Question

A data set includes weights of garbage discarded in one week from 62 different households. Correlation matrix: The paired weiState the conclusion. Because the absolute value of the test statistic is the positive critical value, there sufficient evide

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
15 0.514 0.641
16 0.497 0.623
17 0.482 0.606
18 0.468 0.59
19 0.456 0.575
20 0.444 0.561
25 0.396 0.505
30 0.361 0.463
35 0.335 0.43
40 0.312 0.402
45 0.294 0.378
50 0.279 0.361
60 0.254 0.33
70 0.236 0.305
80 0.22 0.286
90 0.207 0.269
100 0.196 0.256
0 0
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Answer #1

Solution:-

Null Hypothesis H0: The population correlation coefficient is not significantly different from 0.( \rho = 0)

Alternate Hypothesis HA: The population correlation coefficient is significantly different from 0.( \rho \neq 0)

Significance level = 0.05

Degree of freedoms:-

D.F = n - 2

D.F = 60

Test statistics:-

n - 2 t=rX V1-r?

t = 0.968

There are two critical values + 0.254

rCritical = + 0.254

tCritical = + 2.034

p value for test = 0.337

Since p value(0.337) is greater than the significance value so we have to accept the null hypothesis.

Because the value of the test statistics is lies below the positive critical value there is not sufficient evidence to support the claim that there is linear correlation between two variables.

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