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 normal distribution, approximately what percentage of scores lie within one standard deviation of the mean?

  1. A95%
  2. B99.7%
  3. C50%
  4. D68%
2.Research methods in psychologyHard

Patients chosen because they have extremely high anxiety scores show lower scores at a second test, even without treatment. What is the most likely explanation?

  1. AThe Hawthorne effect
  2. BCounterbalancing
  3. CRegression to the mean
  4. DA ceiling effect
3.Descriptive statisticsMedium
101520253035404550

Find the sample standard deviation s of the data: 50, 10, 21, 12, 21, 42.

n = 6
s =
4.Descriptive statisticsHard
z −2−1012

Exam scores are normally distributed with mean 170 and standard deviation 8. Find the z-score z of a score of 182.

μ = 170σ = 8x = 182
z =
5.ProbabilityMedium
P(at least one six) = ??1?2P(six) = 1/6 on each roll

A fair die is rolled 2 times. What is the probability P of getting at least one six?

n = 2
P =
6.ProbabilityHard
xp(x)100.3200.1300.25400.35μ

X has the distribution shown, and Y = −2X + 1. Find the standard deviation σY of Y.

x: 10, 20, 30, 40p(x): 0.3, 0.1, 0.25, 0.35Y = −2X + 1
σY =
7.Normal distributionMedium
30.233.2P = ?

The delivery times are normally distributed with mean 32 min and standard deviation 6 min. What fraction lies between 30.2 and 33.2 min? Give P as a decimal.

μ = 32σ = 6x1 = 30.2x2 = 33.2
P =
8.Normal distributionHard
35455565758595pointstop 30%

The exam scores are normally distributed with mean 65 points and standard deviation 10 points. Drag the cut-off so the shaded top 30% is to the right of it. Find the cut-off x.

μ = 65 pointsσ = 10 points30%
x = points
9.Confidence intervalsMedium
x̄E₁ = 8.699%, n = 36x̄E₂ = ?99%, n = 324

A 99% confidence interval for a mean, based on 36 observations and a known σ, has a margin of error of 8.6. What is the margin of error E₂ if the same study uses 324 observations instead?

99%n₁ = 36E₁ = 8.6n₂ = 324
E₂ =
10.Confidence intervalsHard
n = 16Lσ = ?s = 7.195%

The diameters of 16 machined shafts have a sample standard deviation of 7.1 µm. Assuming the diameters are normally distributed, find the upper 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 = 7.1 µmχ² = 6.262, 27.488
σ = µm
11.Hypothesis tests & confidence intervalsMedium
−2.0932.093t = ?df = 19

20 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 −2.41 and the standard deviation of the differences is 7.2. Find the paired test statistic t for H₀: μd = 0.

n = 20d̄ = −2.41sd = 7.2
t =
12.Hypothesis tests & confidence intervalsHard
σ = 10, n = ?x̄ − Ex̄ + Ex̄E = 399%

How large must a sample be so that the 99% confidence interval for a mean has a margin of error of at most 3? The standard deviation is 10. Find the smallest sample size n.

σ = 10E = 399%
n =
13.Linear regression & correlationMedium
xy129223323421521619xy246202530

The table gives 6 paired observations of x = car age in years and y = price in k€. Find Pearson's correlation coefficient r.

n = 6
r =
14.Linear regression & correlationHard
xy352558768982119413111xy5106080100x = 6

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 = 6.

n = 6x = 6
ŷ =
15.Chi-square testsMedium
yesnoTotalmen204060women2872100Total48112160

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 × 2
χ² =
16.Chi-square testsHard
passfailTotalage 18–30441660age 31–50221840age 51+303060Total9664160dfα = 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 = 0.01 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 = 1603 × 2α = 0.01
χ² =

3Solutions

1.Research methods in psychology

Answer: D · 68%

Why this is the answer

About 68% of scores fall within ±1 SD, about 95% within ±2 SD and about 99.7% within ±3 SD.

Answer

D · 68%
2.Research methods in psychology

Answer: C · Regression to the mean

Why this is the answer

When people are selected for extreme scores, measurement error and chance mean their later scores tend to move towards the average. Control groups help separate this from treatment effects.

Answer

C · Regression to the mean
3.Descriptive statistics

Answer: s = 16.33

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 = 156/6 = 26

Squared deviations

Σ(x − x̄)² = 24² + (−16)² + (−5)² + (−14)² + (−5)² + 16² = 1334

Standard deviation

s = √(1334/5) = 16.33

Answer

s = 16.33
4.Descriptive statistics

Answer: z = 1.5

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 = (182 − 170)/8 = 1.5

Answer

z = 1.5
5.Probability

Answer: P = 0.3056

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)2 = 0.6944
P = 1 − 0.6944 = 0.3056

Answer

P = 0.3056
6.Probability

Answer: σY = 24.72

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 = 26.5, Var(X) = 152.8, σX = 12.36

Linear change

σY = |−2|·12.36 = 24.72

Answer

σY = 24.72

The mean does move: E(Y) = −2·26.5 + 1 = −52.

7.Normal distribution

Answer: P = 0.1972

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 = (30.2 − 32)/6 = −0.3
z₂ = (33.2 − 32)/6 = 0.2

Area between

P = Φ(0.2) − Φ(−0.3) = 0.5793 − 0.3821 = 0.1972

Answer

P = 0.1972
8.Normal distribution

Answer: x = 70.2 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) = 0.7

z from the table

z ≈ 0.52

Back to the original units

x = 65 + 0.52·10 = 70.2 points

Answer

x = 70.2 points
9.Confidence intervals

Answer: E₂ = 2.867

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.

Ratio of margins

E₂/E₁ = (2.576/2.576)·√(36/324) = 0.3333

New margin

E₂ = 8.6·0.3333 = 2.867

Answer

E₂ = 2.867
10.Confidence intervals

Answer: σ = 10.99 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 = 15·7.1² = 756.1

Limits for σ²

756.1/27.488 = 27.51
756.1/6.262 = 120.8

Limit for σ

σ = √120.8 = 10.99 µm

Answer

σ = 10.99 µm

95% interval for σ: [5.245, 10.99] µm.

11.Hypothesis tests & confidence intervals

Answer: t = −1.497

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 = 7.2/√20 = 1.61

Test statistic

t = −2.41/1.61 = −1.497

Answer

t = −1.497

With df = 19, the two-sided 5% critical value is 2.093; |t| is smaller, so H₀ is not rejected.

12.Hypothesis tests & confidence intervals

Answer: n = 74

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 = (2.576·10/3)² = 73.73

Round up

n = 74

Answer

n = 74
13.Linear regression & correlation

Answer: r = −0.9

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 = 21/6 = 3.5
ȳ = 136/6 = 22.67

Sums of squares and products

Sxx = Σ(x − x̄)² = 17.5
Sxy = Σ(x − x̄)(y − ȳ) = −29
Syy = Σ(y − ȳ)² = 59.33

Correlation coefficient

r = −29/√(17.5·59.33) = −0.9

Answer

r = −0.9
14.Linear regression & correlation

Answer: ŷ = 65.59

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 = 48/6 = 8
ȳ = 465/6 = 77.5

Sums of squares and products

Sxx = Σ(x − x̄)² = 70
Sxy = Σ(x − x̄)(y − ȳ) = 417

Regression line

b = 417/70 = 5.957
a = 77.5 − 5.957·8 = 29.84
ŷ = 29.84 + 5.957·x

Prediction

ŷ = 29.84 + 5.957·6 = 65.59

Answer

ŷ = 65.59
15.Chi-square tests

Answer: χ² = 0.5079

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: 60·48/160 = 18, 60·112/160 = 42
women: 100·48/160 = 30, 100·112/160 = 70

Chi-square statistic

χ² = (20 − 18)²/18 + (40 − 42)²/42 + (28 − 30)²/30 + (72 − 70)²/70
χ² = 0.5079

Answer

χ² = 0.5079

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

16.Chi-square tests

Answer: χ² = 7.361

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

age 18–30: 36, 24
age 31–50: 24, 16
age 51+: 36, 24

Chi-square statistic

χ² = (44 − 36)²/36 + (16 − 24)²/24 + (22 − 24)²/24 + (18 − 16)²/16 + (30 − 36)²/36 + (30 − 24)²/24
χ² = 7.361

Decision

df = 2, χ²crit = 9.210
7.361 < 9.210

Answer

χ² = 7.361

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

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