The wattle thickness (in millimeters) of 15 randomly selected chickens was measured before and after treatment with phytohemagglutinin (PHA). Does treatment with PHA increase wattle thickness

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