Answer:
At a significance level of 0.05, there is enough evidence to claim that there is a significant relationship between x and y.
P-value = 0.003.
Step-by-step explanation:
If we perform a regression analysis relating x and y, we get the best fitting line with equation:

and a correlation coefficient r:

We have to test the hypothesis, where the alternative hypothesis claims that there is a relationship between these two variables, and the null hypothesis claiming there is no relationship (meaning that the correlation is not significantly different from 0).
This can be written as:

where ρ is the population correlation coefficient for x and y.
The significance level is assumed to be 0.05.
The sample size is n=16.
The degrees of freedom are df=14.

The test statistic can be calculated as:

For a test statistic t=2.05 and 14 degrees of freedom, the P-value is calculated as:

The P-value (0.003) is smaller than the significance level (0.05), so the effect is significant enough.
The null hypothesis is rejected.
At a significance level of 0.05, there is enough evidence to claim that there is a significant relationship between x and y.