Keller MATH 533 Course Project Part B, Hypothesis Testing

Keller MATH 533 Course Project Part B, Hypothesis Testing. Hypotheses 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 value to a company knowing to accept or reject a hypothesis will aid in the decision making process. Information derived from hypothesis testing can aid in more timely decision. Requirement A: The average (mean) sales per week exceeds 41.5 per salesperson. Key Statistics as computed by Minitab Variable N N* Mean SE Mean StDev Minimum Q1 Median Q3 SALES 100 0 42.340 0.417 4.171 32.000 39.250 42.000 45.000 N for Variable Maximum Range IQR Mode Mode SALES 52.000 20.000 5.750 44 12 Step 1 Hypotheses Ho: μ = 41.5 Ha: μ > 41.5 Step 2 Level of Significance The assumed standard deviation = 4.171 Variable N Mean StDev SE Mean 95% CI SALES 100 42.340 4.171 0.417 (41.522, 43.158) 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. Thus 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 42.34. The computed, using Minitab, Z score = 2.01. The computed score is greater than 1.645, reject the null hypothesis. We can be 95% confident that the average sales per week ca will fall within the 95% Confidence Interval (CI) of 41.522, 43.158. As a manager you can expect a preforming salesperson to achieve Step 2 Level of Significance The a = 0.05 is given. Step 3 Identify the statistical test to use. Use 1 Proportion because StDev is known and the sample (n=50) is a large sample (n > 30). 1Proportion score Test – Single Tail One Proportion- Z = -1.01 and a P-Value of 0.157.