## Question 920:

Asked on January 13, 2012

Tags:
1-sample t-test

1

## Answer:

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.

- The Null Hypothesis is that the mean service time is 3.5 minutes.

- 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).
- The SEM is the standard deviation divided by the square root of the sample size : .04/SQRT(49) = .0057.
- So the test statistic is : .1/.0057 = 17.50
- 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.

- This means it is highly unlikely we would observe such a difference given this sample size if the mean really were 3.5 minutes.
- We reject the Null Hypothesis because the p-value is very low and conclude the new ordering system improves service times.