In: Statistics and Probability
Wildlife biologists inspect 150 deer taken by hunters and find 26 of them carrying ticks that test positive for Lyme disease.
a.) Create a 90% confidence interval for the percentage of deer that may carry such ticks.( __%,__%)
b.) If the scientists want to cut the margin of error in half, how many deer must they inspect?
c.) What concerns do you have about this sample?
| sample success x = | 26 | |
| sample size n= | 150 | |
| sample proportion p̂ =x/n= | 0.1730 | |
| std error se= √(p*(1-p)/n) = | 0.0309 | |
| for 90 % CI value of z= | 1.645 | |
| margin of error E=z*std error = | 0.0508 | |
| lower bound=p̂ -E = | 0.1222 | |
| Upper bound=p̂ +E = | 0.2238 | |
| from above 90% confidence interval for population proportion =12.22% , 22.38 % |
b)
since margin of error is inversly proportion to square root of sample size
therefore required sample size =4*150 =600
c)
since above is a sample of deer taken by hunters , this might not represent the population of deer in focus and may give biased results