Question

The Central Company manufactures a certain specialty item once a month in a batch production run. The number of items produced in each run varies from month to month as demand fluctuates. The company is interested in the relationship between the size of the production run (x) and the number of hours of labor (y) required for the run. The company has collected the following data for the ten most recent runs.

Number of items: 40 30 70 90 50 60 70 40 80 70
Labor (hours): 82 60 139 180 99 119 140 73 157 146
A. State the test statistic value t.
B. State the p value

Answer 1

a)
`t=(0.997√(10-2))/(√(1-(0.997)^2))=36.43`

b)The p value for this case can be founded like this:

`p_v = 2*P(t_(8) >36.43) = 3.53*10^(-10)`

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:

`r=(n(\sum xy)-(\sum x)(\sum y))/(√([n\sum x^2 -(\sum x)^2][n\sum y^2 -(\sum y)^2]))`

For our case we have this:

n=10
`\sum x = 600, \sum y = 1195, \sum xy = 78600, \sum x^2 =39400, \sum y^2 =156901`

`r=(10(78600)-(600)(1195))/(√([10(39400) -(600)^2][10(156901) -(39400)^2]))=0.997`

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:
`\rho =0`

Alternative hypothesis:
`\rho \\eq 0`

The statistic is gven by:

`t=(r √(n-2))/(√(1-r^2))`

And is distributed with n-2 degrees of freedom. df=n-2=10-2=8

`t=(0.997√(10-2))/(√(1-(0.997)^2))=36.43`

Part b

The p value for this case can be founded like this:

`p_v = 2*P(t_(8) >36.43) = 3.53*10^(-10)`

Is a very low value so we have enough evidence to reject the null hypothesis.