Answer:
E.Yes, because the sample is large enough to satisfy the normality conditions
Step-by-step explanation:
Given that a random sample of 100 residents was selected, and 97 of those selected indicated that they were in favor of the proposal then;
The percent in favor of the proposal is 97/100 *100 =97%
Thus from this information ;
p=0.97 and q=1-p =1-0.97 =0.03 where p=population proportion
Finding mean and standard deviation will be;
μp =mean of sample proportion
μp=p= 0.97
δp= standard deviation of sample proportion
δp=√{pq/n }= √{(0.97*0.03)/100} where n=100 (sample size)
δp= 0.0171
3δp = 3*0.0171 =0.051
check using ;
{p- 3δp, p+3δp} = { 0.97-0.05,0.97+0.05} ={0.92,1.02}
compare if {0.92,1.02} lies in {0,1)
A sample is large enough if the interval {p- 3δp, p+3δp} lies wholly within the interval {0,1}
This sample wholly lie in the interval of {0,1} thus it is safe to assume p' is approximately normally distributed
