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toothpaste666
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Homework Statement
The following are the number of sales which a sample of nine salespeople of industrial chemicals in California and a sample of six salespeople of industrial chemicals in Oregon made over a certain fixed period of time.California: 59, 68, 44, 71, 63, 46, 69, 54, 48
Oregon: 50, 36, 62, 52, 70, 41
Assuming that the populations sampled can be approximated closely with normal distributions having the same variance, Is there a difference in the number of sales between the California salespeople and the Oregon salespeople? Conduct a hypothesis test at the significance level .01.
The Attempt at a Solution
since both have the same variances and they are normal, we use a t test.
t = [X - Y - δ]/[(Sp)sqrt(1/n1 + 1/n2)]
where Sp^2 = [(n1-1)S1^2 + (n2-1)S2^2]/[n1+n2-2]
H0: δ = μ1 - μ2 = 0
H1: δ≠0
it is a two sided test so tα/2 = t.01/2 = t.005 for n1+n2-2 = 9 + 6 - 2 = 13 degrees of freedom = 3.012
so we reject H0 if t > 3.012 or t < 3.012
using Y (oregon) = ∑x/n2 and X (california) = ∑x/n1
I found Y = 51.8 and X = 58
using the formula S^2 = [∑x^2 - (∑x)^2/n]/[n-1]
we find S1^2 = 109 and S2^2 = 161
using the formula above Sp^2 = [8(109) + 5(161)]/[9+6-2] = 129
Sp = sqrt(129) = 11.36
plugging all this into t test stastic:
t = [59 - 51.8 - 0]/[(11.36)sqrt(1/9 + 1/6)] = 6.2/ 5.99 = 1.036
this falls within the range so we accept the null hypothesis. there is no difference in the number of sales.
Am I doing this correctly?