## Question 920:

1

No answer provided yet.For this question we need to use the 1-sample t-test. We are assuming that the service times are approximately normally distributed.

1. The Null Hypothesis is that the mean service time is 3.5 minutes.
2. First we generate the test statistic which is the difference between the sample mean and test mean : 3.5-3.4 =.1 dividing by the standard error of the mean (SEM).
3. The SEM is the standard deviation divided by the square root of the sample size : .04/SQRT(49) = .0057.
4. So the test statistic is : .1/.0057 = 17.50
5. Now we want to know the probability of obtaining a value 17.50 standard errors from the mean. We can look this value up in a t-table based on our sample size (49-1 =48 degrees of freedom). Most t-tables only go to 3 or 4 standard errors from the mean so we can also use the excel function = TDIST(17.50,48,2) and we get a p-value of less than .00001.
6. This means it is highly unlikely we would observe such a difference given this sample size if the mean really were 3.5 minutes.
7. We reject the Null Hypothesis because the p-value is very low and conclude the new ordering system improves service times.