Question

An agronomist wishes to estimate, to within one percentage point, the proportion of a new variety of seed that will germinate when planted, with 95% confidence. A typical germination rate is 97%. Estimate the minimum size sample required.

Answer 1

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

$\hat{p} = 0.97$

Margin of error = 1% = 0.01

We have to find the minimum sample size for a 95% confidence interval.

Formula for margin of error:

$z_{stat}\times \sqrt{\dfrac{\hat{p}(1-\hat{p})}{n}}$

$z_{critical}\text{ at}~\alpha_{0.05} = 1.96$

Putting values, we get,

$1.96\times \sqrt{\dfrac{0.97(1-0.97)}{n}}\leq 0.01\\\\\sqrt{n} \geq 1.96\times \dfrac{\sqrt{0.97(1-0.97)}}{0.01}\\\\\sqrt{n}\geq 33.4350\\\Rightarrow n \geq 1117.9056$

Rounding off to integer,

$n = 1118$

Thus, the minimum sample size must be 1118 to have margin of error within one percentage point.