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.

New problemsBack to the course

In the print window, choose “Save as PDF” to keep a copy on your device.

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 test is so easy that almost all participants score near the maximum. This is called:

  1. AA ceiling effect
  2. BA practice effect
  3. CA floor effect
  4. DAn order effect
2.Research methods in psychologyHard

Why do many researchers argue that p-values should be reported together with effect sizes and confidence intervals?

  1. AP-values show the probability the hypothesis is true
  2. BA p-value does not show how large or precise the effect is
  3. CEffect sizes make sample size irrelevant to significance
  4. DConfidence intervals are only needed in qualitative research
3.Descriptive statisticsMedium
20253035404550

Find the sample standard deviation s of the data: 45, 34, 21, 16, 50.

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

Exam scores are normally distributed with mean 500 and standard deviation 100. Find the z-score z of a score of 540.

μ = 500σ = 100x = 540
z =
5.ProbabilityMedium
xp(x)00.3510.2520.130.3μ = 1.35

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

x: 0, 1, 2, 3p(x): 0.35, 0.25, 0.1, 0.3
Var(X) =
6.ProbabilityHard
xp(x)100.25200.35300.15400.25μ

X has the distribution shown, and Y = 3X + 10. Find the standard deviation σY of Y.

x: 10, 20, 30, 40p(x): 0.25, 0.35, 0.15, 0.25Y = 3X + 10
σY =
7.Normal distributionMedium
−3−2−10123zshaded area to the right of z = 0.9115

Drag z so that the shaded area under the standard normal curve, to the right of z, is 0.9115. Find z to two decimals.

A = 0.9115
z =
8.Normal distributionHard
35455565758595pointstop 70%

The exam scores are normally distributed with mean 65 points and standard deviation 10 points. Drag the cut-off so the shaded top 70% is to the right of it. Find the cut-off x.

μ = 65 pointsσ = 10 points70%
x = points
9.Confidence intervalsMedium
n = 7, s = 16L = ?Ux̄ = 19E90%

A sample of 7 items has mean 19 and standard deviation s = 16. The population is roughly normal. Find the lower limit of the 90% confidence interval for the mean.

n = 7x̄ = 19s = 1690%t(6) = 1.943
lower limit =
10.Confidence intervalsHard
p = 0.5, n = ?p̂ − Ep̂ + Ep̂E = 0.0399%

Nothing is known in advance about the proportion of defective products. How many products n must be inspected so that a 99% confidence interval for the proportion has a margin of error of at most 0.03?

p = 0.5E = 0.0399%
n =
11.Hypothesis tests & confidence intervalsMedium
−2.2282.228t = ?df = 10

11 employees take a typing test before and after a training course. For each person the difference d = after − before (words per minute) is recorded. The mean difference is 0.84 and the standard deviation of the differences is 1. Find the paired test statistic t for H₀: μd = 0.

n = 11d̄ = 0.84sd = 1
t =
12.Hypothesis tests & confidence intervalsHard
z −1.64501.645

A sample of 16 measurements has mean 194.4. The population standard deviation is 15. Find the upper limit U of the 90% confidence interval for the mean.

x̄ = 194.4σ = 15n = 1690%
U =
13.Linear regression & correlationMedium
xy147252368471570xy24506070

The table gives 5 paired observations of x = hours studied and y = exam score. Find Pearson's correlation coefficient r.

n = 5
r =
14.Linear regression & correlationHard
xy121219314412510xy24101520x = 2.5

The table gives 5 paired observations of x = car age in years and y = price in k€. Use the least-squares line ŷ = a + b·x. Predict ŷ for x = 2.5.

n = 5x = 2.5
ŷ =
15.Chi-square testsMedium
lowmediumhighTotalurban19263580rural16232160Total354956140

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 = 1402 × 3
χ² =
16.Chi-square testsHard
yesnoTotalline 1182240line 24654100line 33268100Total96144240dfα = 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 = 2403 × 2α = 0.05
χ² =

3Solutions

1.Research methods in psychology

Answer: A · A ceiling effect

Why this is the answer

Ceiling effects hide differences between conditions because no one can score higher. Floor effects are the reverse, when a task is too hard.

Answer

A · A ceiling effect
2.Research methods in psychology

Answer: B · A p-value does not show how large or precise the effect is

Why this is the answer

A p-value depends heavily on sample size and only indicates how surprising the data are under the null. Effect sizes show magnitude and intervals show precision.

Answer

B · A p-value does not show how large or precise the effect is
3.Descriptive statistics

Answer: s = 14.72

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 = 166/5 = 33.2

Squared deviations

Σ(x − x̄)² = 11.8² + 0.8² + (−12.2)² + (−17.2)² + 16.8² = 866.8

Standard deviation

s = √(866.8/4) = 14.72

Answer

s = 14.72
4.Descriptive statistics

Answer: z = 0.4

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 = (540 − 500)/100 = 0.4

Answer

z = 0.4
5.Probability

Answer: Var(X) = 1.528

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 = 0·0.35 + 1·0.25 + 2·0.1 + 3·0.3 = 1.35

Mean of the squares

E(X²) = 0·0.35 + 1·0.25 + 4·0.1 + 9·0.3 = 3.35

Variance

Var(X) = 3.35 − 1.35² = 1.528

Answer

Var(X) = 1.528
6.Probability

Answer: σY = 33.41

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.

Spread of X

μ = 24, Var(X) = 124, σX = 11.14

Linear change

σY = |3|·11.14 = 33.41

Answer

σY = 33.41

The mean does move: E(Y) = 3·24 + 10 = 82.

7.Normal distribution

Answer: z = −1.35

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 − 0.9115 = 0.0885

Read the z-table backwards

Find 0.0885 inside the table and read off z.

z = −1.35

Answer

z = −1.35
8.Normal distribution

Answer: x = 59.8 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) = 0.3

z from the table

z ≈ −0.52

Back to the original units

x = 65 − 0.52·10 = 59.8 points

Answer

x = 59.8 points
9.Confidence intervals

Answer: lower limit = 7.25

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 = 6, t = 1.943

Margin of error

E = 1.943·16/√7 = 11.75

Interval

19 ± 11.75 → [7.25, 30.75]

Answer

lower limit = 7.25
10.Confidence intervals

Answer: n = 1844

Confidence interval for a proportion: p̂ ± z·√(p̂(1 − p̂)/n), where p̂ = x/n. To reach a margin E you need n ≥ z²·p(1 − p)/E², rounded up; without a prior estimate use p = 0.5, which gives the largest n.

Solve for n

n = 2.576²·0.5·0.5/0.03² = 1843

Round up

n = 1844

Answer

n = 1844
11.Hypothesis tests & confidence intervals

Answer: t = 2.786

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 of the mean difference

SE = 1/√11 = 0.3015

Test statistic

t = 0.84/0.3015 = 2.786

Answer

t = 2.786

With df = 10, the two-sided 5% critical value is 2.228; |t| is larger, so H₀ is rejected.

12.Hypothesis tests & confidence intervals

Answer: U = 200.6

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 = 1.645·15/√16 = 6.169

Interval limit

U = 194.4 + 6.169 = 200.6

Answer

U = 200.6
13.Linear regression & correlation

Answer: r = 0.9145

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 = 308/5 = 61.6

Sums of squares and products

Sxx = Σ(x − x̄)² = 10
Sxy = Σ(x − x̄)(y − ȳ) = 65
Syy = Σ(y − ȳ)² = 505.2

Correlation coefficient

r = 65/√(10·505.2) = 0.9145

Answer

r = 0.9145
14.Linear regression & correlation

Answer: ŷ = 16.65

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 = 76/5 = 15.2

Sums of squares and products

Sxx = Σ(x − x̄)² = 10
Sxy = Σ(x − x̄)(y − ȳ) = −29

Regression line

b = −29/10 = −2.9
a = 15.2 − (−2.9)·3 = 23.9
ŷ = 23.9 − 2.9·x

Prediction

ŷ = 23.9 + (−2.9)·2.5 = 16.65

Answer

ŷ = 16.65
15.Chi-square tests

Answer: χ² = 1.106

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: 80·35/140 = 20, 80·49/140 = 28, 80·56/140 = 32
rural: 60·35/140 = 15, 60·49/140 = 21, 60·56/140 = 24

Chi-square statistic

χ² = (19 − 20)²/20 + (26 − 28)²/28 + (35 − 32)²/32 + (16 − 15)²/15 + (23 − 21)²/21 + (21 − 24)²/24
χ² = 1.106

Answer

χ² = 1.106

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

16.Chi-square tests

Answer: χ² = 4.583

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: 16, 24
line 2: 40, 60
line 3: 40, 60

Chi-square statistic

χ² = (18 − 16)²/16 + (22 − 24)²/24 + (46 − 40)²/40 + (54 − 60)²/60 + (32 − 40)²/40 + (68 − 60)²/60
χ² = 4.583

Decision

df = 2, χ²crit = 5.991
4.583 < 5.991

Answer

χ² = 4.583

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

ExamProblems · examproblems.com/courses/research-methods-psychology