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

A researcher compares three or more group means. Why is a one-way ANOVA used instead of several t-tests?

  1. AANOVA always finds larger effects than the equivalent set of t-tests would
  2. BMultiple t-tests inflate the overall Type I error rate
  3. Ct-tests cannot handle numbers above 100
  4. Dt-tests need qualitative data
3.Descriptive statisticsMedium
152025303540455055

Find the sample standard deviation s of the data: 37, 58, 22, 15, 28, 47, 57.

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

A quantity is normally distributed with mean 60 and standard deviation 8. What proportion P of values is below 63.2?

μ = 60σ = 8x = 63.2
P =
5.ProbabilityMedium
P(at least one six) = ??1?2?3?4?5P(six) = 1/6 on each roll

A fair die is rolled 5 times. What is the probability P of getting at least one six?

n = 5
P =
6.ProbabilityHard
xp(x)00.3510.120.1530.0540.35μ

X has the distribution shown, and Y = 10X + 1. Find the standard deviation σY of Y.

x: 0, 1, 2, 3, 4p(x): 0.35, 0.1, 0.15, 0.05, 0.35Y = 10X + 1
σY =
7.Normal distributionMedium
167.85192.35P = ?

The heights of adult men are normally distributed with mean 178 cm and standard deviation 7 cm. What fraction lies between 167.85 and 192.35 cm? Give P as a decimal.

μ = 178σ = 7x1 = 167.85x2 = 192.35
P =
8.Normal distributionHard
μ = 65x = ?80%

The exam scores are normally distributed with mean 65 points and standard deviation 10 points. Find the value x below which 80% of them lie (the 80th percentile).

μ = 65σ = 10P = 80%
x = points
9.Confidence intervalsMedium
n = 6, sd​ = 80L = ?Ud̄ = −5.4E99%

The blood pressure of 6 patients is measured before and after a treatment. The differences d = before − after (mmHg) have mean −5.4 and standard deviation 8. Find the lower limit of the 99% confidence interval for the mean difference.

n = 6d̄ = −5.4sd = 899%t(5) = 4.032
lower limit = mmHg
10.Confidence intervalsHard
σ = 39, n = ?x̄ − Ex̄ + Ex̄E = 195%

The population standard deviation is 39. How large a sample n do you need so that a 95% confidence interval for the mean has a margin of error of at most 1?

σ = 39E = 195%
n =
11.Hypothesis tests & confidence intervalsMedium
t −1.99801.998

A sample of 64 students has a mean score of 97.38 and a sample standard deviation of 15. Find the test statistic t for H₀: μ = 100.

μ₀ = 100x̄ = 97.38s = 15n = 64
t =
12.Hypothesis tests & confidence intervalsHard
z −1.9601.96

A sample of 25 measurements has mean 106.8. The population standard deviation is 20. Find the upper limit U of the 95% confidence interval for the mean.

x̄ = 106.8σ = 20n = 2595%
U =
13.Linear regression & correlationMedium
xy24486108171019xy5105101520

The table gives 5 paired observations of x = load in kN and y = extension in mm. Their sums are given. Find Pearson's correlation coefficient r from these sums.

n = 5Σx = 30Σy = 58Σx² = 220Σy² = 830Σxy = 426
r =
14.Linear regression & correlationHard
xy239349447555650760xy246405060x = 3.5

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 (3.5, 43) is made. Find its residual e = y − ŷ.

n = 6(3.5, 43)
e =
15.Chi-square testsMedium
busbikecarTotalstudents3914962staff34213893Total733547155

The contingency table shows observed counts. For a chi-square test of independence, find the expected count E for the cell “students, bike”.

row total = 62column total = 35N = 155
E =
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 9 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 · 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

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

Answer: B · Multiple t-tests inflate the overall Type I error rate

Why this is the answer

Running many pairwise tests multiplies the chance of a false positive. ANOVA tests all groups together and is followed by corrected post hoc tests.

Answer

B · Multiple t-tests inflate the overall Type I error rate
3.Descriptive statistics

Answer: s = 16.97

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 = 264/7 = 37.71

Squared deviations

Σ(x − x̄)² = (−0.7143)² + 20.29² + (−15.71)² + (−22.71)² + (−9.714)² + 9.286² + 19.29² = 1727

Standard deviation

s = √(1727/6) = 16.97

Answer

s = 16.97
4.Descriptive statistics

Answer: P = 0.6554

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 = 0.4

Standard normal table

Φ(0.4) = 0.6554
P = 0.6554

Answer

P = 0.6554
5.Probability

Answer: P = 0.5981

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)5 = 0.4019
P = 1 − 0.4019 = 0.5981

Answer

P = 0.5981
6.Probability

Answer: σY = 17.17

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 = 1.95, Var(X) = 2.948, σX = 1.717

Linear change

σY = |10|·1.717 = 17.17

Answer

σY = 17.17

The mean does move: E(Y) = 10·1.95 + 1 = 20.5.

7.Normal distribution

Answer: P = 0.9063

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 = (167.85 − 178)/7 = −1.45
z₂ = (192.35 − 178)/7 = 2.05

Area between

P = Φ(2.05) − Φ(−1.45) = 0.9798 − 0.07353 = 0.9063

Answer

P = 0.9063
8.Normal distribution

Answer: x = 73.42 points

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) = 0.8 → z = 0.8416

Back to x

x = 65 + 0.8416·10 = 73.42 points

Answer

x = 73.42 points
9.Confidence intervals

Answer: lower limit = −18.57 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 = 5, t = 4.032

Margin of error

E = 4.032·8/√6 = 13.17

Interval

−5.4 ± 13.17 → [−18.57, 7.768]

Answer

lower limit = −18.57 mmHg
10.Confidence intervals

Answer: n = 5844

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 = (1.96·39/1)² = 5843

Round up

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

n = 5844

Answer

n = 5844
11.Hypothesis tests & confidence intervals

Answer: t = −1.397

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 = 15/√64 = 15/8 = 1.875

Test statistic

t = (97.38 − 100)/1.875 = −1.397

Answer

t = −1.397
t0.025; df = 63 = 1.998

|t| < 1.998, so H₀ is not rejected at the 5% level (two-sided).

12.Hypothesis tests & confidence intervals

Answer: U = 114.6

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 = 1.96·20/√25 = 7.84

Interval limit

U = 106.8 + 7.84 = 114.6

Answer

U = 114.6
13.Linear regression & correlation

Answer: r = 0.9836

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 = 426 − 30·58/5 = 78
Sxx = 220 − 30²/5 = 40
Syy = 830 − 58²/5 = 157.2

Correlation coefficient

r = 78/√(40·157.2) = 0.9836

Answer

r = 0.9836
14.Linear regression & correlation

Answer: e = −3.686

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 = 300/6 = 50

Sums of squares and products

Sxx = Σ(x − x̄)² = 17.5
Sxy = Σ(x − x̄)(y − ȳ) = 58

Regression line

b = 58/17.5 = 3.314
a = 50 − 3.314·4.5 = 35.09
ŷ = 35.09 + 3.314·x

Predicted value

ŷ = 35.09 + 3.314·3.5 = 46.69

Residual

e = 43 − 46.69 = −3.686

Answer

e = −3.686

The point lies below the line.

15.Chi-square tests

Answer: E = 14

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 = 62·35/155 = 14

Answer

E = 14
16.Chi-square tests

Answer: χ²crit = 18.48

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

Table

χ²crit(df = 7, α = 0.01) = 18.475

Answer

χ²crit = 18.475

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

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