In the table below, the gross values of milk produced (in millions of Rand) and cattle raised (in millions) in South Africa, for ten years, are given. The relationship between milk produced (x) and cattle raised (y) is investigated.
X
28.7 36.3 41.3 41.1 45.5 47.7 54.7 56.9 71.2 71.8
Y
17.7 21.4 21.2 23.7 23.8 22.3 22.3 22.9 25.2 27.1
The following information is known:
∑ x = 495.2 ∑ y = 227.6 ∑ x2 = 26336.2
∑ y2 = 5237.26 ∑ xy = 11543.89
You have to test the hypothesis that there is no correlation between milk and
cattle raised. The value of the test statistic is
To test the hypothesis that there is no correlation between milk produced and cattle raised, we can use the formula for the test statistic:
r = ∑xy - (∑x)(∑y)/n
_____________________
√(∑x^2 - (∑x)^2/n)(∑y^2 - (∑y)^2/n)
where r is the sample correlation coefficient, ∑xy is the sum of the products of x and y, ∑x is the sum of x values, ∑y is the sum of y values, ∑x^2 is the sum of squared x values, ∑y^2 is the sum of squared y values, and n is the number of observations.
Using the given values:
∑x = 495.2
∑y = 227.6
∑x^2 = 26336.2
∑y^2 = 5237.26
∑xy = 11543.89
n = 10
We can plug in these values into the formula to calculate the correlation coefficient:
r = (11543.89 - (495.2)(227.6)/10) / (√((26336.2 - (495.2)^2/10)(5237.26 - (227.6)^2/10)))
r = (11543.89 - 112630.72/10) / (√((26336.2 - 24520.24/10)(5237.26 - 51910.24/10)))
r = (11543.89 - 11263.07) / (√((26336.2 - 2452.02)(5237.26 - 519.10)))
r = 280.82 / (√(23884.18 * 4718.16))
r = 280.82 / (√112795207.9728)
r = 280.82 / 10629.92
r ≈ 0.0265
The sample correlation coefficient is approximately 0.0265.
To test the hypothesis, we can compare the absolute value of the correlation coefficient (|r|) to a critical value. Let's assume a significance level of 0.05. The critical value for a two-tailed test with n=10 is approximately 0.685.
Since |0.0265| < 0.685, we fail to reject the null hypothesis. Therefore, there is not enough evidence to suggest a significant correlation between milk produced and cattle raised.
To test the hypothesis that there is no correlation between milk produced and cattle raised, we can calculate the test statistic using the given information.
The formula to calculate the test statistic for a correlation coefficient is:
t = (r * sqrt(n-2)) / sqrt(1 - r^2)
Where:
- r is the sample correlation coefficient
- n is the sample size
In this case, the sample correlation coefficient (r) is calculated as:
r = ( ∑ xy - (∑ x * ∑ y) / n ) / sqrt( (∑ x^2 - (∑ x)^2 / n) * (∑ y^2 - (∑ y)^2 / n) )
Given:
- ∑ x = 495.2
- ∑ y = 227.6
- ∑ x^2 = 26336.2
- ∑ y^2 = 5237.26
- ∑ xy = 11543.89
- n = sample size = 10
Let's plug these values into the formula to calculate the sample correlation coefficient (r):
r = ( 11543.89 - (495.2 * 227.6) / 10 ) / sqrt( (26336.2 - (495.2)^2 / 10) * (5237.26 - (227.6)^2 / 10) )
Simplifying this gives:
r = ( 11543.89 - 112701.92 / 10 ) / sqrt( (26336.2 - 245052.04 / 10) * (5237.26 - 51809.76 / 10) )
r = ( 11543.89 - 11270.192 ) / sqrt( (26336.2 - 24505.204) * (5237.26 - 5180.976) )
r = ( 2684.698 ) / sqrt( (1831.996) * (45.8296) )
r = ( 2684.698 ) / sqrt( 83898.204 )
r = 0.1112809301
Now we can calculate the test statistic (t) using the calculated correlation coefficient (r) and the sample size (n):
t = ( r * sqrt(n-2) ) / sqrt(1 - r^2)
t = ( 0.1112809301 * sqrt(10-2) ) / sqrt(1 - (0.1112809301)^2)
t = ( 0.1112809301 * sqrt(8) ) / sqrt(1 - 0.0124018373)
t = ( 0.1112809301 * 2.828427125 ) / sqrt(0.9875981627)
t = 0.314058326 / 0.9937540582
t = 0.3160591619
Therefore, the value of the test statistic is 0.3160591619.
To test the hypothesis that there is no correlation between milk produced and cattle raised, we can use the test statistic formula:
r = (∑xy - (∑x * ∑y) / N) / sqrt((∑x^2 - (∑x)^2 / N) * (∑y^2 - (∑y)^2 / N))
Where:
- r is the correlation coefficient
- ∑xy is the sum of the product of x and y
- ∑x and ∑y are the sums of x and y
- ∑x^2 and ∑y^2 are the sums of the squares of x and y
- N is the number of observations
Given the provided information, we can substitute the values into the formula:
r = (11543.89 - (495.2 * 227.6) / 10) / sqrt((26336.2 - (495.2)^2 / 10) * (5237.26 - (227.6)^2 / 10))
Now, calculating the test statistic:
r = (11543.89 - (495.2 * 227.6) / 10) / sqrt((26336.2 - 245201.44 / 10) * (5237.26 - 51925.76 / 10))
r = (11543.89 - 112712 / 10) / sqrt((26336.2 - 24520.144) * (5237.26 - 5192.576))
r = (11543.89 - 11271.2) / sqrt((26336.2 - 24520.144) * (5237.26 - 5192.576))
r = 271.69 / sqrt((1816.056) * (44.684))
r = 271.69 / sqrt(81172.37)
r = 271.69 / 284.932
r ≈ 0.953
Therefore, the value of the test statistic is approximately 0.953.