Answer:
The minimum sample size must be 1118 to have margin of error within one percentage point.
Step-by-step explanation:
We are given the following in the question:
Germination rate = 97% = 0.97
[tex]\hat{p} = 0.97[/tex]
Margin of error = 1% = 0.01
We have to find the minimum sample size for a 95% confidence interval.
Formula for margin of error:
[tex]z_{stat}\times \sqrt{\dfrac{\hat{p}(1-\hat{p})}{n}}[/tex]
[tex]z_{critical}\text{ at}~\alpha_{0.05} = 1.96[/tex]
Putting values, we get,
[tex]1.96\times \sqrt{\dfrac{0.97(1-0.97)}{n}}\leq 0.01\\\\\sqrt{n} \geq 1.96\times \dfrac{\sqrt{0.97(1-0.97)}}{0.01}\\\\\sqrt{n}\geq 33.4350\\\Rightarrow n \geq 1117.9056[/tex]
Rounding off to integer,
[tex]n = 1118[/tex]
Thus, the minimum sample size must be 1118 to have margin of error within one percentage point.
