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

A researcher studies the effect of gender on memory scores. Why is this a quasi-experiment rather than a true experiment?

  1. AThere are only two conditions, and a true experiment needs at least three levels of the IV
  2. BThe DV is not measured
  3. CThe IV is a pre-existing characteristic, so participants cannot be randomly allocated to it
  4. DThe study takes place in a laboratory
2.Research methods in psychologyHard

Increasing the sample size in a study most directly:

  1. AGuarantees a statistically significant result if any effect exists
  2. BRemoves all bias from the sample
  3. CIncreases the effect size
  4. DIncreases statistical power to detect a real effect
3.Descriptive statisticsMedium
152025303540455055

Find the sample standard deviation s of the data: 58, 34, 21, 15, 36, 33, 33.

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

A quantity is normally distributed with mean 500 and standard deviation 60. What proportion P of values is above 560?

μ = 500σ = 60x = 560
P =
5.ProbabilityMedium
red: 4blue: 8

A bag holds 4 red and 8 blue balls. You draw two balls without putting the first one back. What is the probability P that both are red?

4 / 8
P =
6.ProbabilityHard
n = 11, p = 0.25: P(X = 5) = ?k01234567891011

A test has 11 multiple-choice questions. A student who guesses gets each one right with probability 0.25, independently. What is the probability P of exactly 5 correct answers?

n = 11p = 0.25k = 5
P =
7.Normal distributionMedium
−3−2−10123zshaded area between −z and z = 0.7580

Drag z so that the shaded area between −z and z under the standard normal curve is 0.7580. Find z to two decimals.

A = 0.7580
z =
8.Normal distributionHard
μ = 250x = ?75%

The bag weights are normally distributed with mean 250 g and standard deviation 5 g. Find the value x below which 75% of them lie (the 75th percentile).

μ = 250σ = 5P = 75%
x = g
9.Confidence intervalsMedium
μ₁ − μ₂0LUx̄₁ − x̄₂E = ?99%

The battery life of two phone models is compared. Model 1: 60 phones, mean 60.5 h, known standard deviation 20 h. Model 2: 60 phones, mean 68.5 h, known standard deviation 10 h. Find the margin of error E of the 99% confidence interval for μ₁ − μ₂.

n₁ = 60x̄₁ = 60.5 hσ₁ = 20 hn₂ = 60x̄₂ = 68.5 hσ₂ = 10 h99%
E = h
10.Confidence intervalsHard
n = 11Lσ = ?s = 3.695%

The diameters of 11 machined shafts have a sample standard deviation of 3.6 µm. Assuming the diameters are normally distributed, find the upper limit σ of the 95% confidence interval for the population standard deviation. With df = 10, 2.5% of the χ² distribution lies below 3.247 and 2.5% lies above 20.483.

n = 11s = 3.6 µmχ² = 3.247, 20.483
σ = µm
11.Hypothesis tests & confidence intervalsMedium
−2.2622.262t = ?df = 9

10 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 4.75 and the standard deviation of the differences is 7.9. Find the paired test statistic t for H₀: μd = 0.

n = 10d̄ = 4.75sd = 7.9
t =
12.Hypothesis tests & confidence intervalsHard
z −1.9601.96

A sample of 49 measurements has mean 78.6. The population standard deviation is 12. Find the lower limit L of the 95% confidence interval for the mean.

x̄ = 78.6σ = 12n = 4995%
L =
13.Linear regression & correlationMedium
xy274761181710221223xy510101520

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

n = 6
r =
14.Linear regression & correlationHard
xy229337434541649750xy246304050x = 2.5

The table gives 6 paired observations of x = advertising spend in k€ and y = sales in k€. Use the least-squares line ŷ = a + b·x. Predict ŷ for x = 2.5.

n = 6x = 2.5
ŷ =
15.Chi-square testsMedium
coughno coughTotalsmokers164460non-smokers4773120Total63117180

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 = 1802 × 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 8 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 · The IV is a pre-existing characteristic, so participants cannot be randomly allocated to it

Why this is the answer

In a quasi-experiment the IV is a naturally occurring difference between people. Without random allocation, other differences between groups may confound the results.

Answer

C · The IV is a pre-existing characteristic, so participants cannot be randomly allocated to it
2.Research methods in psychology

Answer: D · Increases statistical power to detect a real effect

Why this is the answer

Larger samples reduce standard error, increasing power. They do not change the true effect size or remove systematic bias.

Answer

D · Increases statistical power to detect a real effect
3.Descriptive statistics

Answer: s = 13.56

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 = 230/7 = 32.86

Squared deviations

Σ(x − x̄)² = 25.14² + 1.143² + (−11.86)² + (−17.86)² + 3.143² + 0.1429² + 0.1429² = 1103

Standard deviation

s = √(1103/6) = 13.56

Answer

s = 13.56
4.Descriptive statistics

Answer: P = 0.1587

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 = 1

Standard normal table

Φ(1) = 0.8413
P = 1 − 0.8413 = 0.1587

Answer

P = 0.1587
5.Probability

Answer: P = 0.09091

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.

First and second draw

P = 4/12 · 3/11 = 12/132 = 0.09091

Answer

P = 0.09091
6.Probability

Answer: P = 0.0803

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 = 5) = 462·0.255·0.756 = 462·0.0009766·0.178 = 0.0803

Answer

P = 0.0803
7.Normal distribution

Answer: z = 1.17

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 + 0.7580)/2 = 0.8790

Read the z-table backwards

Find 0.8790 inside the table and read off z.

z = 1.17

Answer

z = 1.17
8.Normal distribution

Answer: x = 253.4 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) = 0.75 → z = 0.6745

Back to x

x = 250 + 0.6745·5 = 253.37 g

Answer

x = 253.37 g
9.Confidence intervals

Answer: E = 7.436 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 = 60.5 − 68.5 = −8

Standard error

SE = √(20²/60 + 10²/60) = 2.887

Margin of error

E = 2.576·2.887 = 7.436

Answer

E = 7.436 h

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

10.Confidence intervals

Answer: σ = 6.318 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 = 10·3.6² = 129.6

Limits for σ²

129.6/20.483 = 6.327
129.6/3.247 = 39.91

Limit for σ

σ = √39.91 = 6.318 µm

Answer

σ = 6.318 µm

95% interval for σ: [2.515, 6.318] µm.

11.Hypothesis tests & confidence intervals

Answer: t = 1.901

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 = 7.9/√10 = 2.498

Test statistic

t = 4.75/2.498 = 1.901

Answer

t = 1.901

With df = 9, the two-sided 5% critical value is 2.262; |t| is smaller, so H₀ is not rejected.

12.Hypothesis tests & confidence intervals

Answer: L = 75.24

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 = 1.96·12/√49 = 3.36

Interval limit

L = 78.6 − 3.36 = 75.24

Answer

L = 75.24
13.Linear regression & correlation

Answer: r = 0.972

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 = 87/6 = 14.5

Sums of squares and products

Sxx = Σ(x − x̄)² = 70
Sxy = Σ(x − x̄)(y − ȳ) = 131
Syy = Σ(y − ȳ)² = 259.5

Correlation coefficient

r = 131/√(70·259.5) = 0.972

Answer

r = 0.972
14.Linear regression & correlation

Answer: ŷ = 31.54

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 = 27/6 = 4.5
ȳ = 240/6 = 40

Sums of squares and products

Sxx = Σ(x − x̄)² = 17.5
Sxy = Σ(x − x̄)(y − ȳ) = 74

Regression line

b = 74/17.5 = 4.229
a = 40 − 4.229·4.5 = 20.97
ŷ = 20.97 + 4.229·x

Prediction

ŷ = 20.97 + 4.229·2.5 = 31.54

Answer

ŷ = 31.54
15.Chi-square tests

Answer: χ² = 2.747

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

smokers: 60·63/180 = 21, 60·117/180 = 39
non-smokers: 120·63/180 = 42, 120·117/180 = 78

Chi-square statistic

χ² = (16 − 21)²/21 + (44 − 39)²/39 + (47 − 42)²/42 + (73 − 78)²/78
χ² = 2.747

Answer

χ² = 2.747

df = 1. At α = 0.05 the critical value is 3.841, so H₀ (independence) is not rejected.

16.Chi-square tests

Answer: χ²crit = 14.07

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 = 8 − 1 = 7

Table

χ²crit(df = 7, α = 0.05) = 14.067

Answer

χ²crit = 14.067

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

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