Respuesta :
Answer:
a)[tex]t=\frac{0.997\sqrt{10-2}}{\sqrt{1-(0.997)^2}}=36.43[/tex]
b)The p value for this case can be founded like this:
[tex] p_v = 2*P(t_{8} >36.43) = 3.53*10^{-10}[/tex]
Step-by-step explanation:
We have the following data given:
Number of items (x): 40 30 70 90 50 60 70 40 80 70
Labor (hours) (y): 82 60 139 180 99 119 140 73 157 146
And in order to calculate the correlation coefficient we can use this formula:
[tex]r=\frac{n(\sum xy)-(\sum x)(\sum y)}{\sqrt{[n\sum x^2 -(\sum x)^2][n\sum y^2 -(\sum y)^2]}}[/tex]
For our case we have this:
n=10 [tex] \sum x = 600, \sum y = 1195, \sum xy = 78600, \sum x^2 =39400, \sum y^2 =156901[/tex]
[tex]r=\frac{10(78600)-(600)(1195)}{\sqrt{[10(39400) -(600)^2][10(156901) -(39400)^2]}}=0.997[/tex]
So then the correlation coefficient would be r =0.997
Part a
In order to test the hypothesis if the correlation coefficient it's significant we have the following hypothesis:
Null hypothesis: [tex]\rho =0[/tex]
Alternative hypothesis: [tex]\rho \neq 0[/tex]
The statistic is gven by:
[tex]t=\frac{r \sqrt{n-2}}{\sqrt{1-r^2}}[/tex]
And is distributed with n-2 degrees of freedom. df=n-2=10-2=8
[tex]t=\frac{0.997\sqrt{10-2}}{\sqrt{1-(0.997)^2}}=36.43[/tex]
Part b
The p value for this case can be founded like this:
[tex] p_v = 2*P(t_{8} >36.43) = 3.53*10^{-10}[/tex]
Is a very low value so we have enough evidence to reject the null hypothesis.
