviation is 22. How large a sample n do you need so that a 90% confidence interval for the mean has a margin of error of at most 5?

σ = 22E = 590%
n =
11.Hypothesis tests & confidence intervalsMedium
t −1.99801.998

A sample of 64 students has a mean score of 198.6 and a sample standard deviation of 8. Find the test statistic t for H₀: μ = 200.

μ₀ = 200x̄ = 198.6s = 8n = 64
t =
12.Hypothesis tests & confidence intervalsHard
z −1.64501.645

A sample of 36 measurements has mean 122.8. The population standard deviation is 12. Find the lower limit L of the 90% confidence interval for the mean.

x̄ = 122.8σ = 12n = 3690%
L =
13.Linear regression & correlationMedium
xy126221324422520xy2420222426

The table gives 5 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 = 5
r² =
14.Linear regression & correlationHard
xy2741462282510351238xy51010203040x = 3

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. A new observation (3, 15) is made. Find its residual e = y − ŷ.

n = 6(3, 15)
e =
15.Chi-square testsMedium
coughno coughTotalsmokers164460non-smokers4773120Total63117180

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 = 1802 × 2
χ² =
16.Chi-square testsHard
yesnoTotalline 1184260line 2314980line 3176380Total66154220dfα = 0.05α = 0.0113.8416.63525.9919.21037.81511.34549.48813.277511.07015.086612.59216.812714.06718.475815.50720.090916.91921.6661018.30723.209

The table shows observed counts. Test at α = 0.01 whether the two variables are independent (no continuity correction). Compute the test statistic χ², then compare it with the critical value from the table.

N = 2203 × 2α = 0.01
χ² =

3Solutions

1.Research methods in psychology

Answer: B · All participants receive the same information in the same way

Why this is the answer

Standardisation removes variation in procedure as an extraneous variable and makes the study replicable.

Answer

B · All participants receive the same information in the same way
2.Research methods in psychology

Answer: A · The study may have lacked power, so the result is inconclusive

Why this is the answer

With small samples, even real effects often fail to reach significance (Type II error). Researchers should report the effect size and confidence interval and replicate with a larger sample.

Answer

A · The study may have lacked power, so the result is inconclusive
3.Descriptive statistics

Answer: s = 19.35

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 = 185/6 = 30.83

Squared deviations

Σ(x − x̄)² = 19.17² + (−5.833)² + 29.17² + (−16.83)² + (−10.83)² + (−14.83)² = 1873

Standard deviation

s = √(1873/5) = 19.35

Answer

s = 19.35
4.Descriptive statistics

Answer: P = 0.6554

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 = 0.4

Standard normal table

Φ(0.4) = 0.6554
P = 0.6554

Answer

P = 0.6554
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 = 0.1877

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) = 10·0.251·0.759 = 10·0.25·0.07508 = 0.1877

Answer

P = 0.1877
7.Normal distribution

Answer: z = 0.77

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 + 0.5587)/2 = 0.7794

Read the z-table backwards

Find 0.7794 inside the table and read off z.

z = 0.77

Answer

z = 0.77
8.Normal distribution

Answer: P = 0.7389

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 = 180 − 170 = 10 cm, σ = 15.62 cm

Standardise

z = (0 − 10)/15.62 = −0.64

Probability

P = 1 − Φ(−0.64) = 0.7389

Answer

P = 0.7389
9.Confidence intervals

Answer: lower limit = 0.665 mmHg

Interval for a difference: estimate ± critical value × standard error. For two independent means with known σ: SE = √(σ₁²/n₁ + σ₂²/n₂) and z. With equal unknown variances: pooled s_p and t with n₁ + n₂ − 2 degrees of freedom. For paired data, work with the differences: d̄ ± t·s_d/√n.

Critical value

df = 11, t = 3.106

Margin of error

E = 3.106·7.4/√12 = 6.635

Interval

7.3 ± 6.635 → [0.665, 13.94]

Answer

lower limit = 0.665 mmHg
10.Confidence intervals

Answer: n = 53

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 = (1.645·22/5)² = 52.39

Round up

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

n = 53

Answer

n = 53
11.Hypothesis tests & confidence intervals

Answer: t = −1.4

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 = 8/√64 = 8/8 = 1

Test statistic

t = (198.6 − 200)/1 = −1.4

Answer

t = −1.4
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: L = 119.5

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 = 1.645·12/√36 = 3.29

Interval limit

L = 122.8 − 3.29 = 119.5

Answer

L = 119.5
13.Linear regression & correlation

Answer: r² = 0.5216

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 = 113/5 = 22.6

Sums of squares and products

Sxx = Σ(x − x̄)² = 10
Sxy = Σ(x − x̄)(y − ȳ) = −11
Syy = Σ(y − ȳ)² = 23.2

Coefficient of determination

r² = −11²/(10·23.2) = 0.5216

Answer

r² = 0.5216

About 52.16% of the variation in y is explained by the line.

14.Linear regression & correlation

Answer: e = 4.129

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 = 141/6 = 23.5

Sums of squares and products

Sxx = Σ(x − x̄)² = 70
Sxy = Σ(x − x̄)(y − ȳ) = 221

Regression line

b = 221/70 = 3.157
a = 23.5 − 3.157·7 = 1.4
ŷ = 1.4 + 3.157·x

Predicted value

ŷ = 1.4 + 3.157·3 = 10.87

Residual

e = 15 − 10.87 = 4.129

Answer

e = 4.129

The point lies above the line.

15.Chi-square tests

Answer: χ² = 2.747

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

smokers: 60·63/180 = 21, 60·117/180 = 39
non-smokers: 120·63/180 = 42, 120·117/180 = 78

Chi-square statistic

χ² = (16 − 21)²/21 + (44 − 39)²/39 + (47 − 42)²/42 + (73 − 78)²/78
χ² = 2.747

Answer

χ² = 2.747

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

16.Chi-square tests

Answer: χ² = 5.833

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

line 1: 18, 42
line 2: 24, 56
line 3: 24, 56

Chi-square statistic

χ² = (18 − 18)²/18 + (42 − 42)²/42 + (31 − 24)²/24 + (49 − 56)²/56 + (17 − 24)²/24 + (63 − 56)²/56
χ² = 5.833

Decision

df = 2, χ²crit = 9.210
5.833 < 9.210

Answer

χ² = 5.833

χ² is below the critical value, so H₀ is not rejected: there is no significant evidence of a relationship.

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