## Question 502:

1

The sample taken of 36 administrative staff is reasonably large and above 30 which suggests we can use the standard normal deviate (z-score) to construct the confidence interval, this however would be less conservative. The high precision of the interval (99%) suggests we should take a more conservative approach and construct the interval using the critical value from the t-distribution since it takes into account the size of the sample.  The data here is cycle-time which has a tendency to be not normally distributed and can affect the accuracy of the interval. There is not indication from the wording of the problem that there is a departure from normality.

a. To construct the interval, we first build the standard error of the mean:

1. The standard error of the mean is made up of the standard deviation divided by the square root of the sample size or 5.1/ SQRT(36) = 5.1/6 =  .85
2. We now generate the margin of error times which is the standard error of the mean times the critical value from the t distribution. To find this value we can look it up in a t-table for an alpha of .01 (1-.99) with 35 degrees of freedom or use the Excel function = TINV(.01,35) and get the value 2.72  The margin of error is then .85*2.72 = 2.312
3. Finally, to construct the interval, we add and subtract the margin of error to the mean 97.4- 2.312 and 97.4+2.312 = the 99% confidence interval for the mean is between (95.09 and 99.71 minutes).  This confidence interval represents the most likely interval for the unknown population mean. The population is all administrative staff. If we were to take repeated samples of all administrative staff on average we'd expect the mean to be contained in this interval 99 out of 100 times. Since this is a 2-sided interval, it is unlikely the true mean is greater than 99.7 or less than 95.09, we'd expect to see a mean above or below these intervals about 0.5% of the time (half of 1 percent on each end).

b.  We want to know if the staff are taking LONGER than the 90 minutes so this is a one-sided hypothesis test.

1. The null hypothesis (Ho) is that the staff are working less than or equal to 90 minutes without take a break.
2. The alternative hypothesis is the staffs are taking longer than 90 minutes without a break.
3. We can conduct a 1-sample 1-sided t-test given the mean, sample size and standard deviation. We will reject the null hypothesis if there is less than a 1% chance the difference from 90 minutes we observed was due to chance fluctuations in the sample.
4. The test statistic is derived by subtracting the observed mean from the criteria mean of 90 minutes (97.4-90) = 7.4 minutes.
5. The standard error of the mean is the standard deviation divided by the square root of the sample size = 5.1/SQRT(36) = 5.1/6= .85
6. We now divide the difference by this standard error of the mean 7.4/.85 = 8.7
7. Now we lookup the critical value of t of 8.7 with 35 degrees of freedom for a 1-tailed test using the excel function =TDIST(8.7,35,1). The p value is less than .000001, in other words, the probability the difference of 7.4 minutes was due to chance alone is a fraction of a percent.
8. Or cut-off value of .01 was far exceeded allowing us to reject the null hypothesis and say there is sufficient evidence that admins are not taking a break in 90 minutes or less.
9. A final note, we could have also take the more conservative approach and done a 2 tailed t-test. The difference was so large that even this is statistically significant (p < .000001). We chose to use a 1-tailed test since the question specifically asked about longer than 90 minutes as opposed to greater or less than 90 minutes.  Also note that the boundary of the 99% confidence interval from part 2a also suggested there would be sufficient evidence to reject the null hypothesis since its lower bounds of 95.09 were still ABOVE 90 minutes.