Question

NASA experiment A NASA study compared two methods of determining white blood cell counts in laboratory animals. Table lists results for 42 paired observations. Calculate delta values for each observation and plot these differences as a stemplot. Based on this plot, do you think the methods are interchangeable? Table Data for Exercise White blood cells counts (x1000 dL) by Celdyne method and Unopett method, n-42. CELDYNE UNOPETT CELDYNE UNOPETT 3.0 11.0 12.7 8.1 15.7 9.1 11.3 9.5 7.4 7.9 13.1 10.3 113 11.5 118 2.5 7.5 9.0 9.9

Answer 1

NASA Experiment The stemplot for both the methods are shown below.

Stemplot for CELDYNE											Stemplot for UNOPETT
Stem	Leaf										Stem	Leaf
3	6										2	6
4											3
5	6	7									4
6	1	3									5
7	4	8	9								6	0	3	5	8
8	2	3	8								7	4	5	8	8
9	1	4	5	5	7	8					8	1	4	4	6	8
10	3	3	4	4	4	5	8	9			9	0	1	3	3	3	5	5	6	6	9
11	3	4	5	8							10	3	4	4	5	6
12	5	7	7	9							11	0	0	3	3
13	0	1	6	8							12	3
14	0	1	4								13	0	8	5	7	9
15											14
16	0										15
17											16
18											17
19											18	0
20											19
21											20	4
22											21
23	9										22
											23
											24
											25
											26
											27	6

The back to back stemplot is shown below,

Stemplot for CELDYNE										Stemplot for UNOPETT
										Stem	Leaf
										2	6
									6	3
										4
								7	6	5
								3	1	6	0	3	5	8
							9	8	4	7	4	5	8	8
							8	3	2	8	1	4	4	6	8
				8	7	5	5	4	1	9	0	1	3	3	3	5	5	6	6	9
		9	8	5	4	4	4	3	3	10	3	4	4	5	6
						8	5	4	3	11	0	0	3	3
						9	7	7	5	12	3
						8	6	1	0	13	0	8	5	7	9
							4	1	0	14
										15
									0	16
										17
										18	0
										19
										20	4
										21
										22
									9	23
										24
										25
										26
										27	6

From both the stem plot we can see that, the distribution for both the methods are approximately same hence both the methods can be interchangeable.

2)

a)

The difference in quality of life score are,

QOL_BASE	QOL_3MO	Absolute Difference
2	1	1
4	1	3
3	1	2
4	3	1
5	2	3
6	2	4
4	2	2
4	5	1
3	3	0
3	1	2

b)

Stem	Leaf
0	0	1	1	1	2	2	3	3	4

c)

The t test is performed for the two dependent means,

$t=\frac{\left ( \sum D \right )/N}{\sqrt{\frac{\sum D^2-\left ( \frac{\left (\sum D \right )^2}{}N \right )}{N\left ( N-1 \right )}}}$

QOL_BASE	QOL_3MO	Difference, D	D^2
2	1	1	1
4	1	3	9
3	1	2	4
4	3	1	1
5	2	3	9
6	2	4	16
4	2	2	4
4	5	-1	1
3	3	0	0
3	1	2	4
	Sum	17	49
	Mean	1.7
	Standard Deviation	1.4944341

$t=\frac{\left (17 \right )/10}{\sqrt{\frac{49-\left ( \frac{\left (17 \right )^2}{10} \right )}{10\left ( 10-1 \right )}}}$

$t=3.59726\approx 3.60$

The P-value is obtained using the t distribution table for t = 3.60 and df = 9

$\text{P-value=0.0058}$

$\text{P-value=0.0058}<\alpha=0.05$

Since the P-value is less than 0.05 at 5% significance level, the null hypothesis is rejected. Now we can conclude that there is statistically significant difference between two groups means.