Correlation Hypothesis Testing

Testing whether sample correlation provides evidence of population correlation using critical values

SPEC 5.1 • HYPOTHESIS TESTING

The Product Moment Correlation Coefficient

Definition

The product moment correlation coefficient (PMCC) measures the strength and direction of the linear correlation between two variables.

- 1 \leq r \leq 1

Interpreting r

r = 1

: Perfect positive correlation

r = - 1

: Perfect negative correlation

r = 0

: No linear correlation

Closer to ±1 = stronger correlation

Important

When

r = \pm 1

, all points lie exactly on a straight line. PMCC only measures linear correlation.

r = - 1

r \approx - 0.8

r \approx 0

r \approx 0.8

r = 1

Calculating PMCC Using Your Calculator

You need to know how to calculate the PMCC for bivariate data using your calculator. The formula is not required — use the statistics mode.

Calculator Method (Casio fx-991EX)

Menu → Statistics → 2: A + Bx (linear regression mode)

Enter your x-values in the first column, y-values in the second

Press AC then OPTN → Regression → r

Press = to get the PMCC value

Example Data

x24579y511131722Entering this data and finding r gives: $r = 0.9969$

This indicates strong positive linear correlation.

Exam Note

In exams, you will either be given the value of

r

, or asked to calculate it using your calculator. Make sure you can do this quickly and accurately.

Why Test Correlation?

When we calculate a correlation coefficient from sample data, we get a sample statistic. But does this indicate a real relationship in the whole population?

Key Question

Is the correlation we observe in our sample evidence of genuine correlation in the population, or could it have occurred by chance?

Sample PMCC

Denoted by $r$

Calculated from your data using a calculator

Always a specific number

Population PMCC

Denoted by $ρ$ (Greek letter "rho")

The true correlation for the whole population

Unknown - what we're testing

Exam Tip

Always state your hypotheses using

ρ

(the population parameter), never

r

(the sample statistic).

Setting Up Hypotheses

The null hypothesis always states there is no correlation in the population:

Null Hypothesis

H_{0} : ρ = 0

"There is no linear correlation in the population"

The alternative hypothesis depends on what we're testing for:

Test TypeAlternative HypothesisQuestion Clues

One-tailed (positive)

H_{1} : ρ > 0

"positive correlation", "increases with"

One-tailed (negative)

H_{1} : ρ < 0

"negative correlation", "decreases with"

Two-tailed

H_{1} : ρ \neq = 0

"evidence of correlation", "linear relationship"

Choosing Your Test

If the question asks about "positive" or "negative" correlation specifically → one-tailed. If it asks about "correlation" or "relationship" without specifying direction → two-tailed.

Critical Value Tables

The critical value depends on the sample size

n

and the significance level. The table in your formula booklet shows critical values for one-tailed tests:

Product Moment Coefficient Critical ValuesSample Size

0.100.050.0250.010.005

0.80000.90000.95000.98000.99004

0.68700.80540.87830.93430.95875

0.55090.66940.75450.83290.87457

0.50670.62150.70670.78870.83438

0.47160.58220.66640.74980.79779

0.44280.54940.63190.71550.764610

Critical Rule

Tables give ONE-TAILED values. For a two-tailed test at significance level

α

, use the column for

\frac{α}{2}

.

Example: Two-tailed test at 5% level → use the 0.025 column

The Testing Procedure

5-Step Method

State hypotheses: Write

H_{0} : ρ = 0

and your chosen

H_{1}

Find significance level: Note

α

(e.g., 0.05 for 5%)

Find critical value: Use table with sample size

n

and appropriate column

Compare: Is

∣ r ∣ > critical value

?

Conclude: In context, stating "sufficient evidence" or "not sufficient evidence"

Decision Rule

If

∣ r ∣ > critical value

→ Reject

H_{0}

If

∣ r ∣ \leq critical value

→ Do not reject

H_{0}

Why Absolute Value?

We use

∣ r ∣

because critical values are positive. For negative correlation, we compare the magnitude. A sample

r = - 0.7

has

∣ r ∣ = 0.7

.

Worked Example - One-Tailed (Positive)01

A researcher believes there is a positive correlation between hours of revision and test scores. A sample of 10 students gives $r = 0.62$ . Test at the 5% significance level whether there is evidence of positive correlation.

1

H_{0} : ρ = 0

(no correlation)

•

H_{1} : ρ > 0

(positive correlation) — one-tailed test

2

Significance level

α = 0.05

, sample size

n = 10

3

From tables: critical value = 0.5494 (one-tailed, 5%, n=10)

4

Compare:

∣ r ∣ = 0.62 > 0.5494

✓

5

Since

0.62 > 0.5494

, reject

H_{0}

Conclusion: There is sufficient evidence at the 5% significance level of a positive correlation between hours of revision and test scores.

Worked Example - One-Tailed (Negative)02

A scientist believes there is a negative correlation between the amount of solvent in a solution and the rate of reaction. From 15 samples, $r = - 0.52$ . Test at the 5% significance level.

1

H_{0} : ρ = 0

(no correlation)

•

H_{1} : ρ < 0

(negative correlation) — one-tailed test

2

Significance level

α = 0.05

, sample size

n = 15

3

From tables: critical value = 0.4409 (one-tailed, 5%, n=15)

4

Compare:

∣ r ∣ = ∣ - 0.52∣ = 0.52 > 0.4409

✓

5

Since

0.52 > 0.4409

, reject

H_{0}

Conclusion: There is sufficient evidence at the 5% significance level of a negative correlation between amount of solvent and rate of reaction.

Worked Example - Two-Tailed Test03

A sample of 8 students shows $r = 0.65$ between English and Mathematics marks. Test at the 5% significance level whether there is evidence of a linear relationship.

1

H_{0} : ρ = 0

(no correlation)

•

H_{1} : ρ \neq = 0

(some correlation) — two-tailed test

2

Two-tailed at 5%: use

\frac{0.05}{2} = 0.025

column

3

From tables: critical value = 0.7067 (n=8, 0.025 column)

4

Compare:

∣ r ∣ = 0.65 < 0.7067

✗

5

Since

0.65 < 0.7067

, do not reject

H_{0}

Conclusion: There is not sufficient evidence at the 5% significance level of a linear relationship between English and Mathematics marks.

Worked Example - Different Significance Level04

A researcher claims there is positive correlation between daily temperature and ice cream sales. A sample of 8 days gives $r = 0.75$ . Test at the 1% significance level.

1

H_{0} : ρ = 0

,

H_{1} : ρ > 0

— one-tailed test

2

Significance level

α = 0.01

, sample size

n = 8

3

From tables: critical value = 0.7887 (one-tailed, 1%, n=8)

4

Compare:

∣ r ∣ = 0.75 < 0.7887

✗

5

Since

0.75 < 0.7887

, do not reject

H_{0}

Conclusion: There is not sufficient evidence at the 1% significance level of a positive correlation between temperature and ice cream sales.

The 1% level requires stronger evidence than 5%. The same

r = 0.75

would be significant at 5% but not at 1%.

Worked Example - Using p-values05

A calculator gives $r = 0.35$ with a p-value of 0.087 for a one-tailed test. Test at the 5% significance level for positive correlation.

1

H_{0} : ρ = 0

,

H_{1} : ρ > 0

— one-tailed test

2

Significance level

α = 0.05

3

Given p-value = 0.087

4

Compare: p-value (0.087) >

α

(0.05)

5

Since p-value > 0.05, do not reject

H_{0}

Conclusion: There is not sufficient evidence at the 5% significance level of positive correlation.

If p-value <

α

→ Reject

H_{0}

If p-value ≥

α

→ Do not reject

H_{0}

Worked Example - Full Context06

A biologist measures wing length and body mass of 12 birds. The PMCC is $r = 0.68$ . Test at the 5% level whether there is evidence of a positive correlation. State your hypotheses clearly.

1

H_{0} : ρ = 0

— There is no correlation between wing length and body mass in the population of birds

•

H_{1} : ρ > 0

— There is positive correlation between wing length and body mass

2

One-tailed test at

α = 0.05

,

n = 12

3

From tables: critical value = 0.4973 (one-tailed, 5%, n=12)

4

Compare:

∣ r ∣ = 0.68 > 0.4973

✓

5

Since

0.68 > 0.4973

, reject

H_{0}

Conclusion: There is sufficient evidence at the 5% significance level of a positive correlation between wing length and body mass in birds.

07Worked Example - Interpreting Significance Level

Twelve students sat two Biology tests. A teacher claims students who do well in the theoretical test also tend to do well in the practical test. The PMCC is $r = 0.58$ .

(a) Test at the 5% level. (b) Give an interpretation of the 0.05 significance level in this context.

(a)

1

H_{0} : ρ = 0

,

H_{1} : ρ > 0

— one-tailed (positive correlation claimed)

2

n = 12

,

α = 0.05

, critical value = 0.4973

3

∣ r ∣ = 0.58 > 0.4973

✓ → Reject

H_{0}

Conclusion: There is sufficient evidence at the 5% level that students who do well in the theoretical test tend to do well in the practical test.

(b) Interpretation of 0.05:

The probability of incorrectly rejecting

H_{0}

when it is actually true is 0.05 (5%).

In context: There is a 5% probability of concluding there is positive correlation between theoretical and practical test scores when in fact there is no correlation in the population.

Practice

Question 1: A nutritionist investigates whether there is positive correlation between sugar content and calorie count in breakfast cereals. A sample of 15 cereals gives

r = 0.58

. Test at the 5% significance level.

1

H_{0} : ρ = 0

,

H_{1} : ρ > 0

— one-tailed (positive)

2

n = 15

, one-tailed at 5%: critical value = 0.4409

3

∣ r ∣ = 0.58 > 0.4409

✓ → Reject

H_{0}

Conclusion: There is sufficient evidence at the 5% level of positive correlation between sugar content and calorie count in breakfast cereals.

Practice

Question 2: An economist tests whether there is a linear relationship between unemployment rate and inflation. A sample of 20 countries gives

r = - 0.39

. Test at the 10% significance level.

1

H_{0} : ρ = 0

,

H_{1} : ρ \neq = 0

— two-tailed (linear relationship)

2

Two-tailed at 10%: use 0.05 column.

n = 20

, critical value = 0.3783

3

∣ r ∣ = ∣ - 0.39∣ = 0.39 > 0.3783

✓ → Reject

H_{0}

Conclusion: There is sufficient evidence at the 10% level of a linear relationship between unemployment rate and inflation.

Practice

Question 3: An employee at a weather centre believes there is negative correlation between humidity and visibility. A sample of 6 days from Heathrow gives

r = - 0.58

. Test at the 5% significance level. State your hypotheses clearly.

1

H_{0} : ρ = 0

— There is no correlation between humidity and visibility

•

H_{1} : ρ < 0

— There is negative correlation between humidity and visibility

2

One-tailed at 5%,

n = 6

: critical value = 0.7293

3

∣ r ∣ = ∣ - 0.58∣ = 0.58 < 0.7293

✗ → Do not reject

H_{0}

Conclusion: There is not sufficient evidence at the 5% level of a negative correlation between humidity and visibility at Heathrow.

Summary

Hypothesis Testing for Correlation

StepAction

1. Hypotheses

H_{0} : ρ = 0

always. Choose

H_{1}

based on question.

2. Test typeOne-tailed (>, <) or two-tailed (≠)

3. Critical valueFrom table using

n

and significance level. Two-tailed: halve

α

.

4. Compare

∣ r ∣ > c.v.

→ Reject

H_{0}

5. ConcludeIn context: "sufficient/not sufficient evidence"

Exam Essentials

Hypotheses use $ρ$ (population), not

r

(sample)

Tables give one-tailed values — halve

α

for two-tailed

Use $∣ r ∣$ for comparison (absolute value)

Strict inequality:

∣ r ∣ > c.v.

to reject (not ≥)

Conclusions must be in context

Significance level = probability of incorrectly rejecting

H_{0}