PROJECT PART B
PROJECT PART B: Hypothesis Testing and Confidence Intervals
Math 533 Applied managerial Statistics.
Instructor: Mr. Patrick Mayers.
April 12, 2015
Guillermo H. Quintela.
a. The average (mean) sales per week exceeds 41.5 per salesperson.
The Null Hypothesis: The average (mean) sales per week is greater than or equal to 41.5 per salesperson.
Ho:µ >= 41.5
The alternate Hypothesis: The average (mean) sales per week is less than or equal than 41.5 per salesperson.
Ha:µ < 41.5
For a significance level of α = 0.05 I used the Z-Test of proportion to get the result based on the given hypothesis, being Ho:µ > 41.5.
The critical p-value or this significance level is 0.022, which is less than α = 0.05. Thus, the decision rule is to reject the null Hypothesis.
Descriptive Statistics: SALES
Total
Variable Count Mean SE Mean
SALES 100 42.340 0.417
One-Sample Z
The assumed standard deviation = 4.171
95% Lower N Mean SE Mean Bound Z P
100 42.340 0.417 41.654 2.01 0.022
N Mean SE Mean 95% CI
100 42.340 0.417 (41.522, 43.158)
Summary.
Since the value of the test statistics falls in the rejection region I decided to reject the null hypothesis. In fact, the data se provided gives me enough information to know that the null hypothesis is false, telling me that with a 95% of confidence interval or 0.05 of significance level the statement about the average (mean) sales per week is greater than 41.5 per salesperson, is true.
Since the p-value for this test is smaller than the significance level of 0.05, we may concluding to reject the null hypothesis.
And finally, with an 95% of confidence level in this case is giving me the right answer about the initial statement that the average (mean) sales per week exceeds 41.5 per salesperson.
b. The true population proportion of salespeople that received online training is less than 55%.
The Null
Math 533 Applied managerial Statistics.
Instructor: Mr. Patrick Mayers.
April 12, 2015
Guillermo H. Quintela.
a. The average (mean) sales per week exceeds 41.5 per salesperson.
The Null Hypothesis: The average (mean) sales per week is greater than or equal to 41.5 per salesperson.
Ho:µ >= 41.5
The alternate Hypothesis: The average (mean) sales per week is less than or equal than 41.5 per salesperson.
Ha:µ < 41.5
For a significance level of α = 0.05 I used the Z-Test of proportion to get the result based on the given hypothesis, being Ho:µ > 41.5.
The critical p-value or this significance level is 0.022, which is less than α = 0.05. Thus, the decision rule is to reject the null Hypothesis.
Descriptive Statistics: SALES
Total
Variable Count Mean SE Mean
SALES 100 42.340 0.417
One-Sample Z
The assumed standard deviation = 4.171
95% Lower N Mean SE Mean Bound Z P
100 42.340 0.417 41.654 2.01 0.022
N Mean SE Mean 95% CI
100 42.340 0.417 (41.522, 43.158)
Summary.
Since the value of the test statistics falls in the rejection region I decided to reject the null hypothesis. In fact, the data se provided gives me enough information to know that the null hypothesis is false, telling me that with a 95% of confidence interval or 0.05 of significance level the statement about the average (mean) sales per week is greater than 41.5 per salesperson, is true.
Since the p-value for this test is smaller than the significance level of 0.05, we may concluding to reject the null hypothesis.
And finally, with an 95% of confidence level in this case is giving me the right answer about the initial statement that the average (mean) sales per week exceeds 41.5 per salesperson.
b. The true population proportion of salespeople that received online training is less than 55%.
The Null