Test the significance of the correlation coefficient r at a=0.05 for the data below.
X values 59, 65, 73, 80, 89
Y values 97, 102, 110, 117, 125
A) Accept p=0 because 60.81<3.18
B) Reject p=0 because 61.13>3.18
C) Accept p=0 because 61.13<2.78
D) Reject p=0 because 60.81>2.78
Using an online calculator with the following pairs:
5 data pairs (x,y):
( 59.0 , 97.0 ); ( 65.0 , 102. ); ( 73.0 , 110. ); ( 80.0 , 117. ); ( 89.0 , 125. );
Degrees of freedom = N (number of pairs) - 2 = 3
Correlation coefficient r = 1.00
Check the appropriate table for degrees of freedom and the level of significance to determine whether or not to reject the null.
To test the significance of the correlation coefficient "r" at a significance level of "a = 0.05" for the given data, you need to follow these steps:
Step 1: Calculate the correlation coefficient "r" using the formula:
r = [ Σ((X - X̄) * (Y - Ȳ)) ] / [ √(Σ(X - X̄)² * Σ(Y - Ȳ)²)]
where Σ denotes the sum, X and Y represent the individual values, and X̄ and Ȳ denote the mean values.
Step 2: Determine the degrees of freedom (df), which is equal to the number of pairs of values minus 2. In this case, since there are 5 pairs of values, df = 5 - 2 = 3.
Step 3: Calculate the critical value (t_critical) from the t-distribution table for a two-tailed test, with a significance level of 0.05 and degrees of freedom (df) equal to 3.
Step 4: Calculate the test statistic (t) using the formula:
t = r / [ √((1 - r²) / (n - 2)) ]
where n is the number of pairs of values (in this case, 5).
Step 5: Compare the absolute value of the test statistic (|t|) with the critical value (t_critical). If |t| > t_critical, then reject the null hypothesis (no correlation), and if |t| ≤ t_critical, then accept the null hypothesis (no correlation).
Now let's work through the calculations:
Given X values: 59, 65, 73, 80, 89
Given Y values: 97, 102, 110, 117, 125
Step 1: Calculate the correlation coefficient "r":
X̄ = (59 + 65 + 73 + 80 + 89) / 5 = 73.2
Ȳ = (97 + 102 + 110 + 117 + 125) / 5 = 110.2
Next, calculate the numerator and denominator:
Numerator:
Σ((X - X̄) * (Y - Ȳ)) = (59 - 73.2) * (97 - 110.2) + (65 - 73.2) * (102 - 110.2) +
(73 - 73.2) * (110 - 110.2) + (80 - 73.2) * (117 - 110.2) +
(89 - 73.2) * (125 - 110.2) = -51.8
Denominator:
√(Σ(X - X̄)² * Σ(Y - Ȳ)²) = √[(59 - 73.2)² + (65 - 73.2)² + (73 - 73.2)² +
(80 - 73.2)² + (89 - 73.2)²] *
√[(97 - 110.2)² + (102 - 110.2)² + (110 - 110.2)² +
(117 - 110.2)² + (125 - 110.2)²] = 591.76
Finally, calculate the correlation coefficient "r":
r = -51.8 / 591.76 = -0.0877 (rounded to four decimal places)
Step 2: Determine the degrees of freedom (df):
df = 5 - 2 = 3
Step 3: Lookup the critical value (t_critical):
From the t-distribution table, a two-tailed test with df = 3 at a significance level of 0.05 yields a t_critical value of approximately 3.18.
Step 4: Calculate the test statistic (t):
t = r / [ √((1 - r²) / (n - 2)) ]
t = -0.0877 / [ √((1 - (-0.0877)²) / (5 - 2)) ] = -0.0877 / 0.1861 = -0.4710 (rounded to four decimal places)
Step 5: Compare the absolute value of the test statistic (|t|) with the critical value (t_critical):
|t| = 0.4710
t_critical = 3.18
Since |t| (0.4710) is less than t_critical (3.18), we fail to reject the null hypothesis. Therefore, the correct answer is:
C) Accept p=0 because 61.13<2.78