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

The standard error of the mean becomes smaller when:

  1. AThe sample size increases
  2. BThe sample size decreases
  3. CThe variable is measured at ordinal level
  4. DThe mean increases
2.Research methods in psychologyHard

Effect size measures such as Cohen's d are reported because they:

  1. ATell how many participants dropped out
  2. BReplace the need for any hypothesis
  3. CShow whether the result is statistically significant at the chosen alpha level
  4. DShow how large an effect is, regardless of sample size
3.Descriptive statisticsMedium
303540455055

Find the sample standard deviation s of the data: 56, 39, 48, 28, 34.

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

Exam scores are normally distributed with mean 50 and standard deviation 5. Find the z-score z of a score of 40.

μ = 50σ = 5x = 40
z =
5.ProbabilityMedium
xp(x)10.120.3530.1540.0550.35μ = 3.2

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

x: 1, 2, 3, 4, 5p(x): 0.1, 0.35, 0.15, 0.05, 0.35
σ =
6.ProbabilityHard
n = 7, p = 0.6: P(X = 3) = ?k01234567

A test has 7 multiple-choice questions. A student who guesses gets each one right with probability 0.6, independently. What is the probability P of exactly 3 correct answers?

n = 7p = 0.6k = 3
P =
7.Normal distributionMedium
−3−2−10123zshaded area to the right of z = 0.7611

Drag z so that the shaded area under the standard normal curve, to the right of z, is 0.7611. Find z to two decimals.

A = 0.7611
z =
8.Normal distributionHard
157164171178185192199cmbottom 70%

The heights of adult men are normally distributed with mean 178 cm and standard deviation 7 cm. Drag the cut-off so the shaded bottom 70% is to the left of it. Find the cut-off x.

μ = 178 cmσ = 7 cm70%
x = cm
9.Confidence intervalsMedium
n = 8, s = 5L = ?Ux̄ = 93E95%

A sample of 8 items has mean 93 and standard deviation s = 5. The population is roughly normal. Find the lower limit of the 95% confidence interval for the mean.

n = 8x̄ = 93s = 595%t(7) = 2.365
lower limit =
10.Confidence intervalsHard
n = 20Lσ = ?s = 10.295%

The diameters of 20 machined shafts have a sample standard deviation of 10.2 µm. Assuming the diameters are normally distributed, find the upper limit σ of the 95% confidence interval for the population standard deviation. With df = 19, 2.5% of the χ² distribution lies below 8.907 and 2.5% lies above 32.852.

n = 20s = 10.2 µmχ² = 8.907, 32.852
σ = µm
11.Hypothesis tests & confidence intervalsMedium
−2.1452.145t = ?df = 14

Two teaching methods are compared. Group 1: 10 students, mean score 107, standard deviation 8. Group 2: 6 students, mean score 102.4, standard deviation 4. 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 = 10x̄₁ = 107s₁ = 8n₂ = 6x̄₂ = 102.4s₂ = 4
t =
12.Hypothesis tests & confidence intervalsHard
−1.961.96z = ?

In a trial, 35 of 100 patients on treatment A recover, and 50 of 120 patients on treatment B. Using the pooled proportion, find the test statistic z for H₀: p₁ = p₂.

x₁ = 35n₁ = 100x₂ = 50n₂ = 120
z =
13.Linear regression & correlationMedium
xy244861081410181223xy5101020

The table gives 6 paired observations of x = load in kN and y = extension in mm. Their sums are given. Find Pearson's correlation coefficient r from these sums.

n = 6Σx = 42Σy = 77Σx² = 364Σy² = 1229Σxy = 668
r =
14.Linear regression & correlationHard
xy2641161681810221224xy5101020x = 7

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 (7, 18) is made. Find its residual e = y − ŷ.

n = 6(7, 18)
e =
15.Chi-square testsMedium
lowmediumhighTotalurban1891340rural2211740Total40202080

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 = 802 × 3
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 test of independence is carried out on a contingency table with 2 rows and 6 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 · The sample size increases

Why this is the answer

The standard error estimates how much sample means vary around the population mean. Larger samples give more precise estimates.

Answer

A · The sample size increases
2.Research methods in psychology

Answer: D · Show how large an effect is, regardless of sample size

Why this is the answer

With large samples tiny effects can be significant. Effect sizes express the magnitude of a difference, e.g. d = 0.2 small, 0.5 medium, 0.8 large.

Answer

D · Show how large an effect is, regardless of sample size
3.Descriptive statistics

Answer: s = 11.14

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 = 205/5 = 41

Squared deviations

Σ(x − x̄)² = 15² + (−2)² + 7² + (−13)² + (−7)² = 496

Standard deviation

s = √(496/4) = 11.14

Answer

s = 11.14
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 = (40 − 50)/5 = −2

Answer

z = −2
5.Probability

Answer: σ = 1.47

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 = 1·0.1 + 2·0.35 + 3·0.15 + 4·0.05 + 5·0.35 = 3.2

Mean of the squares

E(X²) = 1·0.1 + 4·0.35 + 9·0.15 + 16·0.05 + 25·0.35 = 12.4

Variance

Var(X) = 12.4 − 3.2² = 2.16
σ = √2.16 = 1.47

Answer

σ = 1.47
6.Probability

Answer: P = 0.1935

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 = 3) = 35·0.63·0.44 = 35·0.216·0.0256 = 0.1935

Answer

P = 0.1935
7.Normal distribution

Answer: z = −0.71

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 − 0.7611 = 0.2389

Read the z-table backwards

Find 0.2389 inside the table and read off z.

z = −0.71

Answer

z = −0.71
8.Normal distribution

Answer: x = 181.6 cm

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) = 0.7

z from the table

z ≈ 0.52

Back to the original units

x = 178 + 0.52·7 = 181.6 cm

Answer

x = 181.6 cm
9.Confidence intervals

Answer: lower limit = 88.82

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 = 7, t = 2.365

Margin of error

E = 2.365·5/√8 = 4.181

Interval

93 ± 4.181 → [88.82, 97.18]

Answer

lower limit = 88.82
10.Confidence intervals

Answer: σ = 14.9 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 = 19·10.2² = 1977

Limits for σ²

1977/32.852 = 60.17
1977/8.907 = 221.9

Limit for σ

σ = √221.9 = 14.9 µm

Answer

σ = 14.9 µm

95% interval for σ: [7.757, 14.9] µm.

11.Hypothesis tests & confidence intervals

Answer: t = 1.301

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 = (9·8² + 5·4²)/14 = 46.86
sp = 6.845

Standard error

SE = 6.845·√(1/10 + 1/6) = 3.535

Test statistic

t = (107 − 102.4)/3.535 = 1.301

Answer

t = 1.301

With df = 14, the two-sided 5% critical value is 2.145; |t| is smaller, so H₀ is not rejected.

12.Hypothesis tests & confidence intervals

Answer: z = −1.011

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 = 35/100 = 0.35
p̂₂ = 50/120 = 0.4167
p̂ = 85/220 = 0.3864

Standard error under H₀

SE = √(0.3864·0.6136·(1/100 + 1/120)) = 0.06593

Test statistic

z = (0.35 − 0.4167)/0.06593 = −1.011

Answer

z = −1.011

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

13.Linear regression & correlation

Answer: r = 0.9935

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.

Centred sums

Sxy = 668 − 42·77/6 = 129
Sxx = 364 − 42²/6 = 70
Syy = 1229 − 77²/6 = 240.8

Correlation coefficient

r = 129/√(70·240.8) = 0.9935

Answer

r = 0.9935
14.Linear regression & correlation

Answer: e = 1.833

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 = 97/6 = 16.17

Sums of squares and products

Sxx = Σ(x − x̄)² = 70
Sxy = Σ(x − x̄)(y − ȳ) = 125

Regression line

b = 125/70 = 1.786
a = 16.17 − 1.786·7 = 3.667
ŷ = 3.667 + 1.786·x

Predicted value

ŷ = 3.667 + 1.786·7 = 16.17

Residual

e = 18 − 16.17 = 1.833

Answer

e = 1.833

The point lies above the line.

15.Chi-square tests

Answer: χ² = 2.4

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: 40·40/80 = 20, 40·20/80 = 10, 40·20/80 = 10
rural: 40·40/80 = 20, 40·20/80 = 10, 40·20/80 = 10

Chi-square statistic

χ² = (18 − 20)²/20 + (9 − 10)²/10 + (13 − 10)²/10 + (22 − 20)²/20 + (11 − 10)²/10 + (7 − 10)²/10
χ² = 2.4

Answer

χ² = 2.4

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

16.Chi-square tests

Answer: χ²crit = 15.09

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 = (2 − 1)·(6 − 1) = 5

Table

χ²crit(df = 5, α = 0.01) = 15.086

Answer

χ²crit = 15.086

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

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