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 correlation, r = 0.40. What proportion of variance in one variable is shared with the other?

  1. A40%
  2. B16%
  3. C4%
  4. D60%
2.Research methods in psychologyHard

A meta-analysis combines 40 studies of a therapy. Why might its overall effect size still be too large?

  1. APublication bias means unpublished null studies may be missing
  2. BMeta-analyses always give smaller effects than the single studies they include
  3. CCombining studies always removes all bias
  4. DMeta-analyses cannot include randomised trials
3.Descriptive statisticsMedium
15202530354045

Find the sample standard deviation s of the data: 32, 11, 16, 42, 32, 48.

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

A quantity is normally distributed with mean 60 and standard deviation 15. What proportion P of values is above 69?

μ = 60σ = 15x = 69
P =
5.ProbabilityMedium
n = 6k = 4N = ?

In how many ways N can a committee of 4 people be chosen from a group of 6? The order does not matter.

n = 6k = 4
N =
6.ProbabilityHard
n = 6, p = 0.6: P(X = 5) = ?k0123456

A test has 6 multiple-choice questions. A student who guesses gets each one right with probability 0.6, independently. What is the probability P of exactly 5 correct answers?

n = 6p = 0.6k = 5
P =
7.Normal distributionMedium
1360 kg1456 kgT = X₁ + … + X₁₆?

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 16 of them is below 1456 kg. Round z to two decimals.

μ = 85 kgσ = 12 kgn = 16< 1456 kg
P =
8.Normal distributionHard
μ = 178x = ?75%

The heights of adult men are normally distributed with mean 178 cm and standard deviation 7 cm. Find the value x below which 75% of them lie (the 75th percentile).

μ = 178σ = 7P = 75%
x = cm
9.Confidence intervalsMedium
n = 12, sd​ = 5.30LUd̄ = 8E = ?90%

The blood pressure of 12 patients is measured before and after a treatment. The differences d = before − after (mmHg) have mean 8 and standard deviation 5.3. Find the margin of error E of the 90% confidence interval for the mean difference.

n = 12d̄ = 8sd = 5.390%t(11) = 1.796
E = mmHg
10.Confidence intervalsHard
n = 16Lσ = ?s = 8.995%

The diameters of 16 machined shafts have a sample standard deviation of 8.9 µm. Assuming the diameters are normally distributed, find the upper limit σ of the 95% confidence interval for the population standard deviation. With df = 15, 2.5% of the χ² distribution lies below 6.262 and 2.5% lies above 27.488.

n = 16s = 8.9 µmχ² = 6.262, 27.488
σ = µm
11.Hypothesis tests & confidence intervalsMedium
t −1.98401.984

A sample of 100 students has a mean score of 101 and a sample standard deviation of 5. Find the test statistic t for H₀: μ = 100.

μ₀ = 100x̄ = 101s = 5n = 100
t =
12.Hypothesis tests & confidence intervalsHard
z −1.9601.96

A sample of 25 measurements has mean 50.2. The population standard deviation is 10. Find the lower limit L of the 95% confidence interval for the mean.

x̄ = 50.2σ = 10n = 2595%
L =
13.Linear regression & correlationMedium
xy151265365471579xy2450607080

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
xy284136198261033xy510102030x = 9

The table gives 5 paired observations of x = load in kN and y = extension in mm. Use the least-squares line ŷ = a + b·x. A new observation (9, 32) is made. Find its residual e = y − ŷ.

n = 5(9, 32)
e =
15.Chi-square testsMedium
lowmediumhighTotalurban15261960rural303753120Total456372180

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 × 3
χ² =
16.Chi-square testsHard
yesnoTotalage 18–305545100age 31–5036440age 51+74680Total16555220dfα = 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 = 0.05 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 = 0.05
χ² =

3Solutions

1.Research methods in psychology

Answer: B · 16%

Why this is the answer

The coefficient of determination is r², here 0.40² = 0.16, so 16% of the variance is shared.

Answer

B · 16%
2.Research methods in psychology

Answer: A · Publication bias means unpublished null studies may be missing

Why this is the answer

If null results are less often published, the pooled effect is inflated. Funnel plots and searches for unpublished data help detect and reduce this.

Answer

A · Publication bias means unpublished null studies may be missing
3.Descriptive statistics

Answer: s = 14.37

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 = 181/6 = 30.17

Squared deviations

Σ(x − x̄)² = 1.833² + (−19.17)² + (−14.17)² + 11.83² + 1.833² + 17.83² = 1033

Standard deviation

s = √(1033/5) = 14.37

Answer

s = 14.37
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: N = 15

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(6, 4) = (6·5·4·3)/(4·3·2·1) = 15

Answer

N = 15
6.Probability

Answer: P = 0.1866

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 = 5) = 6·0.65·0.41 = 6·0.07776·0.4 = 0.1866

Answer

P = 0.1866
7.Normal distribution

Answer: P = 0.9772

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 = 1360 kg, σT = 12·√16 = 48 kg

Standardise

z = (1456 − 1360)/48 = 2

Probability

P = Φ(2) = 0.9772

Answer

P = 0.9772
8.Normal distribution

Answer: x = 182.7 cm

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 for the percentile

Φ(z) = 0.75 → z = 0.6745

Back to x

x = 178 + 0.6745·7 = 182.72 cm

Answer

x = 182.72 cm
9.Confidence intervals

Answer: E = 2.748 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 = 1.796

Margin of error

E = 1.796·5.3/√12 = 2.748

Answer

E = 2.748 mmHg
10.Confidence intervals

Answer: σ = 13.77 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 = 15·8.9² = 1188

Limits for σ²

1188/27.488 = 43.22
1188/6.262 = 189.7

Limit for σ

σ = √189.7 = 13.77 µm

Answer

σ = 13.77 µm

95% interval for σ: [6.575, 13.77] µm.

11.Hypothesis tests & confidence intervals

Answer: t = 2

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 = 5/√100 = 5/10 = 0.5

Test statistic

t = (101 − 100)/0.5 = 2

Answer

t = 2
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: L = 46.28

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 = 1.96·10/√25 = 3.92

Interval limit

L = 50.2 − 3.92 = 46.28

Answer

L = 46.28
13.Linear regression & correlation

Answer: r² = 0.9135

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 = 331/5 = 66.2

Sums of squares and products

Sxx = Σ(x − x̄)² = 10
Sxy = Σ(x − x̄)(y − ȳ) = 62
Syy = Σ(y − ȳ)² = 420.8

Coefficient of determination

r² = 62²/(10·420.8) = 0.9135

Answer

r² = 0.9135

About 91.35% of the variation in y is explained by the line.

14.Linear regression & correlation

Answer: e = 2.75

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 = 30/5 = 6
ȳ = 99/5 = 19.8

Sums of squares and products

Sxx = Σ(x − x̄)² = 40
Sxy = Σ(x − x̄)(y − ȳ) = 126

Regression line

b = 126/40 = 3.15
a = 19.8 − 3.15·6 = 0.9
ŷ = 0.9 + 3.15·x

Predicted value

ŷ = 0.9 + 3.15·9 = 29.25

Residual

e = 32 − 29.25 = 2.75

Answer

e = 2.75

The point lies above the line.

15.Chi-square tests

Answer: χ² = 3.348

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

urban: 60·45/180 = 15, 60·63/180 = 21, 60·72/180 = 24
rural: 120·45/180 = 30, 120·63/180 = 42, 120·72/180 = 48

Chi-square statistic

χ² = (15 − 15)²/15 + (26 − 21)²/21 + (19 − 24)²/24 + (30 − 30)²/30 + (37 − 42)²/42 + (53 − 48)²/48
χ² = 3.348

Answer

χ² = 3.348

df = 2. At α = 0.05 the critical value is 5.991, so H₀ (independence) is not rejected.

16.Chi-square tests

Answer: χ² = 39.2

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

age 18–30: 75, 25
age 31–50: 30, 10
age 51+: 60, 20

Chi-square statistic

χ² = (55 − 75)²/75 + (45 − 25)²/25 + (36 − 30)²/30 + (4 − 10)²/10 + (74 − 60)²/60 + (6 − 20)²/20
χ² = 39.2

Decision

df = 2, χ²crit = 5.991
39.2 > 5.991

Answer

χ² = 39.2

χ² is above the critical value, so H₀ is rejected: the variables appear to be related.

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