Question

Corn: In a prior sample of corn, farmer Carl finds that 14% of the sample has worms but the margin of error for his population estimate was too large. He wants to set the margin of error to be 1.5% at the 95% confidence level. Enter your answers as whole numbers. What is the minimum sample size required for this type of accuracy? What is the minimum sample size needed if we have no prior knowledge of the sample proportion

Answer 1

The minimum sample size required to obtain this type of accuracy is 740

What is the margin of error for large samples?

Suppose that we have:

Sample size n > 30

Sample standard deviation = s

Population standard deviation = $\sigma$

Level of significance = $\alpha$

Then the margin of error(MOE) is obtained as

Case 1: Population standard deviation is known

Margin of Error = MOE = $Z_{\alpha/2}\dfrac{\sigma}{\sqrt{n}}$

Case 2: Population standard deviation is unknown

MOE = $Z_{\alpha/2}\dfrac{s}{\sqrt{n}}$

where Z is the critical value of the test statistic at the level of significance

As per the given, we have

The prior estimate of population proportion:

Margin or error : 1.5% = 0.015

Critical value for 95% confidence = 1.96

Formula to find the sample size:-

$Z_{\alpha/2}\dfrac{\sigma}{\sqrt{n}}$

= $n = p(1-p) \dfrac{Z_{\alpha/2}}{E}^2\\\\n = 0.14(1-0.14) \dfrac{1.96}{0.015}^2\\\\n = 740$

Hence, the minimum sample size required to obtain this type of accuracy is 740.

Learn more about MOE;

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