Question

(c) Actually, the 10,000 larvae in each group came from a series of 50 different runs of the experiment, with 200 larvae in each group for each run. The researchers believe that conditions differ from run to run, and thus it makes sense to treat each run as a case (rather than each fly). In this analysis, we are looking at paired data, and the response variable would be the difference in the number of larvae surviving between the choice group and the no choice group, for each of the 50 runs. The counts (Choice and NoChoice and difference (Choice − NoChoice) in number of surviving larva are stored in MateChoice. Using the single variable of differences, calculate the p-value for testing whether the average difference is greater than 0

Choice	NoChoice	Difference
114	131	-17
126	136	-10
140	130	10
121	122	-1
123	130	-7
123	144	-21
137	133	4
114	124	-10
118	110	8
123	135	-12
116	128	-12
128	132	-4
141	145	-4
131	142	-11
125	179	-54
138	115	23
111	140	-29
128	135	-7
126	142	-16
111	105	6
122	124	-2
123	111	12
142	123	19
127	98	29
132	135	-3
133	112	21
98	106	-8
117	103	14
98	116	-18
132	96	36
104	116	-12
125	114	11
118	114	4
117	108	9
116	104	12
110	114	-4
132	111	21
106	101	5
110	115	-5
139	131	8
101	99	2
125	110	15
117	112	5
118	111	7
116	109	7
105	95	10
129	120	9
124	106	18
113	92	21
124	112	12

Answer 1

here claim is the average difference is greater than 0

where difference (d )= (Choice − NoChoice)

so the null and alternative hypothesis is

Ho:ud =0

Ha:ud >0 (one tailed)

using minitab>stat>basic stat>paired t

we have

Paired T-Test and CI: Choice, NoChoice

Paired T for Choice - NoChoice

N Mean StDev SE Mean
Choice 50 121.34 11.05 1.56
NoChoice 50 119.52 16.51 2.33
Difference 50 1.82 15.82 2.24

95% lower bound for mean difference: -1.93
T-Test of mean difference = 0 (vs > 0): T-Value = 0.81 P-Value = 0.210

the p-value for testing whether the average difference is greater than 0 is 0.210