The post anaesthesia care area (recovery room) at St. Luke's Hospital in Pflugerville, Ontario, was recently enlarged. The hope was that with the enlargement the mean number of patients per day would be more than 25. A random sample of 15 days from an approximately normal population revealed the following numbers of patients:
25 27 25 26 25 28 28 27 24 26 25 29 25 27 24
a) What is the null hypothesis?
b) What is the alternative hypothesis?
c) What test will you use to test the alternative hypothesis?
d) Find the p-value and interpret its meaning.
e) At the ? = 0.05 significance level, can we conclude that the mean number of patients per day is more than 25? From Statdisk 95% C/I 25.273 < mean< 26.916
f) Is this a right-tail test, left-tail test, or two-tail test?
The solution provides step by step method for the calculation of testing of hypothesis. Formula for the calculation and Interpretations of the results are also included. Interactive excel sheet is included. The user can edit the inputs and obtain the complete results for a new set of data.
Franklin Park Mall Hypothesis Testing
The owners of the Franklin Park Mall wished to study customer shopping habits. From earlier
studies the owners are under the impression that a typical shopper spends 0.75 hours at the mall,
with a standard deviation of 0.10 hours. Recently the mall owners added some specialty restaurants designed to keep shoppers in the mall longer. The consulting firm, Brunner and Swanson Marketing Enterprises, has been hired to evaluate the effects of the restaurants. A sample of 45 shoppers by Brunner and Swanson revealed that the mean time spent in the mall had increased to 0.80 hours.
a. Develop a test of hypothesis to determine if the mean time spent in the mall is more than
0.75 hours. Use the .05 significance level.
b. Suppose the mean shopping time actually increased from 0.75 hours to 0.77 hours. What is
the probability this increase would not be detected?
c. When Brunner and Swanson reported the information in part (b) to the mall owners, the owners
were upset with the statement that a survey could not detect a change from 0.75 to 0.77
hours of shopping time. How could this probability be reduced?