called previously about their bills is now no larger than $20.00. Other bank managers at
the meeting suggested that this statement may be in error and that it might be
worthwhile to conduct a test to see if there is statistical support for the call center
manager’s statement. The file called Bank Call Center contains data for a random
sample of 67 customers from the call center population. Assuming that the population
standard deviation for past due amounts is known to be $60.00, what should be
concluded based on the sample data? Test using α = 0.10.
A) Because p-value = 0.4121 > alpha = 0.10, we do not reject the null hypothesis.
The sample data do not provide sufficient evidence to reject the call center manager’s
statement that the mean past due amount is $20.00 or less.
B) Because p-value = 0.4121 > alpha = 0.10, we reject the null hypothesis.
The sample data provide sufficient evidence to reject the call center manager’s
statement that the mean past due amount is $20.00 or less.
C) Because p-value = 0.2546 > alpha = 0.10, we do not reject the null hypothesis.
The sample data do not provide sufficient evidence to reject the call center manager’s
statement that the mean past due amount is $20.00 or less.
D) Because p-value = 0.2546 > alpha = 0.10, we reject the null hypothesis.
The sample data provide sufficient evidence to reject the call center manager’s
statement that the mean past due amount is $20.00 or less.
Damage to homes caused by burst piping can be expensive to repair. By the time the
leak is discovered, hundreds of gallons of water may have already flooded the home.
Automatic shutoff valves can prevent extensive water damage from plumbing failures.
The valves contain sensors that cut off water flow in the event of a leak, thereby
preventing flooding. One important characteristic is the time (in milliseconds) required
for the sensor to detect the water leak. Sample data obtained for four different shutoff
valves are contained in the file entitled Waterflow.
Use the Tukey-Kramer multiple comparison technique to discover any differences in
the average detection time. Use a significance level of 0.05.