A dog food company is developing a new dog food for mature animals. The goal of
ID: 2921380 • Letter: A
Question
A dog food company is developing a new dog food for mature animals. The goal of the new product is to maintain the body fat level of dogs within a healthy level, while at the same time meeting all of their nutritional needs. Twenty dogs were assigned randomly to either the control diet or the new diet. The response variable measured was body fat (lbs) adjusted to a dog weight of 50 lbs.
After feeding the dogs their respective diets for 90 days, the control group had a mean body fat content of 10.9 lbs. with a standard deviation of 2.8, while the dogs that received the new diet had a mean body fat content of 10.0 with a standard deviation of 3.5.
At alpha=0.05 is there sufficient statistical evidence to say that the two dog foods differ in their ability to control fat?
Explanation / Answer
Given that,
mean(x)=10.9
standard deviation , 1 =2.8
number(n1)=20
y(mean)=10
standard deviation, 2 =3.5
number(n2)=20
null, Ho: u1 = u2
alternate,two dog foods differ in their ability to control fat H1: 1 != u2
level of significance, = 0.05
from standard normal table, two tailed z /2 =1.96
since our test is two-tailed
reject Ho, if zo < -1.96 OR if zo > 1.96
we use test statistic (z) = (x-y)/sqrt(s.d1^2/n1)+(s.d2^2/n2)
zo=10.9-10/sqrt((7.84/20)+(12.25/20))
zo =0.9
| zo | =0.9
critical value
the value of |z | at los 0.05% is 1.96
we got |zo | =0.898 & | z | =1.96
make decision
hence value of | zo | < | z | and here we do not reject Ho
p-value: two tailed ( double the one tail ) - Ha : ( p != 0.9 ) = 0
hence value of p0.05 > 0,here we reject Ho
ANSWERS
---------------
null, Ho: u1 = u2
alternate, H1: 1 != u2
test statistic: 0.9
critical value: -1.96 , 1.96
decision: do not reject Ho
p-value: 0
we have evidence that two dog foods are similar in their ability to control fat
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