Question

Scientists examined data on mean summer sea‑surface temperatures (in degrees Celsius) and mean coral growth (in millimeters per year) over a several‑year period at locations in the Red Sea.

Temperature (∘C) 29.55 29.96 30.04 30.28 30.34 30.47 30.81
Growth (mm/y) 2.64 2.37 2 .74 2.52 2.34 2.46 2.31

Required:
Use your calculator to find the mean and standard deviation of both sea surface temperature x and growth y and the correlation r between x and y . Then use these basic measures to find the slope and intercept of the least‑squares line for predicting y from x .

Answer 1

Kindly check explanation

Explanation:

Given the data:

Temperature (∘C) 29.55 29.96 30.04 30.28 30.34 30.47 30.81

Growth (mm/y) 2.64 2.37 2.74 2.52 2.34 2.46 2.31

Using calculator and treating temperature and growth data as samples :

Temperature :

Mean : ΣX/n

n = sample size

Mean temperature (m) = 211.45 / 7

Mean temperature = 30.21°C

Standard deviation (s) :

Sqrt(Σ(X - m)²/n - 1)

s = 0.403

Growth:

Mean : ΣX/n

n = sample size

Mean growth (m) = 17.38 / 7

Mean growth = 2.483 mm/yr

Standard deviation (s) :

Sqrt(Σ(X - m)²/n - 1)

s = 0.161

Using the correlation Coefficient a calculator : the correlation Coefficient is - 0.6278, This shows that a negative relationship exists between both variables.

The intercept of the graph = 17.812

The slope of the relationship : -0.504