5 treats divided by six groups of one dog, or in easier terms, five dog treats divided up for 6 dogs. As a fraction, just remove the words except for 5, divided by, and 6, to give you 5/6. Hope this helped.
Answer:
We cannot say that the mean wake time are different before and after the treatment, with 98% certainty. So the zopiclone doesn't appear to be effective.
Step-by-step explanation:
The goal of this analysis is to determine if the mean wake time before the treatment is statistically significant. The question informed us the mean wake time before and after the treatment, the number of subjects and the standard deviation of the sample after treatment. So using the formula, we can calculate the confidence interval as following:
![IC[\mu ; 98\%] = \overline{y} \pm t_{0.99,n-1}\sqrt{\frac{Var(y)}{n}}](https://tex.z-dn.net/?f=IC%5B%5Cmu%20%3B%2098%5C%25%5D%20%3D%20%5Coverline%7By%7D%20%5Cpm%20t_%7B0.99%2Cn-1%7D%5Csqrt%7B%5Cfrac%7BVar%28y%29%7D%7Bn%7D%7D)
Knowing that
:
![IC[\mu ; 98\%] = 98.9 \pm 2.602\frac{42.3}{4} \Rightarrow 98.9 \pm 27.516](https://tex.z-dn.net/?f=IC%5B%5Cmu%20%3B%2098%5C%25%5D%20%3D%2098.9%20%5Cpm%202.602%5Cfrac%7B42.3%7D%7B4%7D%20%5CRightarrow%2098.9%20%5Cpm%2027.516)
![IC[\mu ; 98\%] = [71.387 ; 126,416]](https://tex.z-dn.net/?f=IC%5B%5Cmu%20%3B%2098%5C%25%5D%20%3D%20%5B71.387%20%3B%20126%2C416%5D)
Note that
so we cannot say, with 98% confidence, that the mean wake time before treatment is different than the mean wake time after treatment. So the zopiclone doesn't appear to be effective.
Answer:
0.5
Step-by-step explanation: