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
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.
1Formulas and methods
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
- Convert the value to a z-score: z = (x − μ)/σ.
- Look up Φ(z), the area to the left of z.
- For the area to the right use 1 − Φ(z); for an interval subtract the two left areas.
- 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
- Decide between z (population σ known) and t (only the sample s, with n − 1 degrees of freedom).
- Find the critical value for the confidence level.
- Margin of error: E = critical value × standard deviation/√n.
- 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
- State H₀ and H₁ and decide whether the test is one- or two-sided.
- Calculate the test statistic: (x̄ − μ₀)/(σ/√n), or with s and t.
- Compare with the critical value, or find the p-value.
- 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 field experiment, the researcher:
- AManipulates the IV in a real-world setting
- BManipulates the IV in a laboratory
- CAnalyses data collected by others
- DOnly observes without manipulating anything
2.Research methods in psychologyHard
A researcher runs 20 analyses on a data set and reports only the one with p < 0.05. This practice is called:
- ATriangulation
- BPreregistration
- CCounterbalancing
- DP-hacking
3.Descriptive statisticsMedium
Find the sample standard deviation s of the data: 25, 48, 37, 23, 42, 12.
n = 6
s =
4.Descriptive statisticsHard
Exam scores are normally distributed with mean 100 and standard deviation 10. Find the z-score z of a score of 106.
μ = 100σ = 10x = 106
z =
5.ProbabilityMedium
A fair die is rolled 3 times. What is the probability P of getting at least one six?
n = 3
P =
6.ProbabilityHard
A test has 11 multiple-choice questions. A student who guesses gets each one right with probability 0.5, independently. What is the probability P of exactly 3 correct answers?
n = 11p = 0.5k = 3
P =
7.Normal distributionMedium
Drag z so that the shaded area under the standard normal curve, to the right of z, is 0.3264. Find z to two decimals.
A = 0.3264
z =
8.Normal distributionHard
The exam scores are normally distributed with mean 65 points and standard deviation 10 points. Drag the cut-off so the shaded bottom 10% is to the left of it. Find the cut-off x.
μ = 65 pointsσ = 10 points10%
x = points
9.Confidence intervalsMedium
The battery life of two phone models is compared. Model 1: 40 phones, mean 107.5 h, known standard deviation 16 h. Model 2: 60 phones, mean 109.5 h, known standard deviation 10 h. Find the margin of error E of the 90% confidence interval for μ₁ − μ₂.
n₁ = 40x̄₁ = 107.5 hσ₁ = 16 hn₂ = 60x̄₂ = 109.5 hσ₂ = 10 h90%
E = h
10.Confidence intervalsHard
The diameters of 16 machined shafts have a sample standard deviation of 6.9 µm. Assuming the diameters are normally distributed, find the lower 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 = 6.9 µmχ² = 6.262, 27.488
σ = µm
11.Hypothesis tests & confidence intervalsMedium
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
A sample of 100 measurements has mean 99. The population standard deviation is 20. Find the lower limit L of the 90% confidence interval for the mean.
x̄ = 99σ = 20n = 10090%
L =
13.Linear regression & correlationMedium
The table gives 5 paired observations of x = advertising spend in k€ and y = sales in k€. Their sums are given. Find Pearson's correlation coefficient r from these sums.
n = 5Σx = 25Σy = 255Σx² = 135Σy² = 13091Σxy = 1302
r =
14.Linear regression & correlationHard
The table gives 5 paired observations of x = temperature in °C and y = ice creams sold. Use the least-squares line ŷ = a + b·x. Predict ŷ for x = 19.5.
n = 5x = 19.5
ŷ =
15.Chi-square testsMedium
The contingency table shows observed counts. For a chi-square test of independence, find the expected count E for the cell “rural, high”.
row total = 62column total = 41N = 136
E =
16.Chi-square testsHard
A chi-square goodness-of-fit test checks whether counts in 9 categories follow a Poisson distribution whose mean is estimated from the same data. How many degrees of freedom df does the test have?
k = 9
df =
3Solutions
1.Research methods in psychology
Answer: A · Manipulates the IV in a real-world setting
Why this is the answer
Field experiments, such as staged helping situations in the street, increase ecological validity but give less control over extraneous variables and raise consent issues.
Answer
A · Manipulates the IV in a real-world setting
2.Research methods in psychology
Answer: D · P-hacking
Why this is the answer
P-hacking exploits flexibility in analysis to obtain significance, raising the false-positive rate well above 5%. Preregistration and transparent reporting reduce it.
Answer
D · P-hacking
3.Descriptive statistics
Answer: s = 13.47
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 = 187/6 = 31.17
Squared deviations
Σ(x − x̄)² = (−6.167)² + 16.83² + 5.833² + (−8.167)² + 10.83² + (−19.17)² = 906.8
Standard deviation
s = √(906.8/5) = 13.47
Answer
s = 13.47
4.Descriptive statistics
Answer: z = 0.6
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 = (106 − 100)/10 = 0.6
Answer
z = 0.6
5.Probability
Answer: P = 0.4213
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)3 = 0.5787
P = 1 − 0.5787 = 0.4213
Answer
P = 0.4213
6.Probability
Answer: P = 0.08057
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) = 165·0.53·0.58 = 165·0.125·0.003906 = 0.08057
Answer
P = 0.08057
7.Normal distribution
Answer: z = 0.45
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 − 0.3264 = 0.6736
Read the z-table backwards
Find 0.6736 inside the table and read off z.
z = 0.45
Answer
z = 0.45
8.Normal distribution
Answer: x = 52.2 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) = 0.1
z from the table
z ≈ −1.28
Back to the original units
x = 65 − 1.28·10 = 52.2 points
Answer
x = 52.2 points
9.Confidence intervals
Answer: E = 4.672 h
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.
Point estimate
x̄₁ − x̄₂ = 107.5 − 109.5 = −2
Standard error
SE = √(16²/40 + 10²/60) = 2.84
Margin of error
E = 1.645·2.84 = 4.672
Answer
E = 4.672 h
The interval contains 0, so the data do not show a significant difference.
10.Confidence intervals
Answer: σ = 5.097 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 = 15·6.9² = 714.2
Limits for σ²
714.2/27.488 = 25.98
714.2/6.262 = 114
Limit for σ
σ = √25.98 = 5.097 µm
Answer
σ = 5.097 µm
95% interval for σ: [5.097, 10.68] 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 = 95.71
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 = 1.645·20/√100 = 3.29
Interval limit
L = 99 − 3.29 = 95.71
Answer
L = 95.71
13.Linear regression & correlation
Answer: r = 0.9207
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 = 1302 − 25·255/5 = 27
Sxx = 135 − 25²/5 = 10
Syy = 13091 − 255²/5 = 86
Correlation coefficient
r = 27/√(10·86) = 0.9207
Answer
r = 0.9207
14.Linear regression & correlation
Answer: ŷ = 51.75
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 = 120/5 = 24
ȳ = 318/5 = 63.6
Sums of squares and products
Sxx = Σ(x − x̄)² = 90
Sxy = Σ(x − x̄)(y − ȳ) = 237
Regression line
b = 237/90 = 2.633
a = 63.6 − 2.633·24 = 0.4
ŷ = 0.4 + 2.633·x
Prediction
ŷ = 0.4 + 2.633·19.5 = 51.75
Answer
ŷ = 51.75
15.Chi-square tests
Answer: E = 18.69
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 count
E = 62·41/136 = 18.69
Answer
E = 18.69
16.Chi-square tests
Answer: df = 7
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 = 9 − 1 − 1 = 7
Answer
df = 7
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