Question

1. A Doctor walking through town to his job hospital is a bird watcher, and records the number of storks nest he sees each year in chimneys. After 10 years he notices the number of births and stork sightings seems oddly correlated. Here are the summary statistics between births and stork nest sightings:

N= 10

Xbar = 8.8

Ybar= 33.6

Sxx = 61.6

Syy = 1312.4

Sxy= 250.2

SSE= 296.1656

1. Test if the correlation is > 0 at the 0.5 level
2. Calculate a 95% prediction interval for the number of births if the number of stork sightings is 30

Answer 1

Sample size, n =	10
x̅ =	8.8
y̅ =	33.6
SSxx =	61.6
SSyy =	1312.4
SSxy =	250.2
SSE	296.1656

a)  Correlation coefficient, r = SSxy/√(SSxx*SSyy) = 0.88
Coefficient of determination, r² = (SSxy)²/(SSxx*SSyy) = 0.7743

Null and alternative hypothesis:

Ho:   ρ = 0  
Ha:   ρ > 0  

df = n-2 = 8   

alphav, α = 0.05  

Test statistic:

t = r*√(n-2)/√(1-r²) = 5.2393

p-value =T.DIST.RT(5.2393, 8) = 0.0004  

As p-value <α, reject the null hypothesis.

There is a positive correlation between x and y at 0.05 significance level.

b) Slope, b = SSxy/SSxx =    4.061688312          
y-intercept, a = y̅ -b* x̅ =    -2.142857143          
              
Regression equation :               
ŷ = -2.1429 + 4.0617 x

Predicted value at X =   30              
ŷ =-2.1429 + 4.0617 * 30 =  119.708

Standard error, se = √(SSE/(n-2)) =   6.0845
  
At α = 0.05 and df = n-2 = 8, critical value, t_c = T.INV.2T(0.05, 8 ) =   2.3060

95% prediction interval for the number of births if the number of stork sightings is 30