Question

Many people believe that the moon influences the actions of some individuals. A study of dementia patients in nursing homes recorded various types of disruptive behaviors every day for 12 weeks. Days were classified as moon days if they were in a three-day period centered at the day of the full moon. For each patient the average number of disruptive behaviors was computed for moon days and for all other days. The data for the 15 subjects whose behaviors were classified as aggressive are presented in the table below. The patients in this study are not a random sample of dementia patients. However, we examine their data in the hope that what we find is not unique to this particular group of individuals and applies to other patients who have similar characteristics.

Patient	Moon days	Other days	Difference	Patient	Moon days	Other days	Difference
1	3.34	0.27	3.07	9	6.01	1.59	4.42
2	3.67	0.59	3.08	10	4.34	0.60	3.74
3	2.68	0.32	2.36	11	3.32	0.65	2.67
4	3.33	0.19	3.14	12	0.68	0.69	-0.01
5	3.33	1.26	2.07	13	1.32	1.26	0.06
6	3.66	0.11	3.55	14	0.33	0.23	0.10
7	4.67	0.30	4.37	15	2.00	0.38	1.62
8	2.67	0.40	2.27

To analyze these paired data, we first subtract the disruptive behaviors for other days from the disruptive behaviors for moon days. These 15 differences form a single sample. They appear in the "Difference" columns in the table. The first patient, for example, averaged 3.34 aggressive behaviors on moon days but only 0.27 aggressive behaviors on other days. The difference

3.34 − 0.27 = 3.07

is what we will use in our analysis.

Next, we examine the distribution of these differences. Below is a stemplot of the differences. This plot indicates that there are three patients with very small differences, but there are no indications of extreme outliers or strong skewness. We will proceed with our analysis using the Normality-basedmethods of this section. (Round each value to one decimal place. Enter numbers from smallest to largest separated by spaces. Enter NONE for stems with no values.)

The 15 differences have the following sample mean and sample standard deviation. (Round your answers to three decimal places.)

Answer 1

data in ascending order when rounded of to 1 decimal place :

0.0 0.0 0.1 1.6 2.0 2.2 2.3 2.6 3.0 3.0 3.1 3.5 3.7 4.3 4.4

we see the above stem and leaf plot satisfy the result given in the question.

For sample mean and sample standard deviation, in minitab go to stat -> basic statistics -> display descriptive statistics. select the required measures and click ok.

mean	st dev	median
2.434	1.461	2.67