Answer:
a) [6.741, 17.259]
b) 2,305
Step-by-step explanation:
a)
Let p be the proportion of underweight rats and n the sample size.
n=30
p= 12/30 = 0.4
Since, both np>10 and n(1-p)>10 and the sample size = 30 is big enough, we can approximate the distribution with a Normal curve of mean 12 and standard deviation
[tex]\large s =\sqrt{np(1-p)}=\sqrt{30*0.4*0.6}=2.683[/tex]
We then find the values a, b such that the area of the curve between [a, b] equals 95%=0.95.
That values can be easily worked out with a spreadsheet and we have the 95% confidence interval [6.741, 17.259]
See picture attached
b)
Using Simple Random Sampling in an infinite population (this is such a large population that we do not know the exact number) we have that the sample size should be the nearest integer to
[tex]\large \frac{Z^2p(1-p)}{e^2}[/tex]
where
Z= the z-score corresponding to the confidence level, in this case 95%, so Z=1.96 (this means that the area under the Normal N(0,1) between [-1.96,1.96] is 95%=0.95)
p= the proportion of underweight rats = 0.4
q = 1-p = 0.6
e = the error proportion = 0.02
Crunching the numbers
[tex]\large \frac{Z^2pq}{e^2}=\frac{(1.96)^2*0.4*0.6}{(0.02)^2}=2,304.96\approx 2,305[/tex]
The sample size should be at least 2,305.
