Answer:
Explanation:
Dr. Chung, a renowned nutritionist, has consistently proclaimed the benefits of a balanced breakfast. To substantiate her claim, she asks her participants to go without breakfast for one week. The following week he asks the same participants to make sure they eat a complete breakfast. Following each week, Dr. Chung asks the supervisors of each participant to rate their productivity for that week.
Participant      Performance with Breakfast(X)    Performance Without Breakfast
(Y)
#1 Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â 8 Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â 6
#2 Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â 6 Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â 6
#3 Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â 8 Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â 5
#4 Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â 8 Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â 5
Using α = .05, do a one-tailed test of the doctor's hypothesis.
What is your T=
Performance with breakfast
mean, X = (8+6+8+8)/4 = 7.5
using the formular, [tex]S = \sqrt{\frac{sum(xi-X)^{2} }{N-1} }[/tex]
standard deviation, S =√[ [(8-7.5)² + (6-7.5)² + (8-7.5)² + (8-7.5)²]/(4-1)]
S = 1
Number of sample, N = 4
Performance without breakfast
mean, x = (6+5+6+5)/4 = 5.5
Using the formula as above, standard deviation Â
[tex]s = \sqrt{\frac{sum(xi-x)^{2} }{n-1} }[/tex]
s =√[ [(6-5.5)² + (6-5.5)² + (5-5.5)² + (5-5.5)²]/4-1]
s = 0.58
n = 4
T is given as follows
[tex]T = \frac{X- x}{\sqrt({\frac{S^{2} }{N} }+\frac{s^{2} }{n} )}[/tex]
T = [(X - x)/[(S²/N) + (s²/n)]]
T = [(7.5 - 5.5)/[(1²/4) + (0.58²/4)]]
 T = 3.46
