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

Which design uses different participants in each condition?

  1. AMatched pairs design with the same people
  2. BRepeated measures design
  3. CIndependent groups design
  4. DWithin-subjects design
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
20253035404550

Find the sample standard deviation s of the data: 50, 48, 16, 35, 19, 35, 39.

n = 7
s =
4.Descriptive statisticsHard
z −2−1012

A quantity is normally distributed with mean 170 and standard deviation 10. What proportion P of values is above 150?

μ = 170σ = 10x = 150
P =
5.ProbabilityMedium
xp(x)00.3520.2540.160.3μ = 2.7

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

x: 0, 2, 4, 6p(x): 0.35, 0.25, 0.1, 0.3
Var(X) =
6.ProbabilityHard
xp(x)00.110.220.1530.2540.3μ

X has the distribution shown, and Y = 5X + 5. Find the standard deviation σY of Y.

x: 0, 1, 2, 3, 4p(x): 0.1, 0.2, 0.15, 0.25, 0.3Y = 5X + 5
σY =
7.Normal distributionMedium
−3−2−10123zshaded area between −z and z = 0.9676

Drag z so that the shaded area between −z and z under the standard normal curve is 0.9676. Find z to two decimals.

A = 0.9676
z =
8.Normal distributionHard
35455565758595pointsbottom 80%

The exam scores are normally distributed with mean 65 points and standard deviation 10 points. Drag the cut-off so the shaded bottom 80% is to the left of it. Find the cut-off x.

μ = 65 pointsσ = 10 points80%
x = points
9.Confidence intervalsMedium
n = 11, sd​ = 6.10LUd̄ = 3.6E = ?90%

The blood pressure of 11 patients is measured before and after a treatment. The differences d = before − after (mmHg) have mean 3.6 and standard deviation 6.1. Find the margin of error E of the 90% confidence interval for the mean difference.

n = 11d̄ = 3.6sd = 6.190%t(10) = 1.812
E = mmHg
10.Confidence intervalsHard
σ = 17, n = ?x̄ − Ex̄ + Ex̄E = 390%

The population standard deviation is 17. 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 3?

σ = 17E = 390%
n =
11.Hypothesis tests & confidence intervalsMedium
t −1.9901.99

A sample of 81 students has a mean score of 77.58 and a sample standard deviation of 8. Find the test statistic t for H₀: μ = 75.

μ₀ = 75x̄ = 77.58s = 8n = 81
t =
12.Hypothesis tests & confidence intervalsHard
σ = 10, n = ?x̄ − Ex̄ + Ex̄E = 2.590%

How large must a sample be so that the 90% confidence interval for a mean has a margin of error of at most 2.5? The standard deviation is 10. Find the smallest sample size n.

σ = 10E = 2.590%
n =
13.Linear regression & correlationMedium
xy21041562382610331239xy51010203040

The table gives 6 paired observations of x = load in kN and y = extension in mm. Find the coefficient of determination r² of the least-squares line.

n = 6
r² =
14.Linear regression & correlationHard
xy124223322418516xy2420ŷ = 24

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. For which value x does the line predict ŷ = 24?

n = 5ŷ = 24
x =
15.Chi-square testsMedium
yesnoTotalmen7545120women354580Total11090200

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 = 2002 × 2
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 goodness-of-fit test compares observed counts in 5 categories with fully specified expected proportions (no parameters are estimated from the data). Using the table, find the critical value χ²crit at significance level α = 0.05.

α = 0.05
χ²crit =

3Solutions

1.Research methods in psychology

Answer: C · Independent groups design

Why this is the answer

In an independent groups design each condition has different participants, avoiding order effects but risking participant variables.

Answer

C · Independent groups design
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 = 13.07

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 = 242/7 = 34.57

Squared deviations

Σ(x − x̄)² = 15.43² + 13.43² + (−18.57)² + 0.4286² + (−15.57)² + 0.4286² + 4.429² = 1026

Standard deviation

s = √(1026/6) = 13.07

Answer

s = 13.07
4.Descriptive statistics

Answer: P = 0.9772

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 = −2

Standard normal table

Φ(−2) = 0.02275
P = 1 − 0.02275 = 0.9772

Answer

P = 0.9772
5.Probability

Answer: Var(X) = 6.11

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 + 2·0.25 + 4·0.1 + 6·0.3 = 2.7

Mean of the squares

E(X²) = 0·0.35 + 4·0.25 + 16·0.1 + 36·0.3 = 13.4

Variance

Var(X) = 13.4 − 2.7² = 6.11

Answer

Var(X) = 6.11
6.Probability

Answer: σY = 6.796

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 = 2.45, Var(X) = 1.847, σX = 1.359

Linear change

σY = |5|·1.359 = 6.796

Answer

σY = 6.796

The mean does move: E(Y) = 5·2.45 + 5 = 17.25.

7.Normal distribution

Answer: z = 2.14

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 + 0.9676)/2 = 0.9838

Read the z-table backwards

Find 0.9838 inside the table and read off z.

z = 2.14

Answer

z = 2.14
8.Normal distribution

Answer: x = 73.4 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) = 0.8

z from the table

z ≈ 0.84

Back to the original units

x = 65 + 0.84·10 = 73.4 points

Answer

x = 73.4 points
9.Confidence intervals

Answer: E = 3.333 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 = 10, t = 1.812

Margin of error

E = 1.812·6.1/√11 = 3.333

Answer

E = 3.333 mmHg
10.Confidence intervals

Answer: n = 87

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 = (1.645·17/3)² = 86.89

Round up

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

n = 87

Answer

n = 87
11.Hypothesis tests & confidence intervals

Answer: t = 2.902

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/√81 = 8/9 = 0.8889

Test statistic

t = (77.58 − 75)/0.8889 = 2.902

Answer

t = 2.902
t0.025; df = 80 = 1.99

|t| > 1.99, so H₀ is rejected at the 5% level (two-sided).

12.Hypothesis tests & confidence intervals

Answer: n = 44

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).

Solve for n

n = (1.645·10/2.5)² = 43.3

Round up

n = 44

Answer

n = 44
13.Linear regression & correlation

Answer: r² = 0.9925

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 = 146/6 = 24.33

Sums of squares and products

Sxx = Σ(x − x̄)² = 70
Sxy = Σ(x − x̄)(y − ȳ) = 202
Syy = Σ(y − ȳ)² = 587.3

Coefficient of determination

r² = 202²/(70·587.3) = 0.9925

Answer

r² = 0.9925

About 99.25% of the variation in y is explained by the line.

14.Linear regression & correlation

Answer: x = 1.381

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 = 103/5 = 20.6

Sums of squares and products

Sxx = Σ(x − x̄)² = 10
Sxy = Σ(x − x̄)(y − ȳ) = −21

Regression line

b = −21/10 = −2.1
a = 20.6 − (−2.1)·3 = 26.9
ŷ = 26.9 − 2.1·x

Solve for x

x = (24 − 26.9)/(−2.1) = 1.381

Answer

x = 1.381
15.Chi-square tests

Answer: χ² = 6.818

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

men: 120·110/200 = 66, 120·90/200 = 54
women: 80·110/200 = 44, 80·90/200 = 36

Chi-square statistic

χ² = (75 − 66)²/66 + (45 − 54)²/54 + (35 − 44)²/44 + (45 − 36)²/36
χ² = 6.818

Answer

χ² = 6.818

df = 1. At α = 0.05 the critical value is 3.841, so H₀ (independence) is rejected.

16.Chi-square tests

Answer: χ²crit = 9.488

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 = 5 − 1 = 4

Table

χ²crit(df = 4, α = 0.05) = 9.488

Answer

χ²crit = 9.488

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

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