Egg Production To increase egg production, a farmer decided to increase the amount of time the lights in his hen house were on. Ten hens were randomly selected, and the number of eggs each produced was recorded. After one week of lengthened light time, the same hens were monitored again. The data are given here. At α = 0.05, can it be concluded that the increased light time increased egg production? Hen 1 2 3 4 5 6 7 8 9 10
Before 4 3 8 7 6 4 9 7 6 5
After 6 5 9 7 4 5 10 6 9 6

Step-by-step explanation:

Corresponding egg production before and after the light was put on form matched pairs.

The data for the test are the differences between the egg production before and after the light was put on.

μd = egg production before the light was put on minus egg production after the light was put on.

Before after diff

4 6 - 2

3 5 - 2

8 9 - 1

7 7 0

6 4 2

4 5 - 1

9 10 - 1

7 6 - 1

6 9 - 3

5 6 - 1

Sample mean, xd

= (- 2 - 2 - 1 + 0 + 2 - 1 - 1 - 1 - 3 - 1)/10 =

- 1

xd = - 1

Standard deviation = √(summation(x - mean)²/n

n = 10

Summation(x - mean)² = (- 2 + 1)^2 + (- 2 + 1)^2 + (- 1 + 1)^2 + (0 + 1)^2 + (2 + 1)^2 + (- 1 + 1)^2 + (- 1 + 1)^2 + (- 1 + 1)^2 + (- 3 + 1)^2 + (- 1 + 1)^2 = 16

Standard deviation = √(16/10

sd = 1.26

For the null hypothesis

H0: μd ≥ 0

For the alternative hypothesis

H1: μd < 0

The distribution is a students t. Therefore, degree of freedom, df = n - 1 = 10 - 1 = 9

The formula for determining the test statistic is

t = (xd - μd)/(sd/√n)

t = (- 1 - 0)/(1.26/√10)

t = - 2.51

We would determine the probability value by using the t test calculator.

p = 0.017

Since alpha, 0.05 > than the p value, 0.017, then we would reject the null hypothesis. Therefore, at 5% level of significance, it cannot be concluded that the increased light time increased egg production.

three-fourths of an hour

36÷48=0.75

0.75=