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 
The alternative hypothesis is 
Generally the standard deviation is mathematically evaluated as



The degree of freedom for the one-tail test is



The standard error is evaluated as


The test statistics is evaluated as



The p-value of of t is obtained from the z table, the value is

Given that
then we reject the null hypothesis
Hence the company can conclude that the correlation is positive