Test concerning difference of two means

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The discussion centers on conducting a hypothesis test to determine if there is a significant difference in sales between salespeople in California and Oregon. The t-test is applied, assuming equal variances and normal distributions for both samples. The null hypothesis (H0) states that there is no difference in means, while the alternative hypothesis (H1) suggests a difference exists. After calculating the test statistic, the result falls within the acceptance range, leading to the conclusion that there is no significant difference in sales between the two groups. The calculations appear correct, though some participants suggest verifying the numbers for accuracy.
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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?
 
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toothpaste666 said:

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?

It looks OK, but I have not checked the numbers in detail.
 
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Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

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