 I'm thinking I may have to adjust my staffing schedule at our outlet store on the first floor. I know I need to have more checkout staff on Fridays and Saturdays. I'm wondering if the need for checkout staff is constant across the rest of the week. I collect a data for the number of transactions each day for the last two months. Here is my data. I could run multiple t-tests to see if the mean number of transactions differ across the days. But that would be a lot of tests to run. Instead, I am going to run an ANOVA, an analysis of variance, to see that the means are statistically the same. Or, another way to think about this, is that the test will determine if the daily samples came from the same population. My null hypothesis is that all the means are the same. The alternative hypothesis is that at least one mean is different. Because the consequences of making either a type 1 or a type 2 error are not severe, I will use a 5% level of significance. Although I know how to do this using my calculator, I am going to do it using Excel. Okay, here is my data in an Excel worksheet. The green area is how I capture the data and is how most data files are arranged. I have one record per row, and the values are the three variables in separate columns. But to use the data analysis tool pack in Excel to perform the ANOVA, you have to rearrange the data into the format in the blue section. You need one column for each of the groups. Here the groups are the days. And there must be no gaps between the columns or rows. Notice, you do not have to have a matched set of data. Here, I have 9 weeks of data for Monday and Tuesday, but only 8 for Wednesday and Thursday. Left click on the data ribbon, and go to the right side to find the data analysis tool. If you do not see the data analysis option, you will need to activate the package in your Excel. Left click on data analysis to open the tool dialog box. Scroll until you see the ANOVA, single factor option. I have just one factor, the number of sales transactions. Left click on ANOVA, single factor, and select OK. In the dialog box that opens, click inside the input range window on the up arrow. Select the complete range of data, including the empty cells. Click on the down arrow to go back to the dialog box. Click on grouped by columns. Check the labels in the first row. If you do not have labels, the names of your variables in your data, leave this unchecked. Enter your level of significance, alpha, if it differs from the default 5%. I want the output on this worksheet, so I am clicking in the output options to put the output beginning in cell J1. Then click OK. In the output, I can see the F statistic of 3.24 is greater than the critical value of 2.92, which tells me at least one mean is different. The p-value of 0.035 confirms that there is at least one mean that is different because it is less than my alpha of 5%. Although I could now run t-tests to confirm which mean is different. I think I can just go with my gut and conclude that Monday's average is substantially higher than the other three days. So, I will consider adding more checkout staff on that day. ANOVA is a very useful tool to have, in your manager's belt. Sue, what did you decide to do to reduce the long lines at our checkout stations? Joe, I analyzed the data using a single factor ANOVA and decided that Monday was the real problem day.