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

the answer is 200

step-by-step explanation:

the way the question is worded, it seems to imply the $264 interest is simple interest, not compound interest.

the simple interest for 2 years is $264; the simple interest for 1 year is $132.

$132 interest on a $600 investment means an interest rate of (132/600)*100 = 22%.

the value of $200 at 22% interest compounded annually for three years is

200%281.22%29%5e3

the amount of interest is that value, minus the original $200.

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