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MATH 533 Week 6 Course Project Part B Hypothesis Testing
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MATH 533 Week 6 Course Project Part B Hypothesis Testing
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Math-533Hypothesis Testing
By: Sagar Patel
Professor: Terrance Encalarde
Keller Graduate School of Management
June 10th, 2017
, A hypothesis testing involves the testing of the null hypothesis and the alternative
hypothesis. When testing the hypothesis, either the null hypothesis or the alternative hypotheses
is rejected. The Company can use hypothesis testing to examine if it should keep a current
process, change a current process or start a new process. It can also examine different areas to
see if an employee or manager’s ideas would prove more productive. The hypothesis testing can
quickly be applied or tested saving the Company time and money.
Requirement A: The Average Mean Sales per Week exceeds 41.5 per salesperson.
Key Statistics were computed using Minitab
Descriptive Statistics: Sales (Y)
Variable SALES
NUMBER 100
MEAN 42.55
SE MEAN 0.717
7.1708937
ST.DEVIATION 5
51.421717
VARIANCE 2
MINIMUM 21
Q1 39
MEDIAN 43
Q3 47
MAXIMUM 67
RANGE 46
One-Sample Z: Sales (Y)
The assumed standard deviation = 7.171
Variable N Mean St-Dev SE Mean 95% CI
Sales (Y) 100 42.550 7.171 0.717 (41.145, 43.955)
One-Sample Z: Sales (Y)
Test of mu = 41.5 vs > 41.5
The assumed standard deviation = 7.171
95% Lower
Variable N Mean St-Dev SE Mean Bound Z P
Sales (Y) 100 42.550 7.171 0.717 41.370 1.46 0.072
, Step 1- Hypothesis
Ho: μ = 41.5
Ha: μ> 41.5
Step 2-Level of Significance
The a = 0.05 is given.
Step 3- Identify the statistical test to use
Use z-test because the Standard Deviation is known and the sample (n=100) is a large
sample (n > 30).
The test will let us decide, if we are going to accept or reject that sales will exceed 41.5
per salesperson per week. To test this, a single tail test will be conducted.
Z score Test – Single Tail (Result According to Minitab) One-Sample Z = 1.464 and P = 0.072
Step- 4 Decision Rule
The alternative hypothesis states that mean sales per week exceeds 41.5; this is a one
tailed test to the right. The given a = .05 is to the right of Z = 1.645. Therefore, we reject the
null hypothesis if the Z > 1.645. If the p value is less than the a = .05 then reject the null
hypothesis. Step- 5 Decision Making
The 100 salespersons average mean is 42.55. The computed Z score is = 1.464. The
computed score is less than 1.645, we do not reject the null hypothesis. We can be 95% confident
that the average sales per week will fall within the 95% Confidence Interval (CI) of (41.145,
43.955). The Company can expect a salesperson to achieve sales an average of 42.55 sales a
week.
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