Every morning, Dany jogs the same 2-mile route around her neighborhood. She sele

Question

Every morning, Dany jogs the same 2-mile route around her neighborhood. She selects a random sample of days and records the time it takes her to complete the route as well as her pulse rate just after she gets home. She calculated a Pearson correlation coefficient of r = -0.70 and below is a plot of her data:

a.) List the assumptions of a correlation hypothesis test. Are the assumptions met?

b.) Calculate a 95% confidence interval for the population Pearson correlation coefficient between run time and pulse rate.

c). Based on your interval, can you say that running the 2-mile route faster causes her pulse rate to increase? Why/why not?

Explanation / Answer

a.) List the assumptions of a correlation hypothesis test. Are the assumptions met?

Answer:

The assumptions of a correlation hypothesis test are given as below:

The two variables are normally distributed.

There should be a linear relationship exists between the given two variables.

From the given scatter plot, it is observed that the assumption of linear relationship exists between the given two variables time and pulse rate. The assumption of normality would be check by using normality tests.

b.) Calculate a 95% confidence interval for the population Pearson correlation coefficient between run time and pulse rate.

Here, we have to find 95% confidence interval for the population Pearson correlation coefficient.

We are given

R = -0.70 and

n = 11 (from scatter plot)

Confidence level = 95%

So, Z = 1.96 (by using z-table)

The formula for confidence interval is given as below:

L U

Where,

L = [1+R – (1 – R)*exp(2*Z/sqrt(n – 3)]/[ 1+R + (1 – R)*exp(2*Z/sqrt(n – 3)]

U = [1+R – (1 – R)*exp(-2*Z/sqrt(n – 3)]/[ 1+R + (1 – R)*exp(-2*Z/sqrt(n – 3)]

Now, plug all values in above formula given as below:

L = [1 – 0.70 – (1 + 0.70)*exp(2*1.96/sqrt(11 – 3)]/[ 1 – 0.70 + (1 + 0.70)*exp(2*1.96/sqrt(11 – 3)]

L = [1 – 0.70 – (1 + 0.70)*3.9985]/[ 1 – 0.70 + (1 + 0.70)*3.9985]

L =[ 0.30 - 6.797517977]/[0.30 + 6.797517977]

L = -6.497517977/7.097517977

L = -0.915463405

Now, we have to find upper limit

U = [1+R – (1 – R)*exp(-2*Z/sqrt(n – 3)]/[ 1+R + (1 – R)*exp(-2*Z/sqrt(n – 3)]

U = [1 – 0.70 – (1 + 0.70)*exp(-2*1.96/sqrt(11 – 3)]/[ 1 – 0.70 + (1 + 0.70)*exp(-2*1.96/sqrt(11 – 3)]

U = [1 – 0.70 – (1 + 0.70)*0.25]/[ 1 – 0.70 + (1 + 0.70)*0.25]

U = -0.125155183/0.725155183

U = -0.172590896

Confidence interval = (-0.9155, -0.1726)

Part c

Based on above interval, we can say that running the 2-mile route faster causes her pulse rate to increase because we get the both negative values in the given confidence interval which suggests that faster running increases pulse rate.

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