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. A16%
  2. B40%
  3. C4%
  4. D60%
2.Research methods in psychologyHard

A researcher finds that ice cream sales and drowning deaths are positively correlated. Which explanation is most plausible?

  1. AHot weather increases both ice cream sales and swimming
  2. BIce cream causes drowning
  3. CDrowning increases ice cream sales
  4. DThe correlation must be a calculation error
3.Descriptive statisticsMedium
253035404550

Find the sample standard deviation s of the data: 23, 29, 29, 54, 21.

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

A quantity is normally distributed with mean 70 and standard deviation 15. What proportion P of values is below 40?

μ = 70σ = 15x = 40
P =
5.ProbabilityMedium
P(at least one six) = ??1?2?3?4?5?6?7P(six) = 1/6 on each roll

A fair die is rolled 7 times. What is the probability P of getting at least one six?

n = 7
P =
6.ProbabilityHard
xp(x)10.3520.130.340.150.15μ

X has the distribution shown, and Y = 2X + 10. Find the standard deviation σY of Y.

x: 1, 2, 3, 4, 5p(x): 0.35, 0.1, 0.3, 0.1, 0.15Y = 2X + 10
σY =
7.Normal distributionMedium
6270P = ?

The exam scores are normally distributed with mean 65 points and standard deviation 10 points. What fraction lies between 62 and 70 points? Give P as a decimal.

μ = 65σ = 10x1 = 62x2 = 70
P =
8.Normal distributionHard
μ = 250x = ?20%

The bag weights are normally distributed with mean 250 g and standard deviation 5 g. Find the value x below which 20% of them lie (the 20th percentile).

μ = 250σ = 5P = 20%
x = g
9.Confidence intervalsMedium
μ₁ − μ₂0LUx̄₁ − x̄₂E = ?90%

The battery life of two phone models is compared. Model 1: 50 phones, mean 111.5 h, known standard deviation 7 h. Model 2: 30 phones, mean 119.5 h, known standard deviation 17 h. Find the margin of error E of the 90% confidence interval for μ₁ − μ₂.

n₁ = 50x̄₁ = 111.5 hσ₁ = 7 hn₂ = 30x̄₂ = 119.5 hσ₂ = 17 h90%
E = h
10.Confidence intervalsHard
σ = 25, n = ?x̄ − Ex̄ + Ex̄E = 599%

The population standard deviation is 25. How large a sample n do you need so that a 99% confidence interval for the mean has a margin of error of at most 5?

σ = 25E = 599%
n =
11.Hypothesis tests & confidence intervalsMedium
−2.0642.064t = ?df = 24

Two teaching methods are compared. Group 1: 8 students, mean score 57.5, standard deviation 2.5. Group 2: 18 students, mean score 55.2, 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 = 8x̄₁ = 57.5s₁ = 2.5n₂ = 18x̄₂ = 55.2s₂ = 4
t =
12.Hypothesis tests & confidence intervalsHard
σ = 15, n = ?x̄ − Ex̄ + Ex̄E = 395%

How large must a sample be so that the 95% confidence interval for a mean has a margin of error of at most 3? The standard deviation is 15. Find the smallest sample size n.

σ = 15E = 395%
n =
13.Linear regression & correlationMedium
xy16811910422113251292815131167xy15202530100150

The table gives 6 paired observations of x = temperature in °C and y = ice creams sold. Find Pearson's correlation coefficient r.

n = 6
r =
14.Linear regression & correlationHard
xy24545866288110801291xy5106080x = 7

The table gives 6 paired observations of x = hours studied and y = exam score. Use the least-squares line ŷ = a + b·x. A new observation (7, 66) is made. Find its residual e = y − ŷ.

n = 6(7, 66)
e =
15.Chi-square testsMedium
no defectminormajorTotalday shift12161240night shift363252120Total484864160

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 = 1602 × 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 goodness-of-fit test checks whether counts in 7 categories follow a Poisson distribution whose mean is estimated from the same data. 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 · 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

A · 16%
2.Research methods in psychology

Answer: A · Hot weather increases both ice cream sales and swimming

Why this is the answer

A third variable, temperature, causes both. This illustrates why correlation alone cannot show causation.

Answer

A · Hot weather increases both ice cream sales and swimming
3.Descriptive statistics

Answer: s = 13.24

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 = 156/5 = 31.2

Squared deviations

Σ(x − x̄)² = (−8.2)² + (−2.2)² + (−2.2)² + 22.8² + (−10.2)² = 700.8

Standard deviation

s = √(700.8/4) = 13.24

Answer

s = 13.24
4.Descriptive statistics

Answer: P = 0.02275

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 = 0.02275

Answer

P = 0.02275
5.Probability

Answer: P = 0.7209

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)7 = 0.2791
P = 1 − 0.2791 = 0.7209

Answer

P = 0.7209
6.Probability

Answer: σY = 2.857

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 = 2.6, Var(X) = 2.04, σX = 1.428

Linear change

σY = |2|·1.428 = 2.857

Answer

σY = 2.857

The mean does move: E(Y) = 2·2.6 + 10 = 15.2.

7.Normal distribution

Answer: P = 0.3094

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-scores

z₁ = (62 − 65)/10 = −0.3
z₂ = (70 − 65)/10 = 0.5

Area between

P = Φ(0.5) − Φ(−0.3) = 0.6915 − 0.3821 = 0.3094

Answer

P = 0.3094
8.Normal distribution

Answer: x = 245.8 g

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) = 0.2 → z = −0.8416

Back to x

x = 250 + −0.8416·5 = 245.79 g

Answer

x = 245.79 g
9.Confidence intervals

Answer: E = 5.359 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 = 111.5 − 119.5 = −8

Standard error

SE = √(7²/50 + 17²/30) = 3.258

Margin of error

E = 1.645·3.258 = 5.359

Answer

E = 5.359 h

The interval does not contain 0, so the means differ significantly.

10.Confidence intervals

Answer: n = 166

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 = (2.576·25/5)² = 165.9

Round up

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

n = 166

Answer

n = 166
11.Hypothesis tests & confidence intervals

Answer: t = 1.492

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 = (7·2.5² + 17·4²)/24 = 13.16
sp = 3.627

Standard error

SE = 3.627·√(1/8 + 1/18) = 1.541

Test statistic

t = (57.5 − 55.2)/1.541 = 1.492

Answer

t = 1.492

With df = 24, the two-sided 5% critical value is 2.064; |t| is smaller, so H₀ is not rejected.

12.Hypothesis tests & confidence intervals

Answer: n = 97

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 = (1.96·15/3)² = 96.04

Round up

n = 97

Answer

n = 97
13.Linear regression & correlation

Answer: r = 0.9949

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 = 141/6 = 23.5
ȳ = 745/6 = 124.2

Sums of squares and products

Sxx = Σ(x − x̄)² = 157.5
Sxy = Σ(x − x̄)(y − ȳ) = 880.5
Syy = Σ(y − ȳ)² = 4973

Correlation coefficient

r = 880.5/√(157.5·4973) = 0.9949

Answer

r = 0.9949
14.Linear regression & correlation

Answer: e = −3.5

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 = 417/6 = 69.5

Sums of squares and products

Sxx = Σ(x − x̄)² = 70
Sxy = Σ(x − x̄)(y − ȳ) = 315

Regression line

b = 315/70 = 4.5
a = 69.5 − 4.5·7 = 38
ŷ = 38 + 4.5·x

Predicted value

ŷ = 38 + 4.5·7 = 69.5

Residual

e = 66 − 69.5 = −3.5

Answer

e = −3.5

The point lies below the line.

15.Chi-square tests

Answer: χ² = 3.111

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

day shift: 40·48/160 = 12, 40·48/160 = 12, 40·64/160 = 16
night shift: 120·48/160 = 36, 120·48/160 = 36, 120·64/160 = 48

Chi-square statistic

χ² = (12 − 12)²/12 + (16 − 12)²/12 + (12 − 16)²/16 + (36 − 36)²/36 + (32 − 36)²/36 + (52 − 48)²/48
χ² = 3.111

Answer

χ² = 3.111

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 = 7 − 1 − 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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