Respuesta :
Answer:
Step-by-step explanation:
The question is incomplete. The complete question is:
The wattle thickness (in milimeters) of 15 randomly selected chickens was measured before and after treatment with PHA. Does treatment wih PHA increase wattle thickness?
Chicken Number // Pretreatment // Posttreatment
1 // 1.05 // 3.48
2 // 1.01 // 5.02
3 // 0.78 // 5.37
4 // 0.98 // 5.45
5 // 0.81 // 5.37
6 // 0.95 // 3.92
7 // 1.00 // 6.54
8 // 0.83 // 3.42
9 // 0.78 // 3.72
10 // 1.05 // 3.25
11 // 1.04 // 3.66
12 // 1.03 // 3.12
13 // 0.95 // 4.22
14 // 1.46 // 2.53
15 // 0.78 // 4.39
Solution:
Corresponding wattle thickness before and after treatment form matched pairs.
The data for the test are the differences between the wattle thickness at pretreatment and posttreatment.
μd = wattle thickness at pretreatment minus wattle thickness at posttreatment
Pretreatment. Posttreatment diff
1.05 3.48 -2.43
1.01 5.02 -4.01
0.78 5.37 -4.59
0.98 5.45 -4.47
0.81 5.37 -4.56
0.95 3.92 -2.97
1 6.54 -5.54
0.83 3.42 -2.59
0.78 3.72 -2.94
1.05 3.25 -2.2
1.04 3.66 -2.62
1.03 3.12 -2.09
0.95 4.22 -3.27
1.46 2.53 -1.07
0.78 4.39 -3.61
Sample mean, xd
= (-2.43 - 4.01 - 4.59 - 4.47 - 4.56 - 2.97 - 5.54 - 2.59 - 2.94 - 2.2 - 2.62 - 2.09 - 3.27 - 1.07 - 3.61)/15 = - 3.264
xd = - 3.264
Standard deviation = √(summation(x - mean)²/n
n = 15
Summation(x - mean)² = (- 2.43 + 3.264)^2 + (-4.01 - 3.264)^2 + (-4.59 - 3.264)^2+ (-4.47 - 3.264)^2 + (-4.56 - 3.264)^2 + (-2.97 - 3.264)^2 + (-5.54 - 3.264)^2 + (-2.59 - 3.264)^2 + (-2.94 - 3.264)^2 + (-2.2 - 3.264)^2 + (-2.62 - 3.264)^2 + (-2.09 - 3.264)^2 + (-3.27 - 3.264)^2 + (-1.07 - 3.264)^2 + (-3.61 - 3.264)^2 = 627.32444
Standard deviation = √(627.32444/15
sd = 6.47
For the null hypothesis
H0: μd ≥ 0
For the alternative hypothesis
H1: μd < 0
1) The distribution is a students t. Therefore, degree of freedom, df = n - 1 = 15 - 1 = 14
The formula for determining the test statistic is
t = (xd - μd)/(sd/√n)
t = (- 3.264 - 0)/(6.47/√15)
t = - 1.95
We would determine the probability value by using the t test calculator.
p = 0.036
Assume alpha = 0.05
Since alpha, 0.05 > than the p value, 0.036, then we would reject the null hypothesis. Therefore, we can conclude that at a significance level of 5%, treatment with PHA increase wattle thickness
