Is the correlation between weekly sales and shelf space significant at the .01 l

The marketing manager of a large supermarket chain would like to use shelf space to predict the sales of pet food. For a random sample of 12 similar stores, she gathered the following information regarding the shelf space, in feet, devoted to pet food and the weekly sales in hundreds of dollars. Use these data to answer questions 1 through 3.
Store 1 2 3 4 5 6
Shelf Space 5 5 15 20 20 20
Weekly Sales 1.6 2.2 1.4 1.9 2.4 2.6
Store 7 8 9 10 11 12
Shelf Space 25 25 25 30 30 30
Weekly Sales 2.3 2.7 2.8 2.6 2.9 3.1
Question 1: Compute the value of the sample correlation coefficient between weekly sales and shelf space. Possible answers A. 0.6234 B. 0.827 C. 0.652 D. 0.741 I answered D
----
Using the LinReg function on my TI-84 I get r = 0.3835
===========================
Question 2: Is the correlation between weekly sales and shelf space significant at the .01 level of significance? Possible answers: A. No, the sample correlation coefficient does not exceed the critical value. B. Yes, the p-value of the test for significance is not than .01. C. Yes, the value of the test statistic does exceed the critical value. D. Yes, the computed t-test statistic is less than the critical value.
------------------------------------------

 
Using the A-6 Table for linear correlation I find the critical value
for n = 6 is r = 0.917
Since "r" for the data set is 0.3835 we fail to reject Ho:
Ho: no linear correlation.
Ha: significant linear correlation
Answer: A
=====================================
Question 3: What is the estimated regression equation? A. = 1.45 + 0.044x B. = 1.45 + 0.724x C. = 1.45 - 0.074x D. = 2.63 - 0.174x I answered A.
--
I get y = 0.024x + 1.67

This looks like an elementary stats regression inference question.

\(t=\frac{b}{SE_b}\), where b is the slope of the least-squares regression line (r* sy/sx) and SEb is the standard error of least squares slope, which is given by \(sqrt{\frac{s}{\Sigma (x-\bar{x})\)

The question doesn't specify an expected outcome, so I'm operating under the assumptions that \(H_o:\beta=0 \; \;
H_a: \beta \ne 0\)

I get a t value of about 4.652, which corresponds to a very small area of the curve (p is about .00091). However, you should probably check my work as data entry errors are incredibly[/b] easy to make.

Edit: Oops, I was ninja'd. That being said, I've never heard of the particular technique rsb utilized-- the purpose of regression inference is to determine whether or not the correlation is significant, not whether or not it actually exists. However, without knowing what information table A-6 contains, it's hard to comment. I'm curious as to where n=6 came from as these distributions have the same df of s, n-2.