Respuesta :
Answer:
a) 19.9
b) 1.1099
c) 1232
Step-by-step explanation:
We are given the following in the question:
21, 23, 27, 19, 17, 18, 20, 15, 17, 22
a) estimate for the average weight
[tex]Mean = \displaystyle\frac{\text{Sum of all observations}}{\text{Total number of observation}}[/tex]
[tex]Mean =\displaystyle\frac{199}{10} = 19.9[/tex]
b) standard error of this estimate
Sum of squares of differences = 110.9
[tex]S.D = \sqrt{\dfrac{110.9}{9}} = 3.51[/tex]
Standard Error =
[tex]= \dfrac{s}{\sqrt{n}} = \dfrac{3.51}{\sqrt{10}} = 1.1099[/tex]
c) Sample size for a standard error of 0.1
[tex]0.1 = \dfrac{3.51}{\sqrt{n}} \\\\n = (\dfrac{3.51}{0.1})^2\\\\n = 1232.01[/tex]
Thus, the sample size must be approximately 1232 for a standard error of 0.1.
(a). The average weight of mouse is 19.9
(b). The standard error is 1.1099
(c). To reduce the standard error to 0.1 , number of mice need is 1232
Average of data:
The average weight of mouse is computed as divide sum of data by total number of data.
[tex]Average=\frac{21+23+27+19+17+18+20+15+17+22}{10} \\\\Average=\frac{199}{10}=19.9[/tex]
Standard deviation is,
[tex]S.D=\sqrt{\frac{110.9}{9} }=3.51\\ \\Error=\frac{S.D}{\sqrt{n} }=\frac{3.51}{\sqrt{10} }=1.1099[/tex]
Sample size is,
[tex]0.1=\frac{3.51}{\sqrt{n} } \\\\n=1232[/tex]
Learn more about the standard deviation here:
https://brainly.com/question/475676
