Answer:
a) r = 0.974
b) Critical value = 0.602
Step-by-step explanation:
Given - Two separate tests are designed to measure a student's ability to solve problems. Several students are randomly selected to take both test and the results are give below
Test A | 64 48 51 59 60 43 41 42 35 50 45
Test B | 91 68 80 92 91 67 65 67 56 78 71
To find - (a) What is the value of the linear coefficient r ?
(b) Assuming a 0.05 level of significance, what is the critical value ?
Proof -
A)
r = 0.974
B)
Critical Values for the Correlation Coefficient
n alpha = .05 alpha = .01
4 0.95 0.99
5 0.878 0.959
6 0.811 0.917
7 0.754 0.875
8 0.707 0.834
9 0.666 0.798
10 0.632 0.765
11 0.602 0.735
12 0.576 0.708
13 0.553 0.684
14 0.532 0.661
So,
Critical r = 0.602 for n = 11 and alpha = 0.05
Ok so
Knowing that 1 metre consists of 100cm,
You can conclude and agree that 8 cm is 8/100, correct?
Now, when converting fractions to decimals, it makes it easier out of 100 or 10. Since it's out of 100 already, you take the amount of zero's in 100 (two) and you move the decimal point (2) places to the <em /><u>left<em /></u>
This gives you 0.08
So the fraction is 8/100, which then can be simplified to 4/50, then to 2/25
and the decimal is 0.08.
$10.00 - 3($0.65)
$10.00 - $1.95
= $8.05