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The new director of special programs in XYZ Corporation felt the customers were waiting too long to receive and complete forms needed to enroll in special programs. After collecting some data, Ms. Jones determined the mean wait time was 28 minutes. She felt this time period was in excess, and she instituted new procedures to streamline the process. One month later, a sample of 127 parents was selected. The mean wait time recorded was 26.9 minutes and the standard deviation of the sampling was 8 minutes. Using the .02 level of significance, conduct a 5-step hypothesis testing procedure to determine if the new processes significantly reduced the wait time.
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The solution contains detailed explanation of testing hypotheses using a 5-step hypothesis testing procedure.
- BSc , Wuhan Univ. China
- MA, Shandong Univ.
- "Your solution, looks excellent. I recognize things from previous chapters. I have seen the standard deviation formula you used to get 5.154. I do understand the Central Limit Theorem needs the sample size (n) to be greater than 30, we have 100. I do understand the sample mean(s) of the population will follow a normal distribution, and that CLT states the sample mean of population is the population (mean), we have 143.74. But when and WHY do we use the standard deviation formula where you got 5.154. WHEN & Why use standard deviation of the sample mean. I don't understand, why don't we simply use the "100" I understand that standard deviation is the square root of variance. I do understand that the variance is the square of the differences of each sample data value minus the mean. But somehow, why not use 100, why use standard deviation of sample mean? Please help explain."
- "excellent work"
- "Thank you so much for all of your help!!! I will be posting another assignment. Please let me know (once posted), if the credits I'm offering is enough or you ! Thanks again!"
- "Thank you"
- "Thank you very much for your valuable time and assistance!"
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