# Homework Help: Test concerning difference of two means

1. Nov 21, 2015

### toothpaste666

1. The problem statement, all variables and given/known dataThe 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.

3. 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?

2. Nov 21, 2015

### Ray Vickson

It looks OK, but I have not checked the numbers in detail.

3. Nov 22, 2015

thanks