Complete Question
Pennsylvania Refining Company is studying the relationship between the pump price of gasoline and the number of gallons sold. For a sample of 14 stations last Tuesday, the correlation was 0.65.At the 0.01 significance level Can the company conclude that the correlation is positive
Answer:
Yes the company conclude that the correlation is positive
Step-by-step explanation:
From the question we are told that
The sample size is n = 14
The correlation is r = 0.65
The null hypothesis is ![H_o : r < 0](https://tex.z-dn.net/?f=H_o%20%3A%20%20r%20%3C%20%200)
The alternative hypothesis is ![H_1 : r > 0](https://tex.z-dn.net/?f=H_1%20%3A%20%20r%20%3E%200)
Generally the standard deviation is mathematically evaluated as
![Sr = \sqrt{1- r}](https://tex.z-dn.net/?f=Sr%20%3D%20%20%5Csqrt%7B1-%20r%7D)
![Sr = \sqrt{1- 0.65}](https://tex.z-dn.net/?f=Sr%20%3D%20%20%5Csqrt%7B1-%200.65%7D)
![Sr = 0.616](https://tex.z-dn.net/?f=Sr%20%3D%200.616)
The degree of freedom for the one-tail test is
![df = n- 2](https://tex.z-dn.net/?f=df%20%3D%20%20n-%202)
![df = 14- 2](https://tex.z-dn.net/?f=df%20%3D%20%2014-%202)
![df = 12](https://tex.z-dn.net/?f=df%20%3D%2012)
The standard error is evaluated as
![SE = \frac{0.616}{ \sqrt{12} }](https://tex.z-dn.net/?f=SE%20%20%3D%20%20%5Cfrac%7B0.616%7D%7B%20%5Csqrt%7B12%7D%20%7D)
![SE =0.1779](https://tex.z-dn.net/?f=SE%20%20%3D0.1779)
The test statistics is evaluated as
![t = \frac{r }{SE}](https://tex.z-dn.net/?f=t%20%3D%20%20%5Cfrac%7Br%20%7D%7BSE%7D)
![t = \frac{0.65 }{0.1779}](https://tex.z-dn.net/?f=t%20%3D%20%20%5Cfrac%7B0.65%20%7D%7B0.1779%7D)
![t = 3.654](https://tex.z-dn.net/?f=t%20%3D%20%203.654)
The p-value of of t is obtained from the z table, the value is
![p-value = P(t < 3.654) = 0.00012909](https://tex.z-dn.net/?f=p-value%20%3D%20%20P%28t%20%3C%20%203.654%29%20%3D%20%200.00012909)
Given that
then we reject the null hypothesis
Hence the company can conclude that the correlation is positive