Question

question #7
determine the pearson correlation coefficient for n = 59 and α = 0.01.
critical values of the pearson
correlation coefficient r
n | α = .05 | α = .01
4 | .950 | .999
5 | .878 | .959
6 | .811 | .917
7 | .754 | .875
8 | .707 | .834
9 | .666 | .798
10 | .632 | .765
11 | .602 | .735
12 | .576 | .708
13 | .553 | .684
14 | .532 | .661
15 | .514 | .641
16 | .497 | .623
17 | .482 | .606
18 | .468 | .590
19 | .456 | .575
20 | .444 | .561
25 | .396 | .505
30 | .361 | .463
35 | .335 | .430
40 | .312 | .402
45 | .294 | .378
50 | .279 | .361
60 | .254 | .330
70 | .236 | .305
80 | .220 | .286
90 | .207 | .269
100 | .196 | .256
options: .254, .279, .361, .330

Explanation:

Step1: Identify n and α

We have \( n = 59 \) and \( \alpha = 0.01 \). We need to find the critical value of the Pearson correlation coefficient \( r \) from the given table.

Step2: Locate n in the table

Looking at the table, for \( n = 60 \), the value under \( \alpha = 0.01 \) is \( 0.330 \). Since \( n = 59 \) is very close to \( n = 60 \), we use the value for \( n = 60 \) (as the table doesn't have \( n = 59 \) explicitly, and \( 59 \) is near \( 60 \)).

Answer:

\( 0.330 \) (corresponding to the option with \( 0.330 \))