Test concerning difference of two means

Click For Summary
SUMMARY

The discussion focuses on conducting a hypothesis test to determine if there is a significant difference in sales between salespeople in California and Oregon, using a two-sample t-test. The null hypothesis (H0) states that there is no difference in means (δ = μ1 - μ2 = 0), while the alternative hypothesis (H1) posits that a difference exists (δ≠0). The calculated t-statistic is 1.036, which falls within the acceptance range, leading to the conclusion that there is no significant difference in sales at the 0.01 significance level.

PREREQUISITES
  • Understanding of hypothesis testing and significance levels
  • Familiarity with two-sample t-tests and their assumptions
  • Knowledge of calculating sample means and variances
  • Ability to perform statistical calculations using formulas for t-tests
NEXT STEPS
  • Study the assumptions of the two-sample t-test in detail
  • Learn how to calculate confidence intervals for the difference of means
  • Explore the use of statistical software like R or Python for hypothesis testing
  • Investigate alternative tests for non-normal distributions, such as the Mann-Whitney U test
USEFUL FOR

Students in statistics courses, data analysts, and researchers conducting comparative studies in sales or similar fields will benefit from this discussion.

toothpaste666
Messages
516
Reaction score
20

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?
 
Physics news on Phys.org
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.
 
  • Like
Likes   Reactions: toothpaste666
thanks
 

Similar threads

  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 11 ·
Replies
11
Views
2K
  • · Replies 14 ·
Replies
14
Views
2K
  • · Replies 21 ·
Replies
21
Views
3K
Replies
14
Views
2K
Replies
1
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
Replies
1
Views
4K